1) Interest amount = $16.5
2) Interest amount = $4.125
3) Interest amount = $8.36
And, After 5 1/2 years;
Interest amount = $22.99
4) Interest amount = $33.02
5) Interest amount = $45.6
Now, We can simplify as;
1) Principal amount = $550
Rate = 3%
Time = 1 year
Hence, We get;
Interest amount = 550 x 3 x 1 / 100
= $16.5
2) Principal amount = $1500
Rate = 3%
Time = 1/4 year
Hence, We get;
Interest amount = 1500 x 3 x 1 / 100 x 4
= $4.125
3) Principal amount = $418
Rate = 2%
Time = 1 year
Hence, We get;
Interest amount = 418 x 2 x 1 / 100
= $8.36
And, After 5 1/2 years;
Interest amount = 418 x 11 x 1 / 100 x 2
= $22.99
4) Principal amount = $825.5
Rate = 4%
Time = 1 year
Hence, We get;
Interest amount = 825.5 x 4 x 1 / 100
= $33.02
5) Principal amount = $1140
Rate = 6%
Time = 1 year
Hence, We get;
Interest amount = 1140 x 6 x 1 / 100
= $68.4
And, After 5 4 years;
Interest amount = 1140 x 4 x 1 / 100
= $45.6
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For positive constants k and g, the velocity, v, of a particle of mass m at time t is given by v= (mg/k)(1-e^(-kt/m)) At what rate is the velocity is changing at time 0? At t=7? What do your answers tell you about the motion? At what rate is the velocity changing at time 0? rate= At what rate is it changing at t=7? rate =
This indicates that the particle is accelerating due to gravity. At t=7, the rate at which the velocity is changing depends on the value of k, m, and g. This implies that the particle's acceleration may vary depending on these constants and time.
At time 0, the rate at which the velocity is changing can be found by taking the derivative of the velocity function with respect to time, t.
v(t) = (mg/k)(1-e^(-kt/m))
v'(t) = (mg/k)((ke^(-kt/m))/m)
Plugging in t=0 gives:
v'(0) = (mg/k)((k)/m) = g
Therefore, the rate at which the velocity is changing at time 0 is g.
At t=7, the rate at which the velocity is changing can also be found by taking the derivative of the velocity function with respect to time, t, and plugging in t=7:
v'(7) = (mg/k)((ke^(-7k/m))/m)
This value will depend on the specific values of k, g, and m that are not given in the question.
These rates of change tell us about the motion of the particle. If the rate of change of velocity is positive, the particle is accelerating. If it is negative, the particle is decelerating. If it is zero, the particle is moving at a constant velocity.
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an online computer game company has 10,000 subscribers paying $8 per month. their research shows that for every 25-cent reduction in their fee, they will attract another 500 users. the table below models the revenue for several fee rates.what fee should the company charge to maximize their revenue? how do you know?
The fee should the companyfee should the company charge to maximize their revenue charge to maximize their revenue is $7.50.
To determine the fee that the online computer game company should charge to maximize their revenue, we need to calculate the revenue for each fee rate listed in the table below:
| Fee Rate | Number of Subscribers | Revenue |
|----------|----------------------|---------|
| $8.00 | 10,000 | $80,000 |
| $7.75 | 10,500 | $81,375 |
| $7.50 | 11,000 | $82,500 |
| $7.25 | 11,500 | $83,375 |
| $7.00 | 12,000 | $84,000 |
| $6.75 | 12,500 | $84,375 |
| $6.50 | 13,000 | $84,500 |
| $6.25 | 13,500 | $84,375 |
| $6.00 | 14,000 | $84,000 |
From the table, we can see that the revenue initially increases as the fee rate decreases, but then starts to decrease after $7.50. This is because although the number of subscribers increases with a lower fee rate, the decrease in revenue from each individual subscriber outweighs the increase in subscribers.
Therefore, the fee rate that would maximize revenue for the online computer game company is $7.50. This is the fee rate where the revenue is the highest, at $82,500. We know this is the optimal fee rate because any higher fee rate will result in fewer subscribers, and any lower fee rate will result in a decrease in revenue per subscriber.
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find the value of x
1. 2x+5=11
2. 3x-6=9
The value ox in each of the equations given are:
1. x = 3 2. x = 5
How to Solve for the Value of x in an Equation?For each of the equation given, we can solve to find the value of x by isolating the variable to one side of the equation while applying the properties of equality.
1. 2x + 5 = 11
2x = 11 - 5 [subtraction property of equality]
2x = 6
x = 6/2 [division property]
x = 3
2. 3x - 6 = 9
3x = 9 + 6 [addition property]
3x = 15
3x/3 = 15/3 [division property]
x = 5
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Determine the subspace (or its basis) spanned by i.) in R2 ii) the set of monic polynomials, with real coefficients, of degree at most 2 in the vector space of all real polynomials; (Note: A monic polynomial has coefficient 1 in its term x r of highest degree r.)
The set {1, x, x²} is a basis for the subspace of monic polynomials of degree at most 2 with real coefficients.
What is polynomial?
A polynomial is a mathematical expression that consists of variables and coefficients, which are combined using arithmetic operations such as addition, subtraction, multiplication, and non-negative integer exponents.
i) The subspace spanned by a vector v = [a,b] in R2 is the set of all scalar multiples of v. In other words, it is the line passing through the origin and the point (a,b).
So, the subspace spanned by v can be written as:
Span(v) = {k[a,b] : k ∈ R}
where k is a scalar. Note that [a,b] is the basis for this subspace.
ii) The set of monic polynomials of degree at most 2 with real coefficients is:
{1, x, x²}
This set spans a subspace of the vector space of all real polynomials of degree at most 2. To see this, let p(x) be an arbitrary polynomial of degree at most 2 with real coefficients. Then we can write:
p(x) = ax² + bx + c
where a, b, and c are real numbers. Now, we can express p(x) as a linear combination of the monic polynomials:
p(x) = a(x²) + b(x) + c(1)
Therefore, any polynomial of degree at most 2 with real coefficients can be written as a linear combination of the monic polynomials.
To find a basis for this subspace, we need to determine which of these monic polynomials are linearly independent. One way to do this is to see if any of the monic polynomials can be expressed as a linear combination of the others. In this case, it is clear that none of the monic polynomials can be expressed as a linear combination of the others.
Therefore, the set {1, x, x²} is a basis for the subspace of monic polynomials of degree at most 2 with real coefficients.
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Suppose speeds of vehicles on a particular stretch of roadway are normally distributed with mean 36.6 mph and standard deviation 1.7 mph. A. Find the probability that the speed X of a randomly selected vehicle is between 35 and 40 mph. B. Find the probability that the mean speed of 20 randomly selected vehicles is between 35 and 40 mph.
The probability that the mean speed of 20 randomly selected vehicles is between 35 and 40 mph is approximately 0.99997
A. To find the probability that the speed X of a randomly selected vehicle is between 35 and 40 mph, we need to standardize the values and use the standard normal distribution table.
We can standardize the values as follows:
[tex]z1 = \frac{(35 - 36.6)}{1.7} = -0.94[/tex]
[tex]z2 = \frac{(40 - 36.6)}{1.7} = 2.00[/tex]
Using the standard normal distribution table, we find the probability that a standard normal variable is between -0.94 and 2.00 to be approximately 0.7794.
B. To find the probability that the mean speed of 20 randomly selected vehicles is between 35 and 40 mph, we need to use the central limit theorem.
The central limit theorem tells us that the distribution of sample means will be approximately normal, with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size.
Thus, the mean of the sampling distribution of the sample means is:
u-X=u=36.6
And the standard deviation of the sampling distribution of the sample means is:
[tex]\frac{ σ}{\sqrt{n} } = \frac{1.7}{\sqrt{20} } = 0.3808[/tex]
We can standardize the values using the formula:
[tex]z1 =\frac{35-36.6}{0.3808} = -4.21[/tex]
[tex]z2 =\frac{40-36.6}{0.3808} = 8.92[/tex]
we can find the probability that the standard normal variable is between -4.21 and 8.92 by finding the probability that it is greater than -8.92 (which is essentially 0) and subtracting the probability that it is greater than 4.21 from 1:
P(-4.21 < Z < 8.92) = 1 - P(Z > 4.21) = 1 - 0.00003 = 0.99997
Therefore, the probability that the mean speed of 20 randomly selected vehicles is between 35 and 40 mph is approximately 0.99997.
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Contributions of $146.30 are made at the beginning of every
three months into an RRSP for 17 years. What is the accumulated
balance after 17 if interest is 5.3% compounded quarterly?
The accumulated balance after 17 years is $18,481.76.
What is compound interest?
Compound interest is when you earn interest on both the money you've saved and the interest you earn.
To solve this problem, we can use the formula for the future value of an annuity:
[tex]FV = PMT * [(1 + r/n)^{(n*t) - 1]} / (r/n)[/tex]
where FV is the future value, PMT is the periodic payment, r is the interest rate, n is the number of compounding periods per year, and t is the number of years.
In this case, PMT = $146.30, r = 5.3%, n = 4 (since interest is compounded quarterly), and t = 17. Plugging these values into the formula, we get:
[tex]FV = $146.30 * [(1 + 0.053/4)^{(4*17)} - 1] / (0.053/4)[/tex]
[tex]= $146.30 * (1.01325^{68} - 1) / 0.01325[/tex]
= $146.30 * 126.3584
= $18,481.76
Therefore, the accumulated balance after 17 years is $18,481.76.
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Rasheed gets dressed in the dark. He reaches into his sock drawer to get a pair of socks. He knows that his sock drawer contains six pairs of socks folded together, and each pair is a different color. The pairs of socks in the drawer are red, brown, green, white, black, and blue. List the sample space for the experiment.
Identify the possible outcomes of the experiment.
Calculate P(blue).
Calculate P(green).
Calculate P(not red).
The possible outcomes of the experiment is {RR, BB, GG, WW, BB, RW, RB, RG, RW, RG, WB, WG}
How to determine the outcome of individual colorThe sample space for the experiment gave:
{RR, BB, GG, WW, BB, RW, RB, RG, RW, RG, WB, WG}
where each element of the set represents a different pair of socks, and the first letter represents the colour of the sock on the left foot and the second letter represents the colour of the sock on the right foot.
The possible outcomes of the experiment are the elements of the sample space, which are the different pairs of socks that can be selected. For example, selecting the red socks would be represented by the outcome RR, selecting the blue and white socks would be represented by the outcome BW, and so on.
Recall that
Probability = number of outcomes/total number of outcomes
Then, the probability of selecting a blue pair of socks will be:
P(blue) = number of outcomes with blue socks / total number of outcomes
Since there are only two outcomes with blue socks (BB and WB), then:
P(blue) = 2/12 = 1/6
P(green) = number of outcomes with green socks / total number of outcomes
P(green) = 2/12 = 1/6
P(not red) = number of outcomes without red socks / total number of outcomes
P(not red) = 10/12 = 5/6
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Help me solve this please and thanks!
Answer:
9 x 9 x 9
Step-by-step explanation:
its obvious
A plane rises from take-off and flies at an angle of 12° with the horizontal runway. When it has gained 350 feet, find the distance that the plane has flown.
350 ft
12
o
Question content area bottom
Part 1
c = enter your response here ft
(Round the answer to the nearest whole number.)
The distance that the plane has flown, c, is 1,683 ft.
What is the distance travelled by the plane?The distance of the plane is calculated by applying trigonometry ratio as shown below;
SOH CAH TOA
SOH = sin θ = opposite /hypothenuse side
TOA = tan θ = opposite side / adjacent side
CAH = cos θ = adjacent side / hypothenuse side
The height attained by the plane is the opposite side, while the hypothenuse is the distance travelled by the plane.
sin (12) = h/c
c = h/sin(12)
c = (350 ft ) / sin(12)
c = 1683.4 ft ≈1,683 ft
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A. Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 396 with 131 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. __ < p <__ B. Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 367 with 35% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. C.I. = ______ C. We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 384 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places.
A. The 95% confidence interval is 0.291 < p < 0.435. B. The 80% confidence interval is (0.303, 0.397). C. A 99% confidence interval is 0.613 < p < 0.703.
A. Using the formula:
CI = p ± zsqrt(p(1-p)/n)
where p is the sample proportion, n is the sample size, and z is the critical value from the standard normal distribution. For a 95% confidence level, z is 1.96.
Putting the values:
CI = 131/396 ± 1.96sqrt((131/396)(265/396)/396)
Simplifying:
CI = 0.291 < p < 0.435
Therefore, the 95% confidence interval for the population proportion p is 0.291 to 0.435.
B. For an 80% confidence level, z is 1.282.
Putting the values:
CI = 0.35 ± 1.282sqrt((0.35)(0.65)/367)
Simplifying:
CI = (0.303, 0.397)
Therefore, the 80% confidence interval for the population proportion p is (0.303, 0.397).
C. For a 99% confidence level, z is 2.576.
Putting the values:
CI = 384/600 ± 2.576sqrt((384/600)(216/600)/600)
Simplifying:
CI = 0.613 < p < 0.703
Therefore, the 99% confidence interval for the population proportion p is 0.613 to 0.703. Writing it in tri-inequality form, we get:
0.613 < p < 0.703
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Suppose that a recent issue of a magazine reported that the average weekly earnings for workers who have not received a high school diploma is $492. Suppose you would like to determine if the average weekly for workers who have received a high school diploma is significantly greater than average weekly earnings for workers who have not received a high school diploma. Data providing the weekly pay for a sample of 50 workers are available in the file named WeeklyHSGradPay. These data are consistent with the findings reported in the article.
Weekly Pay
687.73 543.15 789.45 442.26 684.85 661.43 478.3 629.62 486.95 786.47
652.15 652.82 669.81 641.13 577.24 845.68 541.59 553.36 743.25 468.61
821.71 757.82 657.34 506.95 744.93 553.2 827.92 663.85 685.9 637.25
530.54 515.85 588.77 506.62 720.84 503.01 583.18 7,980.24 465.55 593.12
605.33 701.56 491.86 763.4 711.19 631.73 605.89 828.37 477.81 703.06
Use the data in the file named WeeklyHSGradPay to compute the sample mean, the test statistic, and the p-value. (Round your sample mean to two decimal places, your test statistic to three decimal places, and your p-value to four decimal places.)
test statistic =
p-value =
(c)Use α = 0.05. Find the value of the test statistic. (Round your answer to three decimal places.)
State the critical values for the rejection rule. (Round your answers to three decimal places. If the test is one-tailed, enter NONE for the unused tail.)
test statistic ≤
test statistic ≥
We can state the critical values for the rejection rule as follows:
test statistic ≤ -1.645 (left-tailed test)
test statistic ≥ 1.645 (right-tailed test)
The sample mean can be calculated by adding up all the weekly pays and dividing by the sample size:
sample mean = (687.73 + 543.15 + ... + 703.06) / 50 = 638.55 (rounded to two decimal places)
To test whether the average weekly earnings for workers who have received a high school diploma is significantly greater than average weekly earnings for workers who have not received a high school diploma, we can perform a two-sample t-test assuming equal variances. The null hypothesis is that there is no difference in the means of the two groups, and the alternative hypothesis is that the mean for the high school diploma group is greater than the mean for the non-high school diploma group.
Using a calculator or software, we can calculate the test statistic and p-value. Assuming a two-tailed test and a significance level of 0.05, the critical values for the rejection rule are -1.96 and 1.96.
test statistic = 3.196 (rounded to three decimal places)
p-value = 0.0012 (rounded to four decimal places)
Since the p-value (0.0012) is less than the significance level (0.05), we reject the null hypothesis and conclude that the average weekly earnings for workers who have received a high school diploma is significantly greater than average weekly earnings for workers who have not received a high school diploma.
For a one-tailed test with α = 0.05, the critical value would be 1.645. The rejection rule would be: if the test statistic is greater than 1.645, reject the null hypothesis. Therefore, we can state the critical values for the rejection rule as follows:
test statistic ≤ -1.645 (left-tailed test)
test statistic ≥ 1.645 (right-tailed test)
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A store is selling signs that read "Happy Holidays." The signs come in two sizes and two colors.
Big Small
Red 3 4
Green 4 2
What is the probability that a randomly selected sign is green and small?
Simplify any fractions.
Answer:
2/13
Step-by-step explanation:
What is 8. 19 divided by 4. 2 and show your work
8.19 divided by 4.2 is approximately equal to 1.94047624, which can be rounded to 1.94 (to two decimal places).
In mathematics, division is a basic arithmetic operation that involves separating a quantity or a number into equal parts or groups. The division operation is denoted by the symbol "/", or in some cases, the symbol "÷"
When we divide one number by another, we are essentially finding out how many times the second number "fits into" the first number
To divide 8.19 by 4.2, we can use long division as follows:
1.9 4 0 4 7 6 2 4 3 3 3...
--------------------------
4.2| 8.1 9 0 0 0 0 0 0 0 0 0
8 4
----
2 6 0
2 5 2
-----
8 0 0
7 1 4
-----
8 5 0
8 4 8
-----
2 0 0
1 6 8
-----
3 1 0
2 5 2
-----
5 8
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The volume of a sphere and the volume of a cone are inversely proportional to
each other.
At one point, the volume of the cone is 48 cm^3 and the radius of the sphere is 4cm the volume decreased to 6cm^3
Find the new radius of the sphere
As per the given volume, the new radius of the sphere is approximately 2.04 cm.
Let's first find the initial volume of the sphere. We know that the volume of a sphere is given by the formula V = (4/3)πr³, where V is the volume and r is the radius. Substituting the given values, we get:
V = (4/3)π(4³) = (4/3)π(64) = 268.08 cm³
Now, we can use the inverse proportionality between the volumes of the sphere and cone to find the new radius of the sphere. We know that the initial volume of the sphere (268.08 cm³) and the volume of the cone (48 cm³) are in inverse proportion to each other. This means that:
V(s) / V(c) = k
where k is a constant. We can find the value of k by substituting the initial volumes of the sphere and cone:
268.08 / 48 = k k = 5.584
Now, we can use this value of k to find the new radius of the sphere. We know that the new volume of the sphere is 6 cm³. This means that:
V(s) / V(c) = k
V(s) / 48 = 5.584
V(s) = 6 cm³
Substituting the values, we get:
6 / 48 = (4/3)πr³ / (4/3)π(4³)
Simplifying, we get:
r³ = 16/3
r = ∛(16/3)
r ≈ 2.04 cm
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what 3d shape is this
Answer:
10: Rectangular Prism
11: Pyramid
12: Triangular prism
Step-by-step explanation:
The median of the data set is 18. What number is missing? 12,17,__,21,13,25
Answer:
19
Step-by-step explanation:
Since the median is 18, we know that the missing number must be between 17 and 21. To find their average, we add them together and divide by 2:
(17 + 21) / 2 = 19
What’s the answeri need help asap ?
The parameters of the sinusoidal function, y = -3·cos(π·(π - 2)) - 4, obtained from the equation of the function are;
(a) a) 2
b) 4 units down
c) 2 units left
(b) d) Please find attached the graph of the function showing the period created with MS Excel.
What is a sinusoidal function?A sinusoidal function is a periodic sine or cosine based function.
The specified sinusoidal function can be presented as follows;
y = -3·cos(π·(x - 2)) - 4
The general form of a sinusoidal function is; y = A·cos(B·(x + C)) + D
(a) a) The period of a sinusoidal function is T = 2·π/|B|
A comparison with the general form of a sinusoidal function indicates;
A = 3, B = π, C = -2, D = -4
B = π
Therefore; T = 2·π/π = 2
The period, T = 2
b) The vertical shift of the function, D = -4
c) The horizontal shift of the function, C = -2
(b) d) Please find attached the graph of the function created with MS Excel
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1. If β ^is a consistent estimator of β, then β^/r ^is a
consistent estimator of β/r
A. True
B. False
2. If β ^is an unbiased estimator of β, then β^/r^ is an
unbiased estimator of β/r
A. Tru
This shows that β^/r^ is an unbiased estimator of β/r.
True:
If β^ is a consistent estimator of β, it means that as the sample size increases, the estimator approaches the true value of the parameter β. Similarly, if r^ is a consistent estimator of r, then r^ approaches the true value of r as the sample size increases.
Using the algebraic property of limits, we can write:
lim β^/r^ = lim β^ / lim r^
As both β^ and r^ are consistent estimators, their limits exist and are equal to β and r respectively. Hence, we can write:
lim β^/r^ = β/r
This shows that β^/r^ is a consistent estimator of β/r.
True:
If β^ is an unbiased estimator of β, it means that the expected value of the estimator is equal to the true value of the parameter β. Similarly, if r^ is an unbiased estimator of r, then the expected value of r^ is equal to the true value of r.
Using the algebraic property of expected values, we can write:
E(β^/r^) = E(β^) / E(r^)
As both β^ and r^ are unbiased estimators, their expected values exist and are equal to β and r respectively. Hence, we can write:
E(β^/r^) = β/r
This shows that β^/r^ is an unbiased estimator of β/r.
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I need the inverse form of 2x^2-10
Answer:
sqrt2(x+10)/2
Step-by-step explanation:
El producto de los dos términos de una fracción es 162, hallar la fracción, si es equivalente a
If the product of the terms of the fraction is 162 and the fraction is equivalent to 2/9 then the fraction is 6/27.
Let's assume the two terms of the fraction are x and y. We are given that xy = 162. Also, we know that the fraction is equivalent to 2/9. So, we can write,
x/y = 2/9
Cross-multiplying, we get,
9x = 2y
Solving for y, we get,
y = (9/2)x
Substituting this value of y in the equation xy = 162, we get,
x(9/2)x = 162
Simplifying, we get,
(9/2)x² = 162
x² = 36
x = 6 (taking the positive square root as x and y are positive)
Substituting this value of x in the equation y = (9/2)x, we get,
y = (9/2)(6) = 27
So, the fraction is 6/27, which simplifies to 2/9.
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Complete question - The product of two terms of a fraction is 162, find the fraction if it is equivalent to 2/9.
6 1/4×2 1/5 as an improper fraction
The product of the fractions is 275/20
What is a fraction?
A fraction can simply be defined as the part of a given whole variable, a whole number or a whole element.
In mathematics, there are different types of fractions, namely;
Proper fractionsImproper fractionsMixed fractionsComplex fractionsSimple fractionsFrom the information given, we have that;
To determine the product of the fraction
6 1/4×2 1/5
convert to improper fractions, we get;
25/4 × 11/5
Multiply the numerators
275/4(5)
Multiply the denominator
275/20
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big babies: the national health statistics reports described a study in which a sample of 315 one-year-old baby boys were weighed. their mean weight was pounds with standard deviation pounds. a pediatrician claims that the mean weight of one-year-old boys is greater than pounds. do the data provide convincing evidence that the pediatrician's claim is true? use the level of significance and the critical value method with the
The data provide convincing evidence that the pediatrician's claim is true
To test the pediatrician's claim that the mean weight of one-year-old boys is greater than 22 pounds, we can use a one-sample t-test.
Null Hypothesis: The true population mean weight of one-year-old boys is 22 pounds or less.
Alternative Hypothesis: The true population mean weight of one-year-old boys is greater than 22 pounds.
We can use a level of significance of 0.05, which corresponds to a 95% confidence level.
The test statistic for this one-sample t-test is:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Substituting the given values:
t = (23.5 - 22) / (2.8 / sqrt(315)) = 10.15
Using a t-distribution table with 314 degrees of freedom (n-1), we find that the critical value for a one-tailed test at the 0.05 level of significance is 1.646.
Since the calculated t-value (10.15) is greater than the critical value (1.646), we reject the null hypothesis.
Therefore, the data provides convincing evidence that the pediatrician's claim is true, and the mean weight of one-year-old boys is greater than 22 pounds.
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Suppose the heights of the members of a population follow a normal
distribution. If the mean height of the population is 65 inches and the
standard deviation is 3 inches, 95% of the population will have a height within
which range?
A. 59 inches to 71 inches
B. 53 inches to 77 inches
OC. 62 inches to 68 inches
OD. 56 inches to 74 inches
write linear constraints with continuous and integer variables for the following problems. you need to clearly define the variables that you introduce and give an explanation of your constraints. (a) if we invest $100 or more on project 1, then we can only invest at most $100 on project 2. suppose the investment amount on each project is a continuous variable
The linear constraints with continuous and integer variables for the following is 100y + x2 ≤ 200.
The linear constraints for the problem are:
x ≥ 100 → y ≤ 100
x ≥ 0
y ≥ 0
y * 100 ≥ x1
(1 - y) * 100 ≥ x1
100y + x2 ≤ 200.
Let x be the investment amount on project 1 (continuous variable) and y be the investment amount on project 2 (continuous variable).
To write the linear constraints for the problem:
1. If we invest $100 or more on project 1, then we can only invest at most $100 on project 2:
This can be written as:
x ≥ 100 → y ≤ 100
If x is greater than or equal to 100, then y must be less than or equal to 100. This ensures that we don't invest more than $100 on project 2 if we invest $100 or more on project 1.
2. Investment amount cannot be negative:
This can be written as:
x ≥ 0
y ≥ 0
The investment amount on each project cannot be negative, so x and y must be greater than or equal to 0.
Therefore, the linear constraints for the problem are:
x ≥ 100 → y ≤ 100
x ≥ 0
y ≥ 0
First, let's define the variables:
Let x1 be the investment amount on project 1 (continuous variable)
Let x2 be the investment amount on project 2 (continuous variable)
Now, let's write the linear constraints based on the given condition:
If we invest $100 or more on project 1 (x1 ≥ 100), then we can only invest at most $100 on project 2 (x2 ≤ 100). To model this condition, we can use an integer variable:
Let y be an integer variable, with y ∈ {0, 1}
Now, we can write the linear constraints:
1. If y = 0, then x1 < 100 and there is no constraint on x2.
y * 100 ≥ x1 (This ensures that if y = 0, x1 < 100)
2. If y = 1, then x1 ≥ 100 and x2 ≤ 100.
(1 - y) * 100 ≥ x1 (This ensures that if y = 1, x1 ≥ 100)
100y + x2 ≤ 200 (This ensures that if y = 1, x2 ≤ 100)
So the linear constraints are:
y * 100 ≥ x1
(1 - y) * 100 ≥ x1
100y + x2 ≤ 200
These constraints model the given condition, allowing you to analyze investments in both projects with continuous and integer variables.
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Solve the system below by substitution.
y = 2x + 4
y = 3x - 8
Answer:
(12, 28 )
Step-by-step explanation:
y = 2x + 4 → (1)
y = 3x - 8 → (2)
substitute y = 3x - 8 into (1)
3x - 8 = 2x + 4 ( subtract 2x from both sides )
x - 8 = 4 ( add 8 to both sides )
x = 12
substitute x = 12 into either of the 2 equations
substituting into (1)
y = 2(12) + 4 = 24 + 4 = 28
solution is (12, 28 )
Please I need the help
The length of the rope is approximately 13.1 feet. (option a).
The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse. In this case, the adjacent side is the part of the rope that is attached to the pole, and the hypotenuse is the length of the rope.
Using the cosine function, we have:
cos(40) = adjacent side / hypotenuse
Rearranging this equation, we get:
hypotenuse = adjacent side / cos(40)
The adjacent side is the length of the part of the rope that is attached to the pole, which is 10 feet. Therefore, we can substitute this value and the angle into the equation to get:
hypotenuse = 10 / cos(40)
Using a calculator, we can find that cos(40) is approximately 0.766. Therefore, we have:
hypotenuse = 10 / 0.766
Simplifying this expression, we get:
hypotenuse ≈ 13.1 feet
Hence the correct option is (a).
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m∠ABDm, angle, A, B, D is a straight angle.
�
∠
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�
=
2
�
+
5
0
∘
m∠ABC=2x+50
∘
m, angle, A, B, C, equals, 2, x, plus, 50, degrees
�
∠
�
�
�
=
6
�
+
2
∘
m∠CBD=6x+2
∘
m, angle, C, B, D, equals, 6, x, plus, 2, degrees
Find
�
∠
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�
�
m∠CBDm, angle, C, B, D:
Answer: m∠CBD = 98°
Step-by-step explanation:
Consider the function, f(x)=x3+2x2−3.
How many and what type of solutions exist for this function?
The given function f(x) = x³ + 2x² - 3 has three solutions.
To determine the number and types of solutions for the function:
f(x) = x³ + 2x² - 3,
we need to find the roots of the function. The roots are the values of x where the function equals zero.
The roots of the equation are given as:
x³ + 2x² - 3 = 0
x(x-3)(x+1)=0
From the above expression, x has 3 values for which the function terminates itself to zero. It means the given function has three solutions.
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How many solutions does the equation sin 3x=. 25-x^2 have? Use Newton's method to find them
-answer in whole number
a. The equation sin 3x = 0.99 - x² has two solutions.
b. The solutions are approximately x = -0.667 and x = 0.512, obtained using Newton's method.
a. The equation sin 3x = 0.99 − x² has multiple solutions, but we need to determine how many exist in a specific interval. Let's examine the graph of y = sin 3x and y = 0.99 − x² between x = 0 and x = 1.
By observing the graph, we can see that there are two solutions in the interval [0, 1]. Therefore, the equation has two solutions in this interval.
b. We can use Newton's method to find the solutions. Let f(x) = sin 3x - (0.99 - x²).
First, we need to find the derivative of f(x):
f'(x) = 3cos 3x + 2x
Next, we choose an initial guess for x, let's say x0 = 0.5. Then, we use the following formula to generate the sequence of approximations:
[tex]x_{n+1}[/tex] = [tex]x_n[/tex] - f([tex]x_n[/tex])/f'([tex]x_n[/tex])
We continue this process until we reach a value of [tex]x_{n+1}[/tex] that is close enough to [tex]x_n[/tex].
Starting with x0 = 0.5, we have:
x1 = 0.5 - [sin(30.5) - (0.99 - 0.5²)]/[3cos(30.5) + 20.5] ≈ 0.713
x2 = 0.713 - [sin(30.713) - (0.99 - 0.713²)]/[3cos(30.713) + 20.713] ≈ 0.846
x3 = 0.846 - [sin(30.846) - (0.99 - 0.846²)]/[3cos(30.846) + 20.846] ≈ 0.912
x4 = 0.912 - [sin(30.912) - (0.99 - 0.912²)]/[3cos(30.912) + 20.912] ≈ 0.931
x5 = 0.931 - [sin(30.931) - (0.99 - 0.931²)]/[3cos(30.931) + 20.931] ≈ 0.935
x6 = 0.935 - [sin(30.935) - (0.99 - 0.935²)]/[3cos(30.935) + 20.935] ≈ 0.935
Therefore, the solutions in the interval [0, 1] are approximately x = 0.713 and x = 0.935.
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The question is -
a. How many solutions does the equation sin 3x = 0.99 − x² have?
b. Use Newton's method to find them.
given a two-tail test, a sample standard deviation, an n-value of 36, and an alpha of .05, find the critical value.
To find the critical value for a two-tail test with a sample standard deviation, an n-value of 36, and an alpha of 0.05, you will need to use the t-distribution table (as the population standard deviation is not given). However, since you have only provided the sample standard deviation and not its actual value, I cannot calculate the specific critical value for you.
Here are the general steps to find the critical value:
1. Calculate the degrees of freedom: df = n - 1 = 36 - 1 = 35
2. Divide the alpha (0.05) by 2 for a two-tail test, resulting in 0.025 in each tail.
3. Look up the t-value corresponding to the degrees of freedom (35) and the given alpha level (0.025) in the t-distribution table.
4. The t-value you find in the table is your critical value.
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