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ira [324]
3 years ago
5

Aliza needs to run at a rate faster than 8.2 feet per second in order to exceed her fastest time in a race. After running for 15

minutes, her coach determines that she is running at an average rate of 5.8 miles per hour. He converts the average rate to feet per second as shown below:
He concludes that she is not running fast enough to exceed her fastest time.

What errors did the coach make? Check all that apply.

He used an incorrect time ratio converting hours to minutes.
His units do not cancel.
He used an incorrect distance ratio converting miles to feet.
He incorrectly concluded that she is not running fast enough.
He cannot determine her average rate in miles per hour after only 15 minutes.
Mathematics
2 answers:
marissa [1.9K]3 years ago
8 0

Answer:

A) He used an incorrect time ratio converting hours to minutes.

D) He incorrectly concluded that she is not running fast enough.

Step-by-step explanation:

Got it right on Edge ✿♥‿♥✿

Keith_Richards [23]3 years ago
3 0

Answer: he used incorrect time ratio

He incorrectly concluded that she is not running fast enough

Step-by-step explanation:

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12+10+5

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On a vacation trip to Canada you realize that the speed limit signs are in km/hour. Unfortunately, your speedometer only reads i
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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
3 years ago
The difference of twice a number and five is three. Find the num
Grace [21]

Answer:

add 5 then multiply by 1/2

Step-by-step explanation:

Twice a number  is 2n.

The difference of twice a number and five 2n - 5.

The difference of twice a number and five is three. 2n - 5 = 3

2n - 5 = 3

2n - 5 + 5 = 3 + 5     (1) add 5

2n = 8

1/2*2n = 1/2*8    (2) multiply by 1/2

n = 4

7 0
3 years ago
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