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Savatey [412]
2 years ago
12

Grayson ran 2 1/2 miles around the track .each lap is 5/6 of a mile .how many laps did Grayson run?

Mathematics
1 answer:
andrezito [222]2 years ago
5 0

Given:

Number of miles covered = 2\dfrac{1}{2}

Length of each lap = \dfrac{5}{6} of a mile.

To find:

The number of laps.

Solution:

We know that,

\text{Number of laps}=\dfrac{\text{Number of miles covered}}{\text{Length of each lap}}

\text{Number of laps}=\dfrac{2\dfrac{1}{2}}{\dfrac{5}{6}}

\text{Number of laps}=\dfrac{\dfrac{5}{2}}{\dfrac{5}{6}}

\text{Number of laps}=\dfrac{5}{2}\times \dfrac{6}{5}

\text{Number of laps}=3

Therefore, Grayson ran 3 laps.

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Karleigh walks 5/8 mile to school every day . How far does she walk to school in 5 days
ziro4ka [17]

Hello, thanks for using Brainly!

First, our equation is: 5/8 x 5 = 25/8.

Our first step is to simplify.

When you simplify you get 3 1/8.

This is another way you can get it: (Multiply across. )

Number of the miles = \frac{5}{8} 5 = \frac{25}{8} = 3\frac{1}{8}

Enjoy your day.

8 0
3 years ago
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 76.676.6
faltersainse [42]

Answer:

Step-by-step explanation:

Researchers measured the data speeds for a particular smartphone carrier at 50 airports.

The highest speed measured was 76.6 Mbps.

n= 50

X[bar]= 17.95

S= 23.39

a. What is the difference between​ the carrier's highest data speed and the mean of all 50 data​ speeds?

If the highest speed is 76.6 and the sample mean is 17.95, the difference is 76.6-17.95= 58.65 Mbps

b. How many standard deviations is that​ [the difference found in part​ (a)]?

To know how many standard deviations is the max value apart from the sample mean, you have to divide the difference between those two values by the standard deviation

Dif/S= 58.65/23.39= 2.507 ≅ 2.51 Standard deviations

c. Convert the​ carrier's highest data speed to a z score.

The value is X= 76.6

Using the formula Z= (X - μ)/ δ= (76.6 - 17.95)/ 23.39= 2.51

d. If we consider data speeds that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly​ high, is the​ carrier's highest data speed​ significant?

The Z value corresponding to the highest data speed is 2.51, considerin that is greater than 2 you can assume that it is significant.

I hope it helps!

3 0
3 years ago
Factor completely x3 - y3
stealth61 [152]

Answer:

(x-y)(x^2+xy+y^2)

Step-by-step explanation:

x^3 - y^3=(x-y)(x^2+xy+y^2)

8 0
3 years ago
A
Mkey [24]

Answer:

D is the right answer.

Step-by-step explanation:

5 0
3 years ago
How to solve this problem?
Ganezh [65]

The answer should be 1y and -6y

3 0
3 years ago
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