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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.

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:

Δ PQT ~ Δ QRS .....{S-S-S test for similarity}...Proof is below.

Step-by-step explanation:

Given:

In Δ PQT

PQ = 30 ft

QT = 28 ft

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In Δ QRS

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RS = 14 ft

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To Prove:

Δ PQT ~ Δ QRS

Proof:

First we consider the ratio of the sides

$\frac{PQ}{QR}=\frac{30}{15} = \frac{2}{1}$ ..............( 1 )

$\frac{QT}{RS}=\frac{28}{14} = \frac{2}{1}$ ..............( 2 )

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$\frac{PQ}{QR}=\frac{QT}{RS} = \frac{TP}{SQ}$

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$\frac{PQ}{QR}=\frac{QT}{RS} = \frac{TP}{SQ}$

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∴ Δ PQT ~ Δ QRS .....{S-S-S test for similarity}...Proved

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Answer:

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Width: 21 inches

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Solve the quadratic equation by graphing

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The original dimensions of the piece of metal are

Length: 36 inches

Width: 21 inches

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