Find the area of a triangle with base of
​inches and a height of
inches.
A
100100100 square inches
B
505050 square inches
C
252525 square inches
D
12.512.512.5 square inches

The walkway around the swimming pool should be 8 feet wide.

From the question, we are given:

Length of the pool (l) = 15 feet

Width of the pool (w) = 7 feet

width of walkway (X) = X feet

Total Area =  Area of pool and walkway = 713 square feet

Let's derive equation for the pool and walkaway combined:

length of the pool and walkway = 15 + 2X [walkway is on both sides]

width of the pool and walkway = 7 + 2X

Total Area = (length x width) of the pool and walkway

713 = (15 + 2X) x (7 + 2X)

713 = 15(7 + 2X) + 2X(7 + 2X)

713 = 105 + 30X + 14X + 4X²

713 = 105 + 44X + 4X²

Rearrange the equation

4X² + 44X + 105 = 713

Collect like terms and express in a proper quadratic equation

4X² + 44X - 608 = 0

Dividing both sides by 4, we get

X² + 11X - 152 = 0

Solve this equation quadratically

X =  [tex]\frac{-b \± \sqrt{b^{2} - 4ac}}{2a}[/tex]

where

a = 1

b = 11

c = -152

Plug in the value into the equation

X = [tex]\frac{-11 \± \sqrt{11^{2} - 4(1)(-152)}}{2(1)}[/tex]

X = [tex]\frac{-11 \± \sqrt{121 + 608}}{2(1)}[/tex]

X = [tex]\frac{-11 \± \sqrt{729}}{2}[/tex]

X = [tex]\frac{-11 \± 27}{2}[/tex]

X = [tex]\frac{-11 + 27}{2}[/tex] or [tex]\frac{-11 - 27}{2}[/tex]

X = [tex]\frac{16}{2}[/tex] or [tex]\frac{-38}{2}[/tex]

X = 8 or -19

But since the width of the walkway cannot be negative, the only valid solution is therefore:

X = 8

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The frequencies based on the information are are 100, 40, and 6.

In the formula of the median, we have to calculate cumulative frequencies and have to use cumulative frequency of the interval previous to the median class. Also, in the formula of mode, frequencies are used and not the cumulative frequencies.

140 - y = 100

y = 40

substituting y and x in equation 1 is:

x + y + z = 146

100 + 40 + z = 146

z = 6

Therefore, the missing frequencies are 100, 40, and 6.

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A) A probability of approximately 0.128 or 12.8% that exactly 17 students had type A blood.

B) There is a 93.0% chance that at least 17 students had type A blood in the blood drive held at University High.

Part A: To find the probability that exactly 17 of the 50 students had type A blood, we first need to calculate the probability of a single student having type A blood. Since about 31% of the U.S. population has type A blood, the probability of a student at University High having type A blood is also 31%.

We can use the binomial distribution formula to calculate the probability of exactly 17 students having type A blood in a sample of 50 students. The formula is:

P(X = x) = (n choose k) x pˣ x (1-p)ⁿ⁻ˣ

where P(X = x) is the probability of exactly x successes, n is the sample size (in this case, 50), p is the probability of success (31% or 0.31), (n choose k) is the binomial coefficient or the number of ways to choose k successes from n trials.

Plugging in the values, we get:

P(X = 17) = (50 choose 17) x 0.31¹⁷ x (1-0.31)⁵⁰⁻¹⁷ = 0.128 or 12.8%

Part B: To find the probability that at least 17 of the 50 students had type A blood, we need to calculate the probability of 17, 18, 19,...50 students having type A blood and then add those probabilities together. This is because "at least 17" means 17 or more students, so we need to consider all possibilities from 17 to 50.

Therefore, the probability of at least 17 students having type A blood is:

P(X >= 17) = 1 - P(X < 17)

where P(X < 17) is the probability of less than 17 students having type A blood. We can use the binomial distribution formula to calculate this probability as well:

P(X < 17) = P(X = 0) + P(X = 1) + ... + P(X = 16)

Again, this can be a tedious task to calculate manually. Instead, we can use a binomial calculator or a software program to find this probability.

Using a binomial, we find that the probability of less than 17 students having type A blood is approximately 0.070 or 7.0%. Therefore, the probability of at least 17 students having type A blood is:

P(X >= 17) = 1 - P(X < 17) = 1 - 0.070 = 0.930 or 93.0%

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Answer:

THE ANSWER IS (A)

Step-by-step explanation:

BECAUSE I JUST DID AT

Answer:Find the rational roots of x^4+5x^3+7x^2-3x-10=0

To find the rational roots of a polynomial, we use the Rational Root Theorem, which states that if a polynomial has rational roots, then they must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. In this case, the constant term is -10, and the leading coefficient is 1. The factors of 10 are ±1, ±2, ±5, and ±10, and the factors of 1 are ±1. Therefore, the possible rational roots are:

±1, ±2, ±5, ±10

We can test each of these roots by synthetic division or long division to see if they are actually roots of the polynomial. After trying these roots, we find that only -2 and 1 are roots. Therefore, the answer is A. -2,1.

Find all the zeros of the equation 3x^2-4=-x^4

We can rewrite the equation as x^4+3x^2-4=0. To find the zeros of this equation, we can use the Rational Root Theorem as in the previous question, except that now the factors of the constant term -4 are ±1, ±2, and ±4. The factors of the leading coefficient 1 are ±1. Therefore, the possible rational roots are:

±1, ±2, ±4

Testing these roots, we find that the only roots are x=1, x=-1, x=2i, and x=-2i. Therefore, the answer is D. 1,-1,2i,-2i.

What is a polynomial function in standard form with zeros 1,2,-2, and -3?

If a polynomial has roots a, b, c, and d, then it can be written as (x-a)(x-b)(x-c)(x-d). To put it in standard form, we need to expand this expression:

(x-a)(x-b)(x-c)(x-d) = x^4 - (a+b+c+d)x^3 + (ab+ac+ad+bc+bd+cd)x^2 - (abc+abd+acd+bcd)x + abcd

Substituting the given values of a, b, c, and d, we get:

(x-1)(x-2)(x+2)(x+3) = x^4 + 2x^3 - 7x^2 - 8x + 12

Therefore, the answer is B. x^4+2x^3-7x^2-8x+12.

Which correctly describes the roots of the following cubic equation.

x^3-3x^2+4x-12=0

We can use the Rational Root Theorem again to find the possible rational roots of the polynomial. The factors of the constant term -12 are ±1, ±2, ±3, ±4, ±6, and ±12, and the factors of the leading coefficient 1 are ±1. Therefore, the possible rational roots are:

±1, ±2, ±3, ±4, ±6, ±12

Testing these roots, we find that none of them are roots of the polynomial. Therefore, the answer is B. one real root and two complex roots.

What is the solution of 5^3x=900. Round your answer to the nearest hundredth

We can solve for x by taking the logarithm of both sides with base 5:

log5(

Step-by-step explanation:

The answers to all parts are shown below.

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

1. x⁴ + 5x³ + 7x² -3x - 10 = 0

= (x-1)(x+2)(x² + 4x + 5)

Thus, the rational roots are 1 and -2.

2. 3x² -4 = - x⁴

x⁴ + 3x² -4 = 0

(x-1)(x+1)(x² + 4) =0

Thus, the zeroes are 1, -1, -2i, +2i.

3. We can write the zeroes as

p(x)=(x-1)(x-2)(x+2)(x+3).

So, p(x) = x⁴ + 2x³ -7x² -8x +12

4. x³ -3x² +4x- 12=0

(x-3)(x²+4) =0

x= 3, -2i, 3i

5.  x= 1,2,4,6,8,10,11

y= 0,-1,0,4,8,9,8

So, equation of line

(y- 0) = (-1-0)/(2-1)(x-1)

y = -1(x-1)

y + x +1=0

6. x= 2,5,12,20 and y= 30,12,5,3

k = 30/2

k = 15

So, the relationship is y= 15x.

7. cn = constant(k)

n= 44

then, c= 30/44 = 0.68

and, k = 44 x 0.68

k = 30

So, c = 1.5 dollars per person.

8. 125¹/³

= (5³)¹/³

= 5

9. y = 293(1.06)⁴ = 370

10. log₃ 2187

= log₃ 3⁷

= 7

11. log₄ 22

= 2.29

12 . y= 4/x

The new equation is  y = 4/(x - 7) + 6.

13. (6-x)/ (x² + 2x - 3) / (x² -4x- 12)/ (²x² + 4x + 3)

= - (x+1)/ (x+1)(x+2)

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Option C is the true statement from the data that has been gathered and shown in the image

The interquartile range exemplifies the "dispersion" or width of a set [1] by determining the difference between the top quartile (the 25% highest) and lower quartile (the 25% lowest). In reference to the provided picture:

- The bus drivers' interquartile range is 10, represented by subtracting their distance's lowest point (10) from the highest (20).

20 - 10 = 20

- Similarly, the teachers' interquartile range is 15, which stems from finding the difference between their lowest distance (15) and highest distance (30).

30 - 15 = 15

Therefore, comparing both ranges reflects that the bus drivers have an interquartile range of distances that is 5 miles smaller than the one for the teachers. Consequently, we opt for option C.

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Manuel bought a car for 65% (percentage) of the original price of $7,000, he must have paid $4,550 for the car.

The percentage is a number or ratio that represents a fraction of a whole number or the sum of ratios.

The percentage is computed by dividing one number by another and multiplying the result by 100.

The original price of the car = $7,000

The discount factor = 65% (1 - 35%) or 0.65

The discounted price = $4,550 ($7,000 x 0.65)

Thus, by implication, for receiving a discount of 35% (100% - 65%), Manuel paid $4,550 for the car.

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a) The percent of the scores are between 72 and 87 is 77.45%

b) The probability that a score is greater than 77 is 84.13%

c) The probability that a score is less than 82 or greater than 92 is 52.28%

d) About 5 students scored outside two standard deviations of the mean.

a) To find the percent of scores between 72 and 87, we first need to standardize these scores. We can do this by subtracting the mean (82) and dividing by the standard deviation (5). This gives us:

z = (72 - 82) / 5 = -2

z = (87 - 82) / 5 = 1

We can then use a table of standard normal probabilities (also called a z-table) to find the probability of a score being between -2 and 1. This probability is 0.7745, or 77.45%

b) To find the probability that a score is greater than 77, we again need to standardize the score:

z = (77 - 82) / 5 = -1

Using the z-table, we can find the probability of a score being greater than -1 (which is the same as the probability of a score being less than or equal to 77). This probability is 0.1587, or 15.87% (rounded to two decimal places). To find the probability of a score being greater than 77, we subtract this probability from 1:

P(score > 77) = 1 - P(score <= 77) = 1 - 0.1587 = 0.8413, or 84.13%

c) To find the probability that a score is less than 82 or greater than 92, we can break this into two separate probabilities:

P(score < 82) + P(score > 92)

We can standardize these scores as follows:

z = (82 - 82) / 5 = 0

z = (92 - 82) / 5 = 2

Using the z-table, we can find the probabilities associated with these z-scores:

P(score < 82) = P(z < 0) = 0.5 (from the symmetry of the standard normal distribution)

P(score > 92) = P(z > 2) = 0.0228

Adding these probabilities together gives us:

P(score < 82 or score > 92) = 0.5 + 0.0228 = 0.5228, or 52.28%

d) Two standard deviations from the mean in either direction would be:

82 - (2 x 5) = 72

82 + (2 x 5) = 92

So any score below 72 or above 92 would be considered outside two standard deviations. To find how many students scored outside this range, we need to find the total proportion of scores that fall outside this range and multiply by the total number of scores.

To find the proportion of scores outside the range, we can use the same approach as in part c):

P(score < 72 or score > 92) = P(z < -2 or z > 2) = P(z < -2) + P(z > 2)

Using the z-table, we find:

P(z < -2) = 0.0228

P(z > 2) = 0.0228

Adding these probabilities together gives us:

P(score < 72 or score > 92) = 0.0228 + 0.0228 = 0.0456

So 4.56% of scores fall outside the range of 72 to 92. To find the number of students who scored outside this range, we can multiply this proportion by the total number of scores:

0.0456 x 120 = 5.472 ≈ 5

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The complement and the supplement of an angle of 68º are given as follows:

Hence the complement and the supplement of an angle of 68º are obtained as follows:

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The domain of the function g(x) is given as follows:

B. All values of x such that x ≥ 0.

The domain of a function is the set of all the possible input values that can be assumed by the function.

The function for this problem is defined as follows:

[tex]g(x) = \sqrt{8x}[/tex]

The square root function assumes values that are greater than or equals to zero, hence the domain of the function is obtained as follows:

8x ≥ 0.

x ≥ 0.

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The probability P (4)  and of getting more than 2 questions correct are 0.017 and 0.113 respectively.

The number of ways the student can answer each question is 4 (since there are 4 choices), so the probability of getting any one question correct by guessing is 1/4, and the probability of getting any one question wrong by guessing is 3/4.

We can use the binomial probability formula to find the probability of getting a specific number of questions correct out of the 10:

[tex]P(X = k) = ( ^kC _n) * p^k * (1-p)^(n-k)[/tex]

a) P(4) represents the probability of getting exactly 4 questions correct out of 10.

[tex]P(X = 4) = (^{10}C_4) * (1/4)^4 * (3/4)^{(10-4)} \approx 0.017[/tex]

So the probability of getting exactly 4 questions correct is approximately 0.017.

b) P(more than 2) represents the probability of getting 3, 4, 5, ..., or 10 questions correct out of 10. We can use the complement rule to find this probability:

P(more than 2) = 1 - P(X ≤ 2)

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\\= (^{10} C_0) * (1/4)^0 * (3/4)^{10}+ (^{10} C_1) * (1/4)^1 * (3/4)^9+ (^{10} C_2) * (1/4)^2 * (3/4)^8\\\approx 0.887[/tex]

So,

P(more than 2) = 1 - P(X ≤ 2) ≈ 1 - 0.887 ≈ 0.113

Therefore, the probability of getting more than 2 questions correct is approximately 0.113.

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The rate of change for the interval between 0 and 2 for the quadratic equation is 10.

A quadratic equation is a type of polynomial equation of the second degree, which means it has one or more terms that are raised to the power of two.

The general form is;

ax² + bx + c = 0

Where;

The rate of change for the interval between 0 and 2 for the quadratic equation is calculated as;

f(2) - f(0) = 7 - (-3)

             = 7 + 3

             = 10

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The missing lengths of the following two triangles:

Case 1: x = 50 km

Case 2: x = 40 km

Case 3: Obtuse

Case 4: Acute

Case 5: Right

Case 6: m = 16, n = 8√3

Case 7: x = 3√2, y = 3

In this problem we must determine all missing lenghts in a triangle, this can be done by law of cosine:

x² = a² + b² - 2 · a · b · cos X

Where X is the angle opposite to side x.

Now we proceed to determine the missing lengths of the following two triangles:

Case 1

x = √[(14 km)² + (48 km)² - 2 · (14 km) · (48 km) · cos 90°]

x = 50 km

Case 2

x = √[(24 km)² + (32 km)² - 2 · (24 km) · (32 km) · cos 90°]

x = 40 km

In addition, we can determine if any triangle is acute, obtuse and right also by law of cosine:

cos X = - (x² - a² - b²) / (2 · a · b)

Case 3

cos X = - [(18 in)² - (12 in)² - (9 in)²] / [2 · (12 in) · (9 in)]

cos X = - 0.458 (Obtuse)

Case 4

cos X = - [(15 ft)² - (12 in)² - (14 in)²] / [2 · (12 in) · (14 in)]

cos X = 0.342 (Acute)

Case 5

cos X = - [(5 ft)² - (3 in)² - (4 in)²] / [2 · (3 in) · (4 in)]

cos X = 0 (Right)

Finaly, we determine the missing lengths of two right triangles by trigonometric functions:

sin θ = y / r

cos θ = x / r

tan θ = y / x

Now we find the missing lengths:

Case 6

m = 8 / cos 60°

m = 16

n = 8 · tan 60°

n = 8√3

Case 7

x = 3 / sin 45°

x = 3√2

y = 3 / tan 45°

y = 3

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The quotient of the long division of the polynomial (6x³ - 4x² + 11)/(x+3)  is 6x² - 22x + 66.

The quotient of the polynomial divided by a factor of x + 3 is determined by applying long division method as shown below;

6x³ - 4x² + 11 ÷  x + 3

                   6x² - 22x + 66

              ------------------------

  x + 3   √ 6x³ - 4x² + 11

               - (6x³ + 18x²)

             -------------------------

                         -22x² + 11

                    -   (-22x² - 66x)

                -----------------------------

                                  66x + 11

                            -  (66x + 198)

                 ---------------------------------

                                      -187

Thus, the quotient of the long division of the polynomial is obtained as   6x² - 22x + 66.

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Find the quotient of the polynomial using long division

6x3 − 4x2 + 11/x + 3

Yes, the table represents a proportional relationship.

The constant of proportionality is equal to 9.

In Mathematics and Geometry, a proportional relationship refers to a type of relationship that produces equivalent ratios and it can be modeled or represented by the following mathematical equation:

y = kx

Where:

Next, we would determine the constant of proportionality (k) by using the data points contained in the table as follows:

Constant of proportionality, k = y/x

Constant of proportionality, k = 18/2 = 45/5 = 63/7

Constant of proportionality, k = 9.

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Complete Question:

Tell whether each table represent a proportional relationship. If it does, identify the constant of proportionality.

1. X y 2 18 5 45 7 63​

When it comes to interest, most Certificates of Deposit (CDs) will allow periodic interest payouts. So, correct option is B.

A CD is a type of savings account that allows you to earn a fixed interest rate on your deposit for a specific period of time, called the term. Typically, the longer the term of the CD, the higher the interest rate offered.

However, during the term of the CD, the funds are locked in, meaning that you cannot withdraw the principal without incurring a penalty.

While some CDs may allow for unlimited withdrawals of interest, this is not common, and may still come with restrictions or penalties. Loans against principal are also not typically allowed with CDs, as the funds are meant to be held for a set term.

Therefore, the most common option available for CD holders is to receive periodic interest payouts, which can be monthly, quarterly, or annually, depending on the terms of the CD. This allows the CD holder to earn interest on their deposit while still receiving some income during the term of the CD.

So, correct option is B.

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Answer:

y=mx + b

Step-by-step explanation:

y=3x-4

3 is the slope and -4 is the y intercept

The amount that you would save in annual fuel expenses would be $9, 333. 32

The amount that you would have saved over the five years would be $52, 854.63

The amount saved in fuel when using the hybrid is:

= Fuel usage by SUV - Fuel usage by Hybrid

= ( 30, 000 / 9 x 4 ) - ( 30, 000 / 30 x 4 )

= $9, 333.32

If you saved this amount as an annuity, it would come out to:

=  ( 9, 333.32 / 12 ) x  [ ( 1 + 0. 004333 ) ^ ( 12 x 5 ) - 1 ] / 0. 004333

= 777. 78 x 0. 2957 / 0. 004333

= $ 52, 854.63

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Answer:3.77

2.51 are correct

Step-by-step explanation:We can solve this quadratic equation in cos(θ) by using the substitution u = cos(θ):

-7u^2 + 3u + 7 = 0

Multiplying both sides by -1, we get:

7u^2 - 3u - 7 = 0

We can use the quadratic formula to solve for u:

u = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 7, b = -3, and c = -7. Substituting these values, we get:

u = (3 ± sqrt(9 + 196)) / 14

u = (3 ± 5sqrt(5)) / 14

Therefore, either:

Answer:

Step-by-step explanation:

35.26    rounded to  35.3

8.32   rounded to        8.3

6.78    rounded to       6.8

35.3 + 8.3 = 43.6

43.6 + 6.8 = 50.4

The answer is 50.4.

39.26 rounds to 35.3

8.32 rounds to 8.3

6.78 rounds to 6.8

When you add them together, you get 50.4.

Answer:

[tex] {( \frac{1}{5}) }^{3} (312) = \frac{312}{125} = 2.496[/tex]

The volume of this rectangular prism is 2.496 cubic centimeters.

The volume of the rectangular prism is 2.496 cubic centimeters.

If the rectangular prism has 312 cubes with an edge length of 1/5 cm, we may calculate the total volume by multiplying the number of cubes by the volume of each cube.

V = x3 gives the volume of a cube with edge length x. Because the edge length, in this case, is 1/5 cm, the volume of each cube is (1/5)3 = 1/125 cm3.

We multiply the number of cubes by the volume of each cube to determine the volume of the rectangular prism:

312 cubes (1/125 cm3/cube) = 2.496cm3 volume

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Answer:

E. 108 in.²

Step-by-step explanation:

The piece of wood has the shape of a triangular prism.

SA = lateral area + area of the bases

SA = perimeter × height + 2 × bh/2

SA = (5 + 3 + 4) in. × 8 in. + 3 in. × 4 in.

SA = 108 in.²

David has 15 dimes and 5 quarters.

Given that, David has twenty dimes (d) and quarters (q). These coins total $2.75.

Establishing the system of equations,

d + q = 20

d = 20 - q.............(i)

0.1d + 0.25q = 2.75..........(ii)

Put eq(i) in eq(ii),

0.1(20-q)+0.25q = 2.75

2-0.1q + 0.25q = 2.75

0.15q = 0.75

q = 5

Put q = 5 in eq(i)

d = 20-5

d = 15

Hence, David has 15 dimes and 5 quarters.

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The value of the test statistic used in this test is t = -2.168 and the P-value of this test is P-value is 0.034

To test whether the mean time for Dr. McCoy is different from the mean time for Dr. Turk, we can use a two-sample t-test with unequal variances. The test statistic for this test is given by:

t = (X₁ - X₂) / √(s₁²/n₁ + s₂²/n₂)

where X₁ and X₂ are the sample means, s₁ and s₂ are the sample standard deviations, and n₁ and n₂ are the sample sizes.

Plugging in the given values, we get:

t = (78 - 78.7) / √(2.3²/38 + 4.1²/33)

= -2.168

Therefore, the value of the test statistic used in this test is t = -2.168.

b) To find the P-value of this test, we need to use a t-distribution with degrees of freedom given by:

df = (s₁²/n₁+ s₂²/n₂)² / [ (s₁²/n₁)²/(n₁-1) + (s₂²/n₂)²/(n₂-1) ]

Plugging in the given values, we get:

df = (2.3²/38 + 4.1²/33)² / [ (2.3²/38)²/37 + (4.1²/33)²/32 ] ≈ 65.385

Using a t-distribution table or a calculator with t-distribution function, we find that the P-value for a two-tailed test with t = -2.168 and df = 65.385 is approximately 0.034.

Therefore, the P-value of this test is P-value = 0.034

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The value of a in the equation 60°a = [? ]° is dependent on the value of [? ].

To solve for a in the equation 60°a = [? ]°, we need to isolate a on one side of the equation. We can do this by dividing both sides of the equation by 60°:

60°a / 60° = [? ]° / 60°

This simplifies to:

a = [? ]° / 60°

Since we don't have the value of [? ]°, we cannot determine the exact value of a. However, we can simplify the expression by canceling out the ° units:

a = [? ] / 60

Therefore, the value of a is dependent on the value of [? ].

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The missing parts of the triangle are

BD = 8

AD = 8 sqrt(2)

The missing part BD of the figure is solved using similar triangles

The expression is as follows

h / 8 = 8 / h

h^2 = 8 x 8

h^2 = 64

h sqrt = (64)

h = 8

Solving for AB we use Pythagoras theorem

AB^2 = h^2 + 8^2

AB^2 = 8^2 + 8^2

AB = sqrt(64 + 64)

AB = sqrt (128)

AB = 8 sqrt (2)

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The polynomial 18x^3 + 6x^2y - 9x^2 - 3xy when factored completely is (6x^2 - 3x)(3x + y)

From the question, we have the following parameters that can be used in our computation:

18x^3 + 6x^2y - 9x^2 - 3xy

Factoize the expression

So, we have

6x^2(3x + y) - 3x(3x + y)

Factor out 3x + y

So, we have

(6x^2 - 3x)(3x + y)

Hence, the solution is (6x^2 - 3x)(3x + y)

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