All categories

2 years ago

Whats the inverse if 2 5/8

Mathematics

1 answer:

enot [183]2 years ago

8 0

Answer:

$\frac{8}{21}$

Step-by-step explanation:

Given:

$2\frac{5}{8}=\frac{8*2+5}{8}$

Mutliply :

$2\frac{5}{8} =\frac{16+5}{8}$

Add:

$2\frac{5}{8} =\frac{21}{8}$

Multiplicative inverse:

$2\frac{5}{8} =\frac{8}{21}$

You might be interested in

Find the missing sum of a triangle

melamori03 [73]

Answer:

where you supposed to add a picture or..?

Step-by-step explanation:

3 0

3 years ago

(1 pt)

Karolina [17]

C. they are worth the same amount of money, 1/2 and 2/4 are both 50% or 50 cents

4 0

2 years ago

What term generally describes small chunks of rocks and debris in space? meteorites meteoroids meteors meteorides

deff fn [24]

Answer:

meteoroids

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What does feet mean <br><br> Or what's the answer for this

Nesterboy [21]

The answer to your question is 72 inches.

4 0

3 years 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