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2 years ago

Express 0,mnmnmn.... = 0 ,mn, where m and n are distinct integers,in the form p/q where p and q are integers

Mathematics

1 answer:

Rama09 [41]2 years ago

5 0

The required fraction in the form of a/b is 2343/100

<h3>Fractions</h3>

Fractions are written as a ratio of 2 integers

Since the question is incomplete, we can e<u>xpress the decimal 23.43 as a fraction in the form p/q</u>

Given the decimal number 23.43

23.43 = 2343/100

To get this, you will replace the decimal point by 1 and then add zeros depending on the number of decimal numbers that we have.

Learn more on fractions here: brainly.com/question/78672

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What is the best word to describe the probability that the spinner will stop on 1? justify your reasoning.

White raven [17]

answer: B.Unlikely

( i remember learning this, before the world went to dirt)

5 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.