An aquarium 6 ft long, 4 ft wide, and 2 ft deep is full of water. (Recall that the weight density of water is 62.5 lb/ft3.)(a) Find the hydrostatic pressure on the bottom of the aquarium. (give in answer in lb/ft2)(b) Find the hydrostatic force on the bottom of the aquarium. (give in answer in lb)(c) Find the hydrostatic force on one end of the aquarium. (give in answer in lb)

Answer:700

Step-by-step explanation:

Okay, here are the steps:

* Hassam buys 2 adult tickets and 2 child tickets

* Let's assume:

** Adult ticket price = $50

** Child ticket price = $25

* So total ticket price = 2 * $50 + 2 * $25 = $200

* The booking fee is 5% of the total cost

* 5% of $200 is $10

* So total payment = $200 + $10 = $210

Therefore, the total Hassam pays altogether is $210

Therefore, Don's car would use 67.035 L of gas to travel 735 km.

Cross multiplication is a method used to compare the relative sizes of two fractions. To cross-multiply two fractions, you multiply the numerator of one fraction by the denominator of the other fraction, and then do the same with the other numerator and denominator, putting the two products equal to each other. For example, if you have the fractions 2/3 and 3/4, you can cross-multiply as follows:

[tex]2/3 = x/4[/tex]

[tex]2 x 4 = 3 x x[/tex]

[tex]8 = 3x[/tex]

[tex]x = 8/3[/tex]

So, the two fractions are equivalent, and both are equal to 8/3. Cross-multiplication can also be used to solve equations that involve fractions, by isolating the variable on one side of the equation.

To find out how much gas Don's car would use to travel 735 km, we can use the fact that the car uses 9.1 L of gas every 100 km. We can set up a proportion to solve for the amount of gas used:

9.1 L / 100 km = x L / 735 km

To solve for x, we can cross-multiply and simplify:

[tex]9.1 L * 735 km = 100 km * x L[/tex]

[tex]6703.5 Lkm = 100 km * x L[/tex]

[tex]6703.5 Lkm / 100 km = x L[/tex]

[tex]67.035 L = x L[/tex]

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Answer:6.9 and 8.5

Step-by-step explanation: 7.7-0.8=6.9 hours

7.7+0.8=8.5 hours

To use the Chi Squared test on the resulting data we can use the following statements in order:

D. He takes a random sample of balloons.

B. He samples 36 balloons at most.

C. He expects each color to appear at least 5 times.

A statistical hypothesis test used to examine if a variable is likely to come from a specific distribution is the Chi-square goodness of fit test. To ascertain if sample data is representative of the total population, it is widely utilised.

To let you know if there is a correlation, the Chi-Square test provides a P-value. An assumption is being considered, which we can test later, that a specific condition or statement may be true. Consider this: The data collected and expected match each other quite closely, according to a very tiny Chi-Square test statistic. The data do not match very well, according to a very significant Chi-Square test statistic. The null hypothesis is disproved if the chi-square score is high.

Here in the question,

To use the Chi Squared test on the resulting data we can use the following statements in order:

D. He takes a random sample of balloons.

B. He samples 36 balloons at most.

C. He expects each color to appear at least 5 times.

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The complete question is:

Peter bought a big pack of

360

360360 party balloons. The balloons come in

6

66 different colors which are supposed to be distributed evenly in the pack.

Peter wants to test whether the distribution is indeed even, but he doesn't want to go over the entire pack. So, he plans to take a sample and carry out a

χ

2

χ

2

\chi, squared goodness-of-fit test on the resulting data.

Which of these are conditions for carrying out this test? Choose 3 options.

A. He observes each color at least 5 times.

B. He samples 36 balloons at most.

C. He expects each color to appear at least 5 times.

D. He takes a random sample of balloons.

Answer: the answer is A B D

Step-by-step explanation:

The resulting matrix using the specified row transformation to change the matrix; - 4 times row 1 added to row 2 is 0 -8

To apply the specified row transformation, we need to subtract 4 times the first row from the second row.

So the resulting matrix will be:

[  2              3

8 - 4 ( 2 )   4 - 4 ( 3 ) ]

which simplifies to:

[ 2   3

0   -8 ]

Therefore the resulting matrix for the specified row transformation is 0 and - 8.

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s = −81/4,t = −34 so its

Step-by-step explanation:

The solution of the given initial value problem for r as a vector function of t is [tex]r(t) = 3(t + 1)^{(3/2)} i + (-6 e^{-t} + 6) j + (ln(t + 1) + 1) k[/tex].

A differential equation is an equation that contains one or more terms and the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable).

To solve the given differential equation, we will integrate each component of the differential equation and apply the initial condition.

Differential Equation: dr/dt = [tex]9/2 (t + 1)^{1/2} i + 6 e^{-t} j + 1/(t + 1) k[/tex]
Initial condition: r(0) = k

Step 1: Integrate each component of the differential equation with respect to t:
[tex]r(t) = \int(9/2 (t + 1)^{1/2}) dt \ i + \int(6 e^{-t}) dt \ j + \int(1/(t + 1)) dt \ k[/tex]

Step 2: Solve the integrals:
[tex]r(t) = [3(t + 1)^{(3/2)}] i - [6 e^{-t}] j + [ln(t + 1)] k + C[/tex]

Step 3: Apply the initial condition r(0) = k:
[tex]k = [3(0 + 1)^{(3/2)}] i - [6 e^0] j + [ln(0 + 1)] k + C[/tex]
k = 0 i - 6 j + 0 k + C
C = 6j + k

Step 4: Substitute C back into the expression for r(t):
[tex]r(t) = [3(t + 1)^{(3/2)}] i - [6 e^{-t}] j + [ln(t + 1)] k + (6j + k)[/tex]

So, the vector function r(t) is:
[tex]r(t) = 3(t + 1)^{(3/2)} i + (-6 e^{-t} + 6) j + (ln(t + 1) + 1) k[/tex].

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since it starts at 135 and doubles every week f(n)=135*2^n-1

so for example after 4 weeks it would be f(4)=135*2^4-1=135*2^3=135*2*2*2=1080

On average, the value expected for the t-statistic when the null hypothesis is B. 0.

When the null hypothesis is true, the expected value of the t-statistic is zero (0). This is because the t-statistic is calculated by taking the difference between the sample mean and the hypothesized population mean (under the null hypothesis), and dividing it by the standard error.

If the null hypothesis is true, there should be no difference between the sample mean and the hypothesized population mean, resulting in a t-statistic of zero on average. Therefore, the correct answer is option B, "O", which represents zero.

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We can start by simplifying each factor separately and then combine them.

(a + b)/(a^2 + b^2) can be simplified by multiplying both the numerator and denominator by (a - b):

(a + b)/(a^2 + b^2) * (a - b)/(a - b) = (a^2 - b^2)/(a^3 - b^3)

Next, we simplify a/(a - b) by multiplying both the numerator and denominator by (a + b):

a/(a - b) * (a + b)/(a + b) = a(a + b)/(a^2 - b^2)

Lastly, we simplify (a^4 - b^4)/(a + b)^2 by factoring the numerator and expanding the denominator:

(a^4 - b^4)/(a + b)^2 = [(a^2)^2 - (b^2)^2]/(a + b)^2 = [(a^2 + b^2)(a^2 - b^2)]/(a + b)^2

Now we can combine all three simplified factors:

(a + b)/(a^2 + b^2) * a/(a - b) * (a^4 - b^4)/(a + b)^2 = [(a^2 - b^2)/(a^3 - b^3)] * [a(a + b)/(a^2 - b^2)] * [(a^2 + b^2)(a^2 - b^2)]/(a + b)^2

Simplifying further, we can cancel out the (a^2 - b^2) terms and the (a + b) terms:

= [a(a + b)/(a^3 - b^3)] * [(a^2 + b^2)/(a + b)]

= a(a + b)(a^2 + b^2)/(a + b)(a^3 - b^3)

= a(a^2 + b^2)/(a^3 - b^3)

Therefore, the simplified expression is a(a^2 + b^2)/(a^3 - b^3)

Answer:

11 tables - $145

t tables - 35 + 10t

Step-by-step explanation:

11 tables: $35 + 10*11 tables = 145

Answer:

total earning with 11 tables: $145

Step-by-step explanation:

if he gets a $10 tip per table you would do 10×11=110

and since he gets a guaranteed $35 a day you would do 35+110=145

Answer:

8) The distance the jet traveled is the area under the graph.

9) (1/2)(600)(20 + 25) = 13,500 miles

10) (1/2)(600)(5) = 1,500 miles

The probabilities for:

a. P(48 <X <66) = 0.978.

b. P(X> 69) = 0.0618.

c. P(X <= 65)  = 0.8051

d. P(X < 60) = 0.5

a. P(48 < X < 66) can be approximated using the normal distribution as follows:

mean, μ = np = 150 × 0.4 = 60

standard deviation, σ = [tex]\sqrt{(np(1-p)) }[/tex]= [tex]\sqrt{(150 * 0.4 * 0.6)[/tex] = 5.81

We can standardize using the formula z = (x - μ) / σ to find the area under the standard normal distribution between the z-scores corresponding to x = 48 and x = 66:

z1 = (48 - 60) / 5.81 = -2.06

z2 = (66 - 60) / 5.81 = 1.03

Using a standard normal distribution table, we find the area between these z-scores to be approximately 0.978. Therefore, P(48 < X < 66) ≈ 0.978.

b. P(X > 69) can be approximated using the normal distribution as follows:

mean, μ = np = 150 × 0.4 = 60

standard deviation, σ = [tex]\sqrt{(np(1-p)) }[/tex]= [tex]\sqrt{(150 * 0.4 * 0.6)[/tex] = 5.81

We can standardize using the formula z = (x - μ) / σ to find the area under the standard normal distribution to the right of the z-score corresponding to x = 69:

z = (69 - 60) / 5.81 = 1.55

Using a standard normal distribution table, we find the area to the right of this z-score to be approximately 0.0618. Therefore, P(X > 69) ≈ 0.0618.

c. P(X <= 65) can be approximated using the normal distribution as follows:

mean, μ = np = 150 × 0.4 = 60

standard deviation,  σ = [tex]\sqrt{(np(1-p)) }[/tex]= [tex]\sqrt{(150 * 0.4 * 0.6)[/tex]= 5.81

We can standardize using the formula z = (x - μ) / σ to find the area under the standard normal distribution to the left of the z-score corresponding to x = 65:

z = (65 - 60) / 5.81 = 0.86

Using a standard normal distribution table, we find the area to the left of this z-score to be approximately 0.8051. Therefore, P(X <= 65) ≈ 0.8051.

d. P(X < 60) can be approximated using the normal distribution as follows:

mean, μ = np = 150 × 0.4 = 60

standard deviation, σ = sqrt(np(1-p)) = sqrt(150 × 0.4 × 0.6) = 5.81

We can standardize using the formula z = (x - μ) / σ to find the area under the standard normal distribution to the left of the z-score corresponding to x = 60:

z = (60 - 60) / 5.81 = 0

Using a standard normal distribution table, we find the area to the left of this z-score to be 0.5. Therefore, P(X < 60) ≈ 0.5.

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The final solution to the differential equation is:

[tex]y(x) = (-35/2) e^x + (35/2) e^{(2x)} - 9xe^{(2x)[/tex]

To find y as a function of x, we can use the characteristic equation:

[tex]r^3 - 6r^2 - r + 6 = 0[/tex]

We can try factoring it using rational roots theorem:

Possible rational roots are ±1, ±2, ±3, ±6

Trying them out, we find that r = 1 and r = 2 are roots:

[tex](r - 1)(r - 2)^2 = 0[/tex]

Expanding and solving for r, we get:

r = 1, 2, 2

So the general solution to the differential equation is:

[tex]y(x) = c1 e^x + c2 e^{(2x) }+ c3 xe^{(2x)}[/tex]

To find the values of the constants c1, c2, and c3, we can use the initial conditions:

y(0) = c1 + c2 = 0

y'(0) = c1 + 2c2 + 2c3 = -8

y''(0) = c2 + 4c3 = 35

Solving these equations simultaneously, we get:

c1 = -35/2

c2 = 35/2

c3 = -9

So the final solution to the differential equation is:

[tex]y(x) = (-35/2) e^x + (35/2) e^{(2x) }- 9xe^{(2x)[/tex]

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(a) The number of binary strings of length 10 with at least one 1 is 1023.

(b) The number of binary strings of length 10 with at least one 1 and at least one 0 is 2045.

(a) To count the number of binary strings of length 10 with at least one 1, we can subtract the number of strings with all 0's from the total number of binary strings of length 10.

The total number of binary strings of length 10 is 2^10 = 1024, and the number of strings with all 0's is 1 (namely, 0000000000). Therefore, the number of binary strings of length 10 with at least one 1 is:

1024 - 1 = 1023

(b) To count the number of binary strings of length 10 with at least one 1 and at least one 0, we can use the principle of inclusion-exclusion.

The number of strings with at least one 1 is 1023 (as we calculated in part (a)), and the number of strings with at least one 0 is also 1023 (since the complement of a string with at least one 0 is a string with all 1's, and we calculated the number of strings with all 0's in part (a)).

However, some strings have both no 0's and no 1's, so we need to subtract those from the total count. There is only one such string, namely 1111111111. Therefore, the number of binary strings of length 10 with at least one 1 and at least one 0 is:

1023 + 1023 - 1 = 2045.

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To find the differential dy of the given function y = csc 9x, we can start by using the chain rule of differentiation, which states that the derivative of a composite function is the product of the derivative of the outer function and the derivative of the inner function.

The sine function has a reciprocal called the cosecant function. The cosecant function is the hypotenuse divided by the opposite side, just like the sine function is the opposite side divided by the hypotenuse. The hypotenuse and legs are terms for the two sides that are not the right angle.

dy/dx = dy/du * du/dx

We can find the derivative of y = csc u using the formula for the derivative of the cosecant function:

d/dx (csc x) = -csc x cot x

So we have:

dy/du = d/dx (csc u) = -csc u cot u

To find the derivative of the inner function, we can use the chain rule again:

du/dx = 9

Putting it all together, we have:

dy/dx = dy/du * du/dx = -csc u cot u * 9

Substituting u = 9x, we get:

dy/dx = -csc (9x) cot (9x) * 9

Therefore, the differential dy of y = csc 9x is:

dy = -9csc (9x) cot (9x) dx

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The correct statement when using the Excel Regression tool is:

d. Checking the option Constant is Zero forces the intercept to zero.

Explanation: The Excel Regression tool is used to perform linear regression analysis on a set of data. The independent variable values are specified in the Input X Range, while the dependent variable values are specified in the Input Y Range. The Regression tool can be found in the Data Analysis tab under the Analysis group.

When performing linear regression analysis, the intercept term represents the value of the dependent variable when the independent variable is equal to zero. Checking the option Constant is Zero in the Regression tool forces the intercept term to be zero, which can be useful in certain cases. However, it may also reduce the analysis' fit to the data and should be used with caution.

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The probability of selecting a college with in-state tuition less than $5,000 is 10%.

To calculate the probability that a randomly selected college will have an in-state tuition of less than $5,000, we first need to gather data on the number of colleges with in-state tuition less than $5,000 and the total number of colleges.

Let's assume that there are 500 colleges in the dataset, out of which 50 have in-state tuition less than $5,000.

So, the probability of selecting a college with in-state tuition less than $5,000 can be calculated as:

P(In-state tuition < $5,000) = Number of colleges with In-state tuition < $5,000 / Total number of colleges

P(In-state tuition < $5,000) = 50 / 500

P(In-state tuition < $5,000) = 0.1 or 10%

Therefore, the probability of selecting a college with in-state tuition less than $5,000 is 10%.

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Change in concentration = c(5/6) - c(2/3). To estimate the change in concentration (C) in mg/ml of a certain drug in a person's bloodstream when t changes from 40 to 50 minutes,

we will use the given model: c(t) = 2t / (3 + e^(-0.01t)).

First, let's convert the time from minutes to hours since the model uses hours:
40 minutes = 40/60 = 2/3 hours
50 minutes = 50/60 = 5/6 hours
Next, we will calculate the concentration at these times using the model:
1. For t = 2/3 hours:
c(2/3) = 2(2/3) / (3 + e^(-0.01(2/3)))

2. For t = 5/6 hours:
c(5/6) = 2(5/6) / (3 + e^(-0.01(5/6)))
Now, estimate the change in concentration by subtracting the concentrations at these two times:
Change in concentration = c(5/6) - c(2/3)

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a. The null hypothesis for this statement would be that the number of children the younger sister has is not dependent on the number of children the older sister has.

b. The null hypothesis for this statement would be that the distribution of family sizes for older and younger sisters is the same.

For a, The alternative hypothesis would be that there is a dependency between the two variables. This hypothesis can be tested using a chi-squared test for independence.

For b,The alternative hypothesis would be that the distributions are different. This hypothesis can be tested using a two-sample t-test for comparing means or a chi-squared test for comparing proportions.

Both hypotheses can be true or false independently. It is possible that the number of children the younger sister has is independent of the number of children the older sister has, but the distribution of family sizes could be different for older and younger sisters. Conversely, it is also possible that the number of children the younger sister has is dependent on the number of children the older sister has, but the distribution of family sizes is the same for both.

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Complete Question:  

A sociologist is studying influences on family size. He finds pairs of sisters, both of whom are married, and determines for each sister whether she has 0, 1, or 2 or more children. He wants to compare older and younger sisters. Explain what the following hypotheses mean and how to test them.

   a. The number of children the younger sister has is independent of the number of children the older sister has.

     b. The distribution of family sizes is the same for older and younger sisters. Could one hypothesis be true and the other false? Explain.

The probability of having a travel credit card is 0.5 is about 66.76 million lira or 33,380 euros.

(a) The estimated model equation is:

log(p/1-p) = -3.5561 + 0.0532*income

where p is the probability of having a travel credit card.

(b) The sign of ß (0.0532) is positive, which means that there is a positive relationship between income and the probability of having a travel credit card. As income increases, the probability of having a travel credit card also increases.

(c) To find the income level at which the probability of having a travel credit card is 0.5, we can set p = 0.5 in the model equation and solve for income:

log(0.5/1-0.5) = -3.5561 + 0.0532*income

0 = -3.5561 + 0.0532*income

income = 3.5561/0.0532

income = 66.76

Therefore, according to the estimated model equation, the income level at which the probability of having a travel credit card is 0.5 is about 66.76 million lira or 33,380 euros.
(a) The estimated model equation is given by:

log(p / (1 - p)) = -3.5561 + 0.0532 * Income

where p is the probability of having a travel credit card and Income is the annual income in millions of lira.

(b) The sign of ß (the coefficient of Income) is positive, which indicates that as the annual income increases, the probability of having a travel credit card also increases.

(c) To find the income level where the probability is 0.5, we need to solve the equation:

log(0.5 / (1 - 0.5)) = -3.5561 + 0.0532 * Income

0 = -3.5561 + 0.0532 * Income

Income = 3.5561 / 0.0532 ≈ 66.86 million lira

So, the probability of having a travel credit card is 0.5 when the annual income is approximately 66.86 million lira.

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The value of dot product (6+8)⋅(−10) is -140.

The value of (6u+v)⋅(u−10v) can be found using the distributive property and the dot product formula, which is (6u⋅u)+(v⋅u)-(60v⋅v). Substituting the given values, we get (6(6)²)+(8(6))-(60(4)²) = 92.

In the given problem, we are given the values of dot product, norms of two vectors u and v. We need to find the value of (6+8)⋅(−10) and (6u+v)⋅(u−10v). Using the formula for dot product, we get the value of the first expression as -140. For the second expression, we use the distributive property and the formula for dot product.

After substituting the given values, we simplify the expression to get the answer 92. The dot product is a useful tool in linear algebra and can be used to find angles between vectors, projections of vectors, and more.

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a) The first 4 terms of the sequence 10-n are:

- n=1: 10-1=9

- n=2: 10-2=8

- n=3: 10-3=7

- n=4: 10-4=6

The 10th term of the sequence 10-n is:

- n=10: 10-10=0

b) The first 4 terms of the sequence 6-2n are:

- n=1: 6-2(1)=4

- n=2: 6-2(2)=2

- n=3: 6-2(3)=0

- n=4: 6-2(4)=-2

The 10th term of the sequence 6-2n is:

- n=10: 6-2(10)=-14

The probability of winning the minimum award is 1/16,448.

To calculate this probability, follow these steps:

1. Find the total number of ways to pick two white balls from 58: C(58,2) = 58!/(2!(58-2)!) = 1,653.

2. Find the total number of ways to pick one gold ball from 44: C(44,1) = 44!/(1!(44-1)!) = 44.

3. Multiply the number of ways to pick two white balls and one gold ball: 1,653 * 44 = 72,732.

4. Calculate the total number of possible combinations: 58 * 57 * 56 * 55 * 44 = 1,195,084,680.

5. Divide the number of successful combinations by the total number of combinations: 72,732 / 1,195,084,680 = 1/16,448.

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The given statement "for all n>1. it will require exactly n -1 steps to assemble an n piece jigsaw puzzle from start to finish" is true by strong induction.

To prove that the statement holds true for all n > 1, we will use strong induction.

For n = 2, it takes exactly 1 step to assemble a 2-piece jigsaw puzzle, which satisfies the statement.

Inductive hypothesis, Assume that for some k > 1, it takes exactly k - 1 steps to assemble a k-piece jigsaw puzzle from start to finish. That is, the statement is true for all n = 2, 3, ..., k.

Inductive step, We need to show that the statement is also true for n = k + 1. To assemble a (k + 1)-piece jigsaw puzzle, we can first separate one piece from the puzzle. This leaves us with a k-piece puzzle. By the inductive hypothesis, it takes exactly k - 1 steps to assemble the k-piece puzzle. We can then attach the remaining piece to the completed k-piece puzzle, which takes 1 additional step. Therefore, it takes exactly (k - 1) + 1 = k steps to assemble a (k + 1)-piece jigsaw puzzle.

Since the statement is true for n = 2 and the statement is true for n = k implies that the statement is true for n = k + 1, the statement is true for all n > 1 by strong induction.

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To determine if the normal approximation to the binomial is appropriate, we need to check if both np and n(1-p) are greater than or equal to 10.

(a) For n = 500 and p = 0.33, np = 165 and n(1-p) = 335, both of which are greater than 10. Therefore, the normal approximation to the binomial is appropriate.

(b) For n = 400 and p = 0.01, np = 4 and n(1-p) = 396, which are not both greater than 10. Therefore, the normal approximation to the binomial is not appropriate.

(c) For n = 100 and p = 0.61, np = 61 and n(1-p) = 39, both of which are greater than 10. Therefore, the normal approximation to the binomial is appropriate.
To determine if the normal approximation to the binomial is appropriate in each situation, we can use the following rule of thumb: the normal approximation is suitable when both np and n(1-p) are greater than or equal to 10.
A binomial is a polynomial that is the sum of two terms, each of which is a monomial .It is the simplest kind of a sparse polynomial after the monomials.

(a) n = 500, p = 0.33
np = 500 * 0.33 = 165
n(1-p) = 500 * (1 - 0.33) = 500 * 0.67 = 335
Since both values are greater than 10, the normal approximation is appropriate.
Normal distributions are important in statistics and are often used in the natural to represent real-valued random variables whose distributions are not known. Their importance is partly due to the central limit theorem.

(b) n = 400, p = 0.01
np = 400 * 0.01 = 4
n(1-p) = 400 * (1 - 0.01) = 400 * 0.99 = 396
Since np is less than 10, the normal approximation is not appropriate.

(c) n = 100, p = 0.61
np = 100 * 0.61 = 61
n(1-p) = 100 * (1 - 0.61) = 100 * 0.39 = 39
Since both values are greater than 10, the normal approximation is appropriate.

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