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2 years ago

Help please show work

Mathematics

1 answer:

Romashka [77]2 years ago

8 0

Answer:

Just subtract a tuig from the thung

Step-by-step explanation:

Heres an example

subtract 7 from 12 an you will get 5 for number 1 do that for the rest

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Isosceles triangles have two equal sides. true or false

melomori [17]

True
isosceles triangles
have 2 equal sides

5 0

3 years ago

Officer Brimberry wrote 32 tickets for traffic violations last week, but only 12 tickets this week. What is the percent decrease

kolbaska11 [484]

Answer:

62.5 %

Step-by-step explanation:

you have to subtract 32-12, which equals 20

then divide 20 by the original amount (20/32)

20/32= .625, then you have to multiply to get the final answer...hope this helped!

4 0

1 year ago

Find the percent increase from 275 to 561.

MakcuM [25]

Answer:

104%

Step-by-step explanation:

(275 × p) / 100 = 286

((275 × p) / 100) × 100 = 286 × 100

275p = 28600

275p / 275 = 28600 / 275

p = 104

Percent Increase = 104

7 0

1 year ago

Read 2 more answers

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

Which of the following expressions is not correctly written in scientific notation?

anygoal [31]

C because scientific notation has to be number between 1-10

7 0

2 years ago

Read 2 more answers