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3 years ago

If the perimeter is 560 yards what is the dimensions

1 answer:

4 0

It is impossible to know the dimensions with this information, since you don't know anything about the ratio of the sides or the shape forming the perimeter. Let's just assume that the shape is a form of rectangle and has four sides.

560/4=140
If it is a square each side will be 140 yards.

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How determine the scalr factor of A(3,6) A (4.5,9)

VARVARA [1.3K]

Because 3 / 6 = 0.5 and 4.5 / 9 = 0.5, the scale factor is 0.5 ;

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3 years ago

How do i fond the quotient and the remainder ?

zhenek [66]

The remainder is the integer left over after dividing one integer by another. The quotient is the quantity produced by the division of two numbers.
Hope that helped !

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3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

3 years ago

Charlie runs at a speed of 3 yards per second. About how many miles per hour does Charlie run

enyata [817]

To find the answer, you first must find how many yards Charlie can run in a minute. Since there are 60 seconds in a minute, you multiply 3 times 60, which is 180. Then you must multiply this number by 60 again, because there are 60 minutes in an hour. This equals 10,800, your answer.

Charlie can run 10,800 yards in an hour.

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3 years ago

Read 2 more answers

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marissa [1.9K]

Well when you add $84+$29 you get $113

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3 years ago