The distribution of monthly charges for cellphone plans in the United States is approximately normal with a mean of $62 and a standard deviation of $18. What percentage of plans have charges that are less than $83.60?

Answers

Answer 1

About 88.49% of cellphone plans have charges that are less than $83.60.

How to determine the percentage of plans have charges that are less than $83.60?

To determine the percentage of plans that have charges less than $83.60, we need to find the z-score (z) using the given mean and standard deviation, and then look up the corresponding area under the normal distribution curve.

z = (x – μ) / σ

where x = 83.60, mean, μ =  62 and standard deviation, σ = 18

Thus, the z-score of $83.60 is:

z = (83.60 - 62) / 18 = 1.2

Using a standard normal distribution table, we can find that the area to the left of z = 1.20 is 0.8849 or 88.49% (check image attached).

Therefore, about 88.49% of cellphone plans have charges that are less than $83.60.

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The Distribution Of Monthly Charges For Cellphone Plans In The United States Is Approximately Normal

Related Questions

Please help. Is the answer even there?

Answers

The critical values t₀ for a two-sample t-test is ± 2.0.6

To find the critical values t₀ for a two-sample t-test to test the claim that the population means are equal (i.e., µ₁ = µ₂), we need to use the following formula:

t₀ = ± t_(α/2, df)

where t_(α/2, df) is the critical t-value with α/2 area in the right tail and df degrees of freedom.

The degrees of freedom are calculated as:

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

n₁ = 14, n₂ = 12, X₁ = 6,X₂ = 7, s₁ = 2.5 and s₂ = 2.8

α = 0.05 (two-tailed)

First, we need to calculate the degrees of freedom:

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

= (2.5²/14 + 2.8²/12)² / [(2.5²/14)²/13 + (2.8²/12)²/11]

= 24.27

Since this is a two-tailed test with α = 0.05, we need to find the t-value with an area of 0.025 in each tail and df = 24.27.

From a t-distribution table, we find:

t_(0.025, 24.27) = 2.0639 (rounded to four decimal places)

Finally, we can calculate the critical values t₀:

t₀ = ± t_(α/2, df) = ± 2.0639

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Out of 1000 students who appeared in an examination,60% passed the examination.60% of the failing students failed in mathematics and 50% of the failing students failed in English.If the students failed in English and Mathematics only, find the number of students who failed in both subjects.​

Answers

The value of number of students who failed in both mathematics and English is 40.

Since, Given that;

60% of the 1000 students passed the examination,

Hence, we can calculate the number of students who passed the exam as follows:

60/100 x 1000 = 600

So, 600 students passed the examination.

Now, let's find the number of students who failed the examination.

Since 60% of the students passed, the remaining 40% must have failed. Therefore, the number of students who failed the examination is:

40/100 x 1000 = 400

Of the 400 failing students, we know that 60% failed in mathematics.

So, the number of students who failed in mathematics is:

60/100 x 400 = 240

Similarly, we know that 50% of the failing students failed in English.

So, the number of students who failed in English is:

50/100 x 400 = 200

Now, we need to find the number of students who failed in both subjects.

We can use the formula:

Total = A + B - Both

Where A is the number of students who failed in mathematics, B is the number of students who failed in English, and Both is the number of students who failed in both subjects.

Substituting the values we have, we get:

400 = 240 + 200 - Both

Solving for Both, we get:

Both = 240 + 200 - 400

Both = 40

Therefore, the number of students who failed in both mathematics and English is 40.

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LUUK al uit grapii velow.
Part B
-4
Part A
-3-2
Part C
3
2
-2
-3
Part D
Which part of the graph best represents the solution set to the system of
inequalities y ≥x+1 and y + x>-1? (5 points)

Answers

The solution set of given inequalities are represented by Part A.

The given inequalities are

⇒ y ≥ x + 1 and y + x > -1

Hence, The related equations of both inequalities are

y = x + 1

Put x=0, to find the y-intercept and put y=0, to find x intercept.

y = 0 + 1

y = 1

And, 0 = x + 1

x = - 1

Therefore, x-intercept of the equation is (-1,0) and y-intercept is (0,1).

Similarly, for the second related equation

y + x = - 1

y + 0 = - 1

y = - 1

0 + x = - 1

x = - 1

Therefore x-intercept of the equation is (-1,0) and y-intercept is (0,-1).

Now, join the x and y-intercepts of both lines to draw the line.

Now check the given inequalities by (0,0).

0 ≥ 0 + 1

0 ≥ 1

It is a false statement, therefore the shaded region is in the opposite side of origin.

0 + 0 ≥ - 1

0 ≥ - 1

It is a true statement, therefore the shaded region is about the origin.

Hence, From the below figure we can say that the solution set of given inequalities are represented by Part A.

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You roll a 6-sided number cube and toss a coin. Let event A = Toss a heads.

What outcomes are in event A?

What outcomes are in event AC?

Answers

1. Event A includes the outcomes of H and T,

2. while event AC includes all the possible outcomes of rolling a number cube, which are 1, 2, 3, 4, 5, and 6.

1. Event A is defined as tossing a heads on a coin, regardless of the outcome of rolling a number cube. Therefore, the outcomes in event A are H (heads) and T (tails), since either of these outcomes could occur when rolling a number cube and tossing a coin.

2. Event AC is the complement of event A, i.e., it is the set of outcomes that are not in event A. Since event A contains H and T, the outcomes in event AC are the remaining outcomes that are not in event A, which are all the possible outcomes when rolling a number cube: 1, 2, 3, 4, 5, and 6.

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Instructions: Find the missing probability.

P(B)=1/2P(A|B)=11/25P(AandB)=

Answers

We can use the formula:

P(A and B) = P(B) x P(A|B)

We are given:

P(B) = 1/2

P(A|B) = 11/25

Substituting these values into the formula, we get:

P(A and B) = (1/2) x (11/25) = 11/50

Therefore, P(A and B) = 11/50.

What is the total cost of 20 books at R25 each?

Answers

Answer:

R500

Step-by-step explanation:

20 books x R25 each = R500.

A top travels 8 centimeters each time it is spun. if it is spun 7 times what distance does it travel?

Answers

If a top travels 8 centimeters each time it is spun and it is spun 7 times, the total distance it travels is 56 centimeters.

How the total distance is determined:

The total distance is determined by multiplication of the distance traveled per spin and the number of spins.

Multiplication involves the multiplicand, the multiplier, and the product.

The traveling distance per spun = 8 centimeters

The number of spinning of the top = 7 times

The total distance = 56 centimeters (8 x 7)

Thus, using multiplication, the total distance the top travels after the 7th spin is 56 centimeters.

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How many turning points are in the graph of the polynomial function?
4 turning points
5 turning points
6 turning points
7 turning points

Answers

B 5 turning points. For example: The polynomial function y= ×^6 + 3x^5 +
4x^2 + 12 has six turning points. The degree of the polynomial is 6, which means that it has 6-1=5 turning points. These are the points where the function changes from increasing to decreasing or vice versa. In other words, these are the points where the derivative of the function equals 0. To find the exact locations of the turning points, you can set the derivative equal to 0 and solve for

Teena uses 1/4 cup of oil for a cake. How many cakes can she make if she has 6 cups of oil?

Answers

Answer:

24 cakes.

Step-by-step explanation:

6 cups of oil divided by 1/4 cup oil per cake = 24 cakes

6/(1/4) = 24

or 6/(0.25) = 24

She can make 24 cakes with 6 cups of oil.

The number 1.3 is both a(n) __________ and a(n) __________ number.

Answers

The number 1.3 is both a rational and an irrational number.

What is the number 1.3?

The number 1.3 is a rational number because it can be expressed as the quotient of two integers, namely 13/10.

The number 1.3 an irrational number because it cannot be expressed as the ratio of two integers, without repeating or terminating decimals, and its decimal representation goes on forever without repeating.

So we can conclude that the number 1.3 is both rational and irrational number.

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