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joja [24]
3 years ago
14

The football team lost 4 yards on 2 plays in a row. Which of the following could represent the change in field position?

Mathematics
1 answer:
Over [174]3 years ago
8 0
Lost (⬅key word) 4 is equal to -4
On 2 plays that is times 2
So all together is -4×2
A negative times a positive is negative
-4×2= -8
Hope this helped you☺
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Step-by-step explanation:

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3 years ago
Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati
Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly​ selected,  the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c.   D. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X \sim N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})

P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})

P(X < 76) = P(Z< \dfrac{3}{12.5})

P(X < 76) = P(Z< 0.24)

From the standard normal distribution tables,

P(X < 76) = 0.5948

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b.  If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})

P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})

P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})

P( \overline X < 76) = P(Z< 1.2)

From the standard normal distribution tables,

P(\overline X < 76) = 0.8849

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

In order to determine the probability in part (b);  the  normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0
3 years ago
Given a geometric sequence in the table below, create the explicit formula and list any restrictions to the domain.
Makovka662 [10]

Given:

The geometric sequence is:

n              a_n

1                        -4

2                      20

3                     -100

To find:

The explicit formula and list any restrictions to the domain.

Solution:

The explicit formula of a geometric sequence is:

a_n=ar^{n-1}            ...(i)

Where, a is the first term, r is the common ratio and n\geq 1.

In the given sequence the first term is -4 and the second term is 20, so the common ratio is:

r=\dfrac{a_2}{a_1}

r=\dfrac{20}{-4}

r=-5

Putting a=-4,r=-5 in (i), we get

a_n=-4(-5)^{n-1} where n\geq 1

Therefore, the correct option is B.

5 0
3 years ago
Please help! Which of the following would result in an integer?
podryga [215]

Answer:

B

Step-by-step explanation:

Cube root of 27 is 3, which is in integer

5 0
1 year ago
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