Steve rides his bike at a rate of 1760 feet per minute. What is his rate in miles per hour?

Mathematics

1 answer:

raketka [301]2 years ago

4 0

Answer:

<h2>The answer is 19.36 miles per hours</h2>

Step-by-step explanation:

this problem is a unit conversion problem

In this problem, we are expected to convert from feet/min to miles per hour

given that the rate is 1760 feet per minute

from tables, we are given that

1 foot per minute is 0.011 miles per hour hence

1760 feet per minutes will give x miles per hour

cross multiplying we have

x= 1760*0.011

x= 19.36 miles per hours

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Suppose that a person with a push mower can mow a large lawn in 5 hours, whereas the lawn can be mowed with a riding lawn mower

Reil [10]

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Push Mower
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It takes 5 hours to complete the job
1 hour = 1/5 of the job

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Riding Lawn Mower
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It takes 3 hours to complete the job
1 hour = 1/3 of the job

------------------------------------------------------------------
Push Mower + Riding Lawn Mower
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1 hour = 1/5 + 1/3
1 hour = 3/15 + 5/15
1 hour = 8/15 of the job.

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Calculate time needed to complete
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8/15 of the job takes 1 hour
8/15 of the job = 1 hour

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Divide by 8/15 on both side
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8/15 ÷ 8/15 of the job = 1 ÷ 8/15 hour
Whole of the job = 1 x 15/8 hours
Whole of the job = 15/8 hours
Whole of the job = 1 7/8 hours

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Answer: 1 7/8 hours
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6 0

2 years ago

An experiment was performed to compare the wear of two different laminated materials. Twelve pieces of material 1 were tested by

GenaCL600 [577]

Answer:

At 0.05 level of significance, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Step-by-step explanation:

We hypothesize that mean difference between abrasive wear of material 1 and material 2 is greater than 2.

So we write the null hypothesis $H_0 : \mu_1 - \mu_2 >2$ ,

and the alternative hypothesis $H_1: \mu_1 - \mu_2 \leq 2$ .

We will find the T-score as well as the p-value. If the p-value is less than the level of significance, we will reject the null hypothesis, i.e. we will conclude that the abrasive wear of material 1 is less than that of material 2. Otherwise, we will accept the null hypothesis.

Since the variance is unknown and assumed to be equal, we will use the pooled variance

$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = 20.05$ ,

where $n_1 = 12, n_2 =10, s_1 =4, s_2 = 5$ .

The mean of material 1 and material 2 are $\mu_1 =85, \mu_2=81$ respectively and mean difference $d$ is equal to 4. The hypothesize difference $d_0$ is equal to 2.

To find the T-score, we use the following formula

$T = \frac{d - d_0}{\sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2} }}$

Substituting all the values into the T-score formula gives us $T = 1.04$ , and the respective p-value is equal to 0.31. This means we have enough statistical evidence not to reject the null hypothesis, and at 5% significance level, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.