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3 years ago

Three consecutive integers have a sum of 56. what are the three integers

Mathematics

1 answer:

denis23 [38]3 years ago

8 0

Answer:

impossible

Step-by-step explanation:

X + X + 1 + X + 2 = 56. To solve for X, you first add the integers together and the X variables together. Then you subtract 3 from each side, followed by dividing by 3 on each side. ...

3X + 3 = 56. 3X + 3 - 3 = 56 - 3.

3X = 53. 3X/3 = 53/3.

X = 17 2/3. Since 17 2/3 is not an integer, there is no true answer to this problem.

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

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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.