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4 years ago

Mila jogged 4.6 miles in 48 minutes. if she jogged at a constant rate what was her jogging speed.

Mathematics

2 answers:

butalik [34]4 years ago

8 0

Answer: 5.75 miles per hour

Step-by-step explanation:

Anastaziya [24]4 years ago

3 0

Answer:

You must divide the time by the distance. Have you ever heard of Distance=rt?

Step-by-step explanation:

Basically, the r stands for rate and the t stands for time, so first, the problem would look like this:

4.6=r(48)

So to find the rate, you would do the opposite of multiplying the rate times the time, so you would divide 48 by 4.6 to get the answer, then I believe to round to the nearest whole number.

I'm not positive, but hopefully this helps.

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3 years ago

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Galina-37 [17]

The answer is B, you will need at least 6 spring roll wrapper packages.

The way you get the answer is by turning the word problem into an expression. You can make this into the fraction (17*6+15*3)/(25). 17 is the number of adults and you multiply this by how many spring rolls Jill wants to give each adult, which is 6, you can do the same for the children. Next the denominator is 25 which represents the amount of spring roll wrappers per each package. When you simplify the answer comes around to 5.88, but since you cant buy 88% of a spring role wrap package you have to buy another whole one, which makes the answer six.

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3 years ago

It costs 4 tokens to park in a garage for an hour. how many hours can you park with 30 tokens?

babunello [35]

Read the question carefully: it costs 4 tokens to park in a garage for an hour.

We will apply the unitary method to solve this question

It costs 4 tokens to park in a garage for 1 hour

Find how many hours can park in a garage for 1 token

If it costs 4 token to park in a garage for 1 hour

Then it will cost 1 token to park in a garage for 1/4 hour

Step2:

With 20 token we can park in a garage for (1/4) * 20

= 5 hours

So, we can park for 5 hours with 20 tokens.

Another method

If we take twenty tokens and divide them into groups of four, we will find that we are left with five groups of tokens. Each group of tokens represents an hour of parking time. This will give us five groups, or five hours, total.

So, we can park for 5 hours with 20 tokens

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3 years ago

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4. Phil bought a block of cheese in the shape of a triangular prism measuring 8 inches by 4 inches by 2 inches. What is the volu

makkiz [27]

Answer:

32 inches^3

Step-by-step explanation:

Volume = 1/2 ( base x height x length)

0.5 ( 4 x 2 x 8) = 32 inches^3

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3 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co

Ira Lisetskai [31]

Answer:

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 103-1 = 102$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have $T = 2.63$ .

Now, we need to find the standard deviation of the sample. That is:

$s = \frac{0.56}{\sqrt{103}} = 0.055$

Now, we multiply T and s

$M = T*s = 2.63*0.055 = 0.145$

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

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3 years ago