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3 years ago

By what percent will a fraction decrease if its numerator is increased by 20% and its denominator is increased by 50%?

Mathematics

2 answers:

ss7ja [257]3 years ago

7 0

Answer: 20%

Step-by-step explanation:

Lets call one number x and the other one y

first you have the fraction x/y

When you increase the numerator by 20% you get 1.2

when you increase the denominator by 50% you get 1.5

Then you get the fraction 1.2x/1.5y

since fractions are basically division you divide 1.2 by 1.5 which is 0.8

0.8 as a percent is 80%

you subtract that from 100 and you get 20%

Brums [2.3K]3 years ago

3 0

Lets find out...

our fraction is 1/2.....
when the numerator is increased by 20%......1 + 0.20(1) = 1.2
when the denominator is increased by 50%....2 + 0.50(2) = 3

1/2 = 0.5....original fraction
1.2 / 3 = 0.4.....increased fraction

decrease = (original number - new number) / original number...* 100
= (0.5 - 0.4) / 0.5...* 100
= (0.1 / 0.5)....* 100
= 0.2 * 100
= 20% decrease <====

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"A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean

Arada [10]

Answer:

85.31% probability that their mean rebuild time exceeds 8.1 hours.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 8.4, \sigma = 1.8, n = 40, s = \frac{1.8}{\sqrt{40}} = 0.2846$

If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.1 hours.

This is 1 subtracted by the pvalue of Z when X = 8.1. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.1 - 8.4}{0.2846}$

$Z = -1.05$

$Z = -1.05$ has a pvalue of 0.1469

1 - 0.1469 = 0.8531

85.31% probability that their mean rebuild time exceeds 8.1 hours.

4 0

3 years ago

How does this polynomial identity work on numerical relationships? (y + x) (ax + b)

Serggg [28]

Let us take 'a' in the place of 'y' so the equation becomes

(y+x) (ax+b)

Step-by-step explanation:

Step 1:

(a + x) (ax + b)

Step 2: Proof

Checking polynomial identity.

(ax+b )(x+a) = FOIL

(ax+b)(x+a)

ax^2+a^2x is the First Term in the FOIL

ax^2 + a^2x + bx + ab

(ax+b)(x+a)+bx+ab is the Second Term in the FOIL

Add both expressions together from First and Second Term

= ax^2 + a^2x + bx + ab

Step 3: Proof

(ax+b)(x+a) = ax^2 + a^2x + bx + ab

Identity is Found .

Trying with numbers now

(ax+b)(x+a) = ax^2 + a^2x + bx + ab

((2*5)+8)(5+2) =(2*5^2)+(2^2*5)+(8*5)+(2*8)

((10)+8)(7) =(2*25)+(4*5)+(40)+(16)

(18)(7) =(50)+(20)+(56)

126 =126

3 0

3 years ago

Someone pls help me !

oksian1 [2.3K]

The correct anser is 1 and only 1

3 0

3 years ago

I need to know what goes in the boxes 20 points. :)

Nana76 [90]

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8 0

3 years ago

Read 2 more answers

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mr Goodwill [35]

Answer:

$48.15

Step-by-step explanation:

7% of 45 is 3.15. so 45 + 3.15= 48.15

8 0

3 years ago