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Nady [450]
4 years ago
8

Jason ran 7/8 of a mile in 12 minutes. How many miles can Jason run in 1 hour?

Mathematics
2 answers:
harkovskaia [24]4 years ago
7 0
Jason can run 40 miles in 1 hour
Scrat [10]4 years ago
4 0
Its technically like 7/8 x 60 because 60 mins is an hour
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Answer:

infinitely many solution

Step-by-step explanation:

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Help with 6? I really need all correct answers. Please help!
Lostsunrise [7]

Answer:

A

Step-by-step explanation:

(f - g)(x) = f(x) - g(x)

= \frac{2x+6}{3x} - \frac{\sqrt{x}-8 }{3x}

Since both fractions have a common denominator of 3x, then subtract the numerators leaving the denominator, that is the numerators simplified

2x + 6 - (\sqrt{x} - 8)

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Hence

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Plz help I do not wanna get grounded​
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Answer:

Area truck = 66.25 ft²

Step-by-step explanation:

Area truck = 12.5 x 5.3 = 66.25

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5 0
3 years ago
A bee flies at 9 feet per second directly to a flowerbed from its hive. The bee stays at the flowerbed for 13 ​minutes, and then
Usimov [2.4K]

Answer: 576 feet

Step-by-step explanation:

You have to do a little subtracting and a little interpreting to get the answer. First you must assume that the bee travels in a straight line. This is math so we can assume this is true, although thousands would disagree.

Second you must assume that because it is loaded with nectar that it travels at a slower rate (not an unreasonable assumption).

Third you have to assume that the distance there is the same distance as coming back. Again this is math. We can make this assumption.

Find the travelling time.

The bee spends 12 minutes hunting down the nectar in the flowerbed. Since it was away from the hive for a total of 16 minutes, the travelling time is

16min - 12min = 4 min

Convert the Minutes to seconds.

1 minute = 60 seconds

4 minutes = 4*60 = 240 seconds.

Find the time each way.

The total time is 240 seconds, but it is not divided evenly.

Let the time there = t

Let the time back = 240 - t

The distances are the same.

r = 6 m/s

r1 = 4 m/s

t1 = t

t2 = 240 - t

dthere = dback

d = rate * time

6 m/s *t = 4 m/s (240 - t)      Remove the brackets on the right.

6 * t = 4*240 - 4t                  Combine the like terms on the right.

6t = 960 - 4t                         Add 4t to both sides

6t + 4t = 960 - 4t + 4t          Combine

10t = 960                              Divide both sides by 10

10t/10 = 960/10                    Do the division

t = 96                                    Time going to the flower bed

Find the distance

d = ?

r = 6 m/s

t = 96 second

d = r*t

d = 6 * 96 = 576 feet Answer

5 0
3 years ago
The energy information administration reported that the mean retail price per gallon of regular grade gasoline was $3.43. Suppos
Morgarella [4.7K]

Answer:

68.26% of regular grade gasoline sold between $3.33 and $3.53 per gallon

81.85% of regular grade gasoline sold between $3.33 and $3.63 per gallon

2.28% of regular grade gasoline sold for more than $3.63 per gallon

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 3.43, \sigma = 0.1

What percentage of regular grade gasoline sold between $3.33 and $3.53 per gallon?

This is the pvalue of Z when X = 3.53 subtracted by the pvalue of Z when X = 3.33. So

X = 3.53

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.53 - 3.43}{0.1}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 3.33

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.33 - 3.43}{0.1}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% of regular grade gasoline sold between $3.33 and $3.53 per gallon

What percentage of regular grade gasoline sold between $3.33 and $3.63 per gallon?

This is the pvalue of Z when X = 3.53 subtracted by the pvalue of Z when X = 3.33. So

X = 3.63

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.63 - 3.43}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 3.33

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.33 - 3.43}{0.1}

Z = -1

Z = -1 has a pvalue of 0.1587

0.9772 - 0.1587 = 0.8185

81.85% of regular grade gasoline sold between $3.33 and $3.63 per gallon

What percentage of regular grade gasoline sold for more than $3.63 per gallon?

This is 1 subtracted by the pvalue of Z when X = 3.63. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.63 - 3.43}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% of regular grade gasoline sold for more than $3.63 per gallon

3 0
3 years ago
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