A bank has determined that the monthly balances of the checking accounts of its customers are normally distributed with an average balance of $1,200 and a standard deviation of $250. What is the probability that a randomly selected bank balance will be less than $1,000?

Answer:

$1,593

Step-by-step explanation:

I=PRT

I=$2,276×14%×5

I=1,593.2 to nearest tenth

I=$1,593

Answer:

There are three ways to roll a sum of 4: (1,3), (2,2), and (3,1). There are 36 possible outcomes when rolling two dice because each die has 6 possible outcomes, so we multiply the number of outcomes of each die: 6 x 6 = 36.

Therefore, the probability of rolling a sum of 4 with two standard dice is:

P(D1 + D2 = 4) = number of ways to roll a sum of 4 / total number of outcomes

P(D1 + D2 = 4) = 3 / 36

Simplifying the fraction, we get:

P(D1 + D2 = 4) = 1 / 12

Therefore, the probability of rolling a sum of 4 with two standard dice is 1/12 or approximately 0.083.

Step-by-step explanation:

in answer

Step-by-step explanation:

so now we are going to find the area of larger figure and subctract the rectangle from the figure

Area of triangle = 1/2 bh

= 1/2 (12) (12)

= 72 m2

Area of rectangle = L× W

= 3 × 9

=27 m2

Area of the figure= (72-27) m2

=45m2

so the figure has area of shaded region 45m2

Answer:

The answer to your problem is, 2.04 inches

Step-by-step explanation:

We can assume that the width of the new phone is ‘ x ‘ inches

We know that [tex]\frac{x}{0.84} = \frac{3.4}{1.4}[/tex]

x = [tex]\frac{3.4}{1.4}[/tex] × 0.84

x = 2.04

Thus the answer to your problem is, 2.04 inches

The average rate at which the object falls during the first 3 seconds is given as follows:

-48.

The average rate of change of a function is given by the change in the output of the function divided by the change in the input of the function.

The function for this problem is defined as follows:

h(x) = 400 - 16x².

The numeric values are given as follows:

Thus the average rate of change is obtained as follows:

r = (256 - 400)/(3 - 0)

r = -48.

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The probability that the standard normal random variable z is less than -0.87 is approximately 0.1922.

To find the probability P(Z < -0.87) for a standard normal random variable or z-score, follow these steps:

1. Identify the given value: In this case, z = -0.87.

2. Look up the corresponding probability in a standard normal (Z) table or use a calculator or software with a built-in function for finding probabilities of standard normal random variables.

Using a standard normal table or software, you'll find that P(Z < -0.87) ≈ 0.1922. So, the probability that the standard normal random variable Z is less than -0.87 is approximately 0.1922.

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Yes, that is correct. The Gaussian elimination rules are essentially the same as the three basic row operations, which allow you to algebraically manipulate a matrix's rows.

You can interchange two rows, multiply a row by a constant, or add a row to another row. These rules are essential in solving systems of linear equations and finding the reduced row echelon form of a matrix. By applying these rules, you can transform a matrix into an equivalent matrix that is easier to work with and reveals important information about the system of equations or the matrix itself. The Gaussian elimination rules, also known as the three basic row operations, allow you to algebraically manipulate a matrix in order to solve systems of linear equations. These operations include:
1. Interchanging two rows (R2 ↔ R3)
2. Multiplying a row by a nonzero number (R1 → kR1, where k is a constant)
3. Adding a row to another row (R2 → R2 + 3R1)
These rules help simplify the matrix and ultimately obtain the unique solution for the system of equations.

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The functions x^2 + 2x and 3x^2 + kx are linearly independent if k is not equal to 0 or 1.

We need to use the Wronskian to determine if the two functions x^2 + 2x and 3x^2 + kx are linearly independent. If the Wronskian is nonzero for all x, then the functions are linearly independent.

The Wronskian of two functions f(x) and g(x) is defined as:

W(f,g)(x) = f(x)g'(x) - g(x)f'(x)

Let's find the Wronskian of x^2 + 2x and 3x^2 + kx:

W(x^2 + 2x, 3x^2 + kx)(x) = (x^2 + 2x)(6x + k) - (3x^2 + kx)(2x + 2)

= 6x^3 + 2kx^2 + 12x^2 + 4kx - 6x^3 - 6kx - 6x^2 - 6x^2

= -2kx^2 + 2kx

We want the Wronskian to be nonzero for all x, which means that -2kx^2 + 2kx cannot be zero for any value of x, except possibly at x = 0. Therefore, we need to find the values of k that make -2kx^2 + 2kx = 0 only at x = 0.

Factoring out 2kx, we get:

-2kx(x - 1) = 0

This expression is equal to zero when x = 0 or x = 1. We want it to be zero only when x = 0, so we need to set the factor (x - 1) to a nonzero constant. This means k cannot be equal to zero or 1.

Therefore, the functions x^2 + 2x and 3x^2 + kx are linearly independent if k is not equal to 0 or 1.

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The area of her garden is 6,330 units² (option b).

We know that the area of the trapezoid plus the area of the right triangle is equal to the area of the rectangle:

Area of trapezoid + Area of right triangle = L x W

We can substitute in the formulas we found earlier for the areas of the trapezoid and the right triangle:

(1/2) x hT x L + (1/2) x W x (L - hT) = L x W

Simplifying and solving for hT, we get:

hT = (2W - L) / 2

Now we can plug this value into the formula for the area of the trapezoid:

Area of trapezoid = (1/2) x hT x L

= (1/2) x [(2W - L) / 2] x L

= (W - L/2) x L

To find the area of the garden, we need to subtract the area of the fish pond (which is the area of the right triangle) from the area of the rectangle. We already found the formula for the area of the right triangle:

Area of right triangle = (1/2) x W x (L - hT)

So the area of the garden is:

Area of garden = L x W - Area of right triangle

= L x W - (1/2) x W x (L - hT)

Substituting in the formula we found earlier for hT, we get:

Area of garden = L x W - (1/2) x W x (L - (2W - L)/2)

= L x W - (1/2) x W x (W/2)

Simplifying, we get:

Area of garden = (3/4) x L x W

Now we can substitute in the values given in the problem to find the area of the garden:

L = 60 units

W = 110 units

Area of garden = (3/4) x L x W

= (3/4) x 60 x 110

= 6,330 units²

Hence option (b) is the right one.

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Answer: I think its 2x + c

Hope it helped :D

Let me know if I helped

Sorry if wrong

The general solution of the O.D.E y²q2 + y2 + 22 +1 = y' involves solving for y in terms of x and a constant, represented by "c". To do this, we can use the technique of separation of variables.

First, we rearrange the equation to isolate the derivative term on one side:

y²q2 + y² + 22 + 1 = y'
y²q2 + y² + 1 = y' - 2
(y²q2 + y² + 1)dy = dx

Next, we integrate both sides with respect to their respective variables:

∫(y²q2 + y² + 1)dy = ∫dx
y³/3 + y + y = x + c
y³ + 3y = 3x + c

At this point, we can use the trigonometric substitution u = tan(x/3 + c) to simplify the expression. Then, we can solve for y in terms of u:

u = tan(x/3 + c)
y = √(u² - 1/3)

Finally, we substitute back in the expression for u and simplify to obtain the general solution:

y = √(tan²(x/3 + c) - 1/3)
y = tan(x/3 + c)

Therefore, the answer is (a) y = tan(x/3 + c).

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Once it reaches the end of the journey, the vessel is approximately 13.68 nautical miles away from the port.

To solve this problem, we can use the law of cosines to find the distance from the vessel to the port. Let's label the angles and sides as follows:

- Angle A is the initial heading of 32 degrees
- Angle B is the counterclockwise change in heading of 32 degrees
- Angle C is the angle between sides a and b (the distance from the vessel to the port)
- Side a is the distance traveled in the first leg, which is 5 nautical miles
- Side b is the distance traveled in the second leg, which is 10 nautical miles
- Side c is the distance from the vessel to the port, which we want to find

Using the law of cosines, we have:

c^2 = a^2 + b^2 - 2ab cos(C)

Plugging in the values we know, we get:

c^2 = 5^2 + 10^2 - 2(5)(10) cos(180-32)

Note that we use 180-32 for the angle C because it is the supplement of angle B.

Simplifying, we get:

c^2 = 125 - 100cos(148)

Using a calculator, we find that cos(148) is approximately -0.6235. Plugging this in, we get:

c^2 = 187.35

Taking the square root, we get:

c = 13.68 nautical miles

Therefore, the vessel is approximately 13.68 nautical miles away from the port once it reaches the end of the journey.

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Solving the Inequality will give us x > -1.25

We have the inequality:-

0. 2(x + 20) – 3 > –7 – 6. 2x

Here we are given a set of instructions and need to use them to get the final answer.

First, we need to use the distributive property. Using that, we will eliminate the brackets to get

0.2x + (0.2 X 20) - 3 > - 7 - 6.2x

or, 0.2x + 4 - 3 > - 7 - 6.2x

Now, we need to combine the like terms. Here it would be 4 and -3. Hence we get

0.2x + 1 > - 7 - 6.2x

Next, we need to use the addition property of inequality, which states that for terms a, b, c

if a > b

then, a + c > b + c

Hence here we will add 6.2x to get

0.2x + 1 + 6.2x > - 7 - 6.2x + 6.2x

or, 6.4x + 1 > - 7

Similarly, applying the subtraction property of inequality states

if a > b

then, a - c > b - c

Here we need to subtract 1 to get

6.4x + 1 - 1 > - 7 - 1

or, 6.4x > - 8

Using the division property will give us

x > - 8/6.4

or, x > 1.25

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We have found the measures of CD, CE, HD, GD, HG, and HF in triangle BCD, given that H is the circumcenter of the triangle.

In triangle BCD, the circumcenter H is the point where the perpendicular bisectors of the sides of the triangle intersect. This point is equidistant from the three vertices of the triangle.

Using the properties of the circumcenter, we can find the measures of various sides and angles of the triangle:

CD = 2FD, where FD is the foot of the perpendicular from H to CD.

CE = BE = 26, since H is equidistant from B and C.

HD = HC = 33, since H is equidistant from D and C.

GD = 1/2BD = 1/2(58) = 29, since H is equidistant from B and D.

HG = √HD² - GD² = √33² - 29² = 2√62 ≈ 15.75, using the Pythagorean theorem.

HF = √HD² - FD² = √33² - 32² = √65 ≈ 8.06, using the Pythagorean theorem.

Therefore, we have found the measures of CD, CE, HD, GD, HG, and HF in triangle BCD, given that H is the circumcenter of the triangle.

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If A and B are independent events with P(A) = 0.4 and P(B) = 0.25, then P(A È B) =The answer is d. 0.55.

Since A and B are independent events, we can use the formula for the probability of the union of two independent events: P(A ∪ B) = P(A) + P(B) - P(A)P(B). Given P(A) = 0.4 and P(B) = 0.25, we can calculate P(A ∪ B) as follows:

P(A ∪ B) = 0.4 + 0.25 - (0.4)(0.25) = 0.4 + 0.25 - 0.10 = 0.55

Or we can calculate as follows:

To find P(A È B), we use the formula: P(A È B) = P(A) + P(B) - P(A and B)

Since A and B are independent, P(A and B) = P(A) x P(B) = 0.4 x 0.25 = 0.1

Therefore, P(A È B) = 0.4 + 0.25 - 0.1 = 0.55.

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All the possible values of x are given as follows:

26 < x < 28.

In a triangle, the sum of the lengths of the two smaller sides has to be greater than the length of the greater side.

If 27 is the greater side, we have that:

x + 1 > 27

x > 26.

If x is the greater side, we have that:

x < 27 + 1

x < 28.

Hence the interval of possible values is given as follows:

26 < x < 28.

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The  sum of the vectors u and v is (-1,1).

To find the sum of the vectors u and v, we need to add their corresponding components.

The vector u has initial point (2,0) and terminal point (3,2), which means its components are (3-2, 2-0) = (1,2).

The vector v has initial point (2,2) and terminal point (0,1), which means its components are (0-2, 1-2) = (-2,-1).

To find the sum of these vectors, we simply add their corresponding components:

u + v = (1,2) + (-2,-1) = (1-2, 2-1) = (-1,1).

Therefore, the sum of the vectors u and v is (-1,1).

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The scale factor of dilation from the small figure to the large figure in the question are;

3. 1 : 4

4. 5 : 6

A scale factor is the ratio of the length of a side of an image (obtained from a preimage) to the length of the corresponding side of the preimage

The scale factor of the polygons obtained from diagrams are;

3. 4.5 yd to 18 yd = 1 to 4

The scale factor is 1 to 4

4. The ratio of the corresponding sides pairs of sides on the image and the preimage are;

Ratio on the large triangle; 42 : 18 = 7 : 3

Ratio on the small triangle; 35 : 15 = 7 : 3

The ratio  of the pair of corresponding sides are equivalent, therefore, the triangles are similar.

The scale factor of the small triangle to the large triangle, obtained from the ratio of the corresponding sides is therefore;

35 : 42 = 5 : 6

15 : 18 = 5 : 6

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The volume of the cone is 180π cubic feet (or approximately 565.49 cubic feet if you evaluate π as 3.14159).

To calculate the volume of a cone, you can use the formula V = (1/3)πr^2h, where r is the radius of the base of the cone and h is its height.

Since the diameter of the cone is 12 feet, the radius is half of that, which is 6 feet. And the height is given as 15 feet.

Plugging these values into the formula of volume, we get:

V = (1/3)π[tex](6)^2[/tex](15)

V = (1/3)π(36)(15)

V = (1/3)(540π)

V = 180π

Thus, the answer is 180π.

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No, The relation is not a function.

Domain = {9, 4, 1, 0}

Range = {-5, - 3, - 1, 0, 1, 3, 5}

We have to given that;

The relation is,

{(9,−5),(4,−3),(1,−1),(0,0),(1,1),(4,3),(9,5)}

We know that;

A relation between a set of inputs having one output each is called a function.

Here, Relation have;

(9, - 5) and (9, 5)

Thus, It does not satisfy the definition of function.

And, The value of domain of relation is,

Domain = {9, 4, 1, 0}

And, The value of range of relation is,

Range = {-5, - 3, - 1, 0, 1, 3, 5}

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Answer:

B

Step-by-step explanation:

Sara read for a constant rate for 6 days and got a total of 7 1/2 hours.

Therefore, the situation can be represented by the equation 1 1/4×6=7 1/2

The car's speed is 60 km/hour.

Let's denote the distance from the starting point to the destination by D, and let's denote the car's speed by S.

Using the formula speed = distance / time.

S = d  / t = (D - 200) / 1  ---- (1)

S = d / t = (D - 80) / 3  ----- (2)

We can simplify equation (2) by multiplying both sides by 3:

Expanding the right-hand side:

3S = D - 80

From equation 1 and 2:

3 (D - 200) = D - 80

3D - 600 = D - 80

3D - D = 600 - 80

2D = 520

D = 260

Therefore, the distance from the starting point to the destination is 260 km.

Using equation (1), we can find the car's speed:

S = 260 - 200 / 1

S = 60 m/s

Therefore, the car's speed is 60 km/hour.

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The complete question is:

A car is 200km from its destination after 1 hour and 80km from its destination after 3 hours. At what rate is the car traveling per hour?

When a French historian spends a semester touring museums and historic battlefields in the United States, it affects U.S. exports, imports, and net exports as follows:

- U.S. Exports: The French historian's spending on tourism services (such as accommodations, guided tours, and local transportation) is considered an export of services. As the historian spends money in the U.S., it will lead to an increase in U.S. exports.
- U.S. Imports: There is no direct impact on U.S. imports, as the historian's activities do not involve the U.S. purchasing goods or services from France or any other country.
- Net Exports: Since the French historian's spending increases U.S. exports without affecting imports, this will result in an increase in U.S. net exports (which is the difference between exports and imports).

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The spares should be kept to maintain a 90% or more probability of system success is 35

Let λ be the normal number of component disappointments per year, at that point the disappointment rate (or rate parameter) is given by λ/52 since there are 52 weeks in a year. Let's signify this by μ = λ/52.

The framework victory likelihood can be modeled utilizing the Poisson dissemination since the disappointments happen haphazardly and freely over time. Let X be the number of component disappointments in a year, at that point X takes after a Poisson conveyance with cruel λ.

To preserve a 90% or more likelihood of system victory, we got to guarantee that the number of saves is adequate to cover at the slightest 90% of the potential disappointments. This implies that the likelihood of having more than k disappointments in a year ought to be less than or break even with 0.1, where k is the number of saves.

Let Y be the number of component disappointments amid the two-week conveyance time. At that point, Y too takes after a Poisson conveyance with cruel μ/26, since there are 26 weeks in a half-year (i.e., two quarters). The likelihood of having more than k disappointments amid the conveyance time is given by:

P(Y > k) = 1 - P(Y ≤ k) = 1 - ∑_[tex]{i=0}^k (e^(-μ/26) (μ/26)^i[/tex]/ i!)

where e is the base of the common logarithm.

To preserve a 90% or more likelihood of system victory, we ought to select k such that P(Y > k) ≤ 0.1. Able to solve for k numerically, employing a spreadsheet or a computer program.

For example, utilizing Microsoft Exceed expectations or Go-ogle Sheets, ready to utilize the taking after an equation to compute P(Y > k) for distinctive values of k:

=1-POISSON(k,μ/26,TRUE)

where POISSON is the Poisson total conveyance work, with the moment contention being the cruel and the third contention being Genuine to indicate an aggregate conveyance.

Beginning with k = 26 (i.e., one save per week), able to increment k until we discover the littlest esteem that fulfills P(Y > k) ≤ 0.1. In this case, we discover that k = 35 saves are required to preserve a 90% or more likelihood of framework victory. 

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The volume of the snowball is decreasing at a rate of approximately 5.4 cm³/min when the radius is 12 cm.

The formula for the volume of a sphere is V = (4/3)πr³, where V is the volume and r is the radius. To find the rate of change of the volume with respect to time, we need to take the derivative of this formula with respect to time. Using the chain rule, we get:

dV/dt = (4/3)π(3r²)(dr/dt)

where dV/dt is the rate of change of the volume with respect to time and dr/dt is the rate of change of the radius with respect to time.

Substituting the given values, we get:

dV/dt = (4/3)π(3(12)²)(-0.3)

= -241.9π cm³/min

Since the rate of change of volume cannot be negative, we take the absolute value of the result to get:

|dV/dt| = 241.9π cm³/min ≈ 759.8 cm³/min

Therefore, the volume of the snowball is decreasing at a rate of approximately 5.4 cm³/min when the radius is 12 cm.

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For the given sequence a(26) = 2

To find a(26), we can use the recursive definition of the sequence and work our way up from a(0) and a(1):

a(0) = -1
a(1) = 1

a(2) = a(1) - a(0) = 1 - (-1) = 2
a(3) = a(2) - a(1) = 2 - 1 = 1
a(4) = a(3) - a(2) = 1 - 2 = -1
a(5) = a(4) - a(3) = -1 - 1 = -2
a(6) = a(5) - a(4) = -2 - (-1) = -1
a(7) = a(6) - a(5) = -1 - (-2) = 1
a(8) = a(7) - a(6) = 1 - (-1) = 2

From this pattern, we can see that the sequence repeats with a period of 6, so we can find a(26) by finding the remainder when 26 is divided by 6:

a(26) = a(26 mod 6) = a(2) = 2

Therefore, a(26) = 2.

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The circle has a diameter of 10 and a radius of 5

the radius times itself ( 5 x 5 ) = 25

25 times pi (3.14) = 78.5

so the circle is 78.5, lets say square cm.

and the circumference is 31.41593 or 31.42  

we need to find the center point and the length of the radius. The center point is at 5. 1.  so h = 5  and  k = 1

Now let’s count from the center to a point on the circumference to find the length of the radius. 5

The radius is 5 so  r = 5

Now let’s plug everything into the standard form of a circle.

(x - h)²+ (y - k)² = r²

The use of a stratified random sample over a simple random sample provides a statistical advantage in ensuring that the sample accurately represents the population. Stratified random sampling involves dividing the population into subgroups or strata based on certain characteristics, such as age or income.

This allows for a more representative sample as it ensures that each stratum is represented proportionally in the sample. In contrast, a simple random sample does not take into account any characteristics or strata of the population, which may result in an unrepresentative sample. Therefore, the use of a stratified random sample provides a statistical advantage by reducing the potential for sampling bias and increasing the accuracy of the study's results.
In the context of your study, a statistical advantage of using a stratified random sample over a simple random sample is that it ensures greater representation and accuracy in the results.

In a stratified random sample, the population is first divided into distinct subgroups or strata based on specific characteristics, such as age, gender, or income. Then, a simple random sample is taken from each stratum. This method helps to better represent each subgroup in the sample, which in turn improves the overall accuracy of the results.

On the other hand, a simple random sample involves selecting individuals from the entire population without considering any specific characteristics. This approach may not adequately represent certain subgroups, leading to potential biases and less accurate results. In summary, stratified random sampling provides a statistical advantage over simple random sampling by ensuring a better representation of subgroups, leading to more accurate and reliable results.

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In a randomized block design, total variation in data is decomposed into three components: variation between blocks (SSB), variation between treatments within each block (SSE), and residual variation within each treatment and block combination (SST).

We can express the total sum of squares (Total SS) in a randomized block design as:

Total SS = SSB + SSE + SST

∑Yij² = SST + SSB + SSE and ∑Yij = bYi• + b∑y•j - bY

SSE = ∑(Yij - y•j - Yi• + Y)²

       = ∑Yij² - 2∑Yijy•j - 2∑YijYi• + 2∑YijY + b∑y•j² + b∑Yi•² - 2bY∑y•j - 2bY∑Yi• + bY²

       = SST + SSB + SSE - b∑y•j² - b∑Yi•² + 2bY∑y•j + 2bY∑Yi• - bY²

Rearranging and simplifying terms:

SSE = b∑y•j² + b∑Yi•² - 2bY∑y•j - 2bY∑Yi• + bY²

Multiplying both sides by k, the number of treatments:

kSSE = bk∑y•j² + bk∑Yi•² - 2bkY∑y•j - 2bkY∑Yi• + bkY²

Also, SST = ∑(Yij - Yi•)² and SSB = ∑(Yi• - Y)²

Therefore,

Total SS = SST + SSB + bk∑y•j² + bk∑Yi•² - 2bkY∑y•j - 2bkY∑Yi• + bkY²

Expanding the terms ∑Yi•² and ∑y•j² using the fact that ∑Yij² = SST + SSB + SSE:

Total SS = SST + SSB + bk(SST + SSB + SSE) - 2bkY∑y•j - 2bkY∑Yi• + bkY²

Simplifying the terms:

Total SS = bkSST + bkSSB + bkSSE - 2bkY∑y•j - 2bkY∑Yi• + bkY

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George's friend Clarence, who is concerned about consumers, suggests implementing a price ceiling that is 50% lower than the monopoly price. With this price ceiling in place, the quantity demanded would increase to 65 units. However, the exact quantity supplied at this new price is not provided in your question.

George's friend Clarence is suggesting a price ceiling, which is a government-imposed limit on how high a price can be charged for a good or service. In this case, the price ceiling is set at 50% below the monopoly price. This means that the price charged for the good or service would be significantly lower than what the monopolistic supplier would typically charge.

At this lower price point, Clarence predicts that the quantity demanded would increase to 65 units, indicating that consumers would be more willing to purchase the product due to its lower price. However, it is unclear what the quantity supplied would be at this price point, as that information is missing from the question.

Overall, George's friend Clarence's suggestion of a price ceiling serves to protect consumers by limiting the monopolistic supplier's ability to charge exorbitant prices and instead promoting a more competitive market.

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