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3 years ago

Complete the table and show work

Mathematics

2 answers:

Alinara [238K]3 years ago

5 0

So the formula for this is for every inch of width, you multiply by 2 to get the length.

Here, I made a table for it.

2×2=4

4×2=8

7×2=14

8×2=16 or 16÷2=8

Murrr4er [49]3 years ago

4 0

2:4 4:8 7:14 & 8:16 is all your answers

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What are 3 examples or area

Tamiku [17]

1) L×W=A

2) 6×3=24

3) 8×8=64

4) 2×90 =180

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3 years ago

93% marked up 35? i need it like nowwwww

vaieri [72.5K]

Answer:

How do you calculate a 35% markup?

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3 years ago

Divide £70 in the ratio 4:3

kirill115 [55]

Answer:

Total=4 +3=7

For the 4

4÷7×70=£40

For the 3

3÷7×70=£30

So the division of £70 in ratio 4:3 is £40:£30

7 0

3 years ago

I need help with this

Anestetic [448]

Answer:

5 prime tier

Step-by-step explanation:

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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

4 years ago