All categories

3 years ago

What is the value when 18 times of the difference of 15 and 12divided by 6?

Mathematics

1 answer:

aliya0001 [1]3 years ago

3 0

Answer:

Step-by-step explanation:

forget all this grammar,

let the number be x,

x = {18(15-12)} / 6

x = {18(3)} / 6

x= 54/6

x= 9

value is therefore 9

You might be interested in

What are the next three terms of each sequence?5,7.5

leva [86]

Answer:

The next three terms are 10, 12.5 and 15

Step-by-step explanation:

I. As you can see 5 and 7.5 is came from 5, 5 + 2.5

So given y = 5 + 2.5( x - 1 ) when y is output, x is term

II. Perform x = 3, 4, 5

So y = 5 + 2.5( 3 - 1 )

= 5 + 5

3rd = 10

y = 5 + 2.5( 4 - 1 )

= 5 + 7.5

4th = 12.5

y = 5 + 2.5 (5 - 1)

= 5 + 10

5th = 15

Leave comment below.

3 0

3 years ago

1 poir

podryga [215]

The given options are:

(A)x+y = 20
(B)7 apps and 14 movies
(C)x-y= 20
(D)y=-x+ 20
(E)8 apps and 12 movies
(F)xy= 20

Answer:

(A)x+y = 20
(D)y=-x+ 20
(E)8 apps and 12 movies

Step-by-step explanation:

If Elizabeth has a combined total of 20 apps and movies.

Where:

Number of apps=x

Number of Movies =y

Then:

Their total,

x+y=20 (Option A)

If we subtract x from both sides

x+y-x=-x+20

y=-x+20 (Option D)

In Option E

8 apps and 12 movies add up to 20. Therefore, this could also apply.

4 0

3 years ago

Read 2 more answers

Write an equation of the line that has a slope of -7 and y-intercept 6 in slope intercept form

Vedmedyk [2.9K]

Equation in <span>slope intercept form
y = -7x + 6

hope it helps</span>

3 0

3 years ago

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

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

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

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

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

$1.75 = \frac{X - 3.6}{0.4}$