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3 years ago

The distance between – 15 and - 2 is (Simplify your answer.)

Mathematics

2 answers:

Aleks [24]3 years ago

7 0

Answer:

-17

Step-by-step explanation:

- plus - = -

so if u subtract -15 from -2= -17

liubo4ka [24]3 years ago

3 0

Answer: -17

Step-by-step explanation: If you do -15 - -2 you get -17. Hope this helps!

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From 99° to 70°f what is the percent decrease

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Researchers have concluded that a dry basin began to fill with water in 1880. the water level rose an average of 1.6 millimeters

docker41 [41]

Answer:

the domain is [0,129]
112 mm
1942

Step-by-step explanation:

1. The function is good for years 1880 to 2009, 0 to 129 years after 1880. Values of x can be anything in the domain [0, 129].

2. 1950 -1880 = 70. The year 1950 corresponds to x=70, so the function tells us the water level rose ...

f(70) = 1.6·70 = 112 . . . . . mm

3. We want to find x when f(x) = 100. That will be the solution to ...

100 = 1.6x

100/1.6 = x = 62.5

Then 62.5 years after 1880 is year 1942.5. The water level was 100 mm higher than in 1880 in the year 1942.

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3 years ago

127.5 is what percent of 51?

poizon [28]

Multiply 1.275 with 51 to get the answer (65.025)

8 0

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