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scoray [572]
3 years ago
14

Suppose that the distribution of touchdown passes (in football) is normally distributed with a mean of 250 feet and a standard d

eviation of 50 feet. We randomly sample 49 touchdowns.
What is the probability that the 49 touchdowns traveled an average of less than 245 feet? Please explain how you derived your answer.
Mathematics
1 answer:
Ahat [919]3 years ago
8 0

Answer: 0.2420

Step-by-step explanation:

Given: Mean : \mu = 250 \text{ feet}

Standard deviation : \sigma =50\text{ inch}

Sample size : n=49

The formula to calculate z is given by :-

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x= 245

z=\dfrac{245-250}{\dfrac{50}{\sqrt{49}}}=-0.7

The P Value =P(Z

Hence, the probability that the 49 touchdowns traveled an average of less than 245 feet= 0.2420

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Answer:

16 : 28

Step-by-step explanation:

If 12 out of the 28 pupils are female, we can calculate the number of males by subtracting the number of females from the total number of pupils.

28 - 12 = 16

So, there are 16 males in the class. Now, we know that the ratio of males : total is 16 : 28.

I hope this helps! Have a lovely day!! :)

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<span>Which shows the elements of the set of whole number multiples of four less than 21?

The multiples of 4 are 4, 8, 12, 16, 20, 24, 28,...

Then, the set of the multiples of four less than 21 is {4, 8, 12, 16, 20}

A: {1, 2, 3, 4, 5}
B: {4, 8, 12, 16, 20}
C: all real numbers
D: null set

Which shows the set(s) the number 2.77 is an element of?

2.77 is a real number, it is not a natural number, and it is not a whole number.

A: real numbers

B: real numbers and natural numbers

C: natural numbers and whole numbers

D: null set


Given sets:

D = {5, 6, 7, 8, 9, 10}, E = {1, 3, 5, 7, 9, 11}, F = {3, 6, 9, 12}

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I believe the question is what is D ∩ E, which is the elements that are as much in D as in E. This is {5,7,9}

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8 0
4 years ago
Two brothers leave from the same point driving in opposite​ directions, the first brother driving at 55 miles per hour and the s
Ira Lisetskai [31]

Answer: they would be able to talk to each other for 1.5 hours

Step-by-step explanation:

If each brother has a communication device that allows him to talk to the other for up to a 210​-mile ​range, by the time they are 210 miles apart, they won't be able to communicate to each other. This means that they would have travelled a total distance of 210 miles.

Let t represent the time it takes either brothers to travel his own distance.

the first brother driving at 55 miles per hour. If the first brother has a​ one-hour head start, this means that the time spent by the first brother is (t + 1) hour.

Distance = speed × time

Distance travelled by the first brother is

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50t

Since the total distance covered is 210 miles, then

55t + 55 + 50t = 210

105t = 210 - 55 = 155

t = 155/105 = 1.476 hours

Approximately 1.5 hours

3 0
3 years ago
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