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4 years ago

What the equivalent expression of 1/64

Mathematics

1 answer:

Rashid [163]4 years ago

3 0

2/128
because if you decide each side by 2 it equals 1/64

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If the area of a rectangle in terms of x is x2 + 15x + 26 / 6x2 and its width is x2 - 3x - 10 / 30x3 Find the length of the rect

Veseljchak [2.6K]

Answer:

The length of the rectangle is;

5x(x+13)/(x-5)

Step-by-step explanation:

Mathematically, we know that the area of a rectangle is the product of the length and width of the triangle

To find the length of the rectangle, we will have to divide the area by the width

we have this as;

(x^2 + 15x + 26)/6x^2 divided by (x^2-3x-10)/30x^3

thus, we have ;

(x^2 + 15x + 26)/6x^2 * 30x^3/(x^2-3x-10)

= (x^2+15x+ 26)/(x^2-3x-10) * 5x

But;

(x^2 + 15x + 26) = (x+ 2)(x+ 13)

(x^2-3x-10) = (x+2)(x-5)

Substituting the linear products in place of the trinomials, we have;

(x+2)(x+13)/(x+2)(x-5) * 5x

= 5x(x+13)/(x-5)

4 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.