Four times the sum of a number and two increased by three is at least twenty seven

How to find third coordinate when given midpoint and a point?

IgorLugansk [536]

Example :
u have an endpoint at (6,5)......ad the midpoint is (4,2)....find the other endpoint.

midpoint formula : (x1 + x2) / 2, (y1 + y2) / 2
one endpoint (6,5)....x1 = 6 and y1 = 5
other endpoint (x,y)...x2 = x and y2 = y
sub into the formula
m = (6 + x) / 2, (5 + y) / 2

okay, so the midpoint is (4,2)....so ur x value will equal 4
(6 + x) / 2 = 4...multiply both sides by 2
6 + x = 4 * 2
6 + x = 8
x = 8 - 6
x = 2 <==

midpoint is (4,2)....so the y value will equal 2
(5 + y) / 2 = 2 ...multiply both sides by 2
5 + y = 2 * 2
5 + y = 4
y = 4 - 5
y = -1 <==

so ur other endpoints are (2,-1)

8 0

2 years ago

Find the volume of a pyramid with a square base, where the perimeter of the base is

Kobotan [32]

Answer:

6.8 cubic feet

Step-by-step explanation:

7.3/4=1.825, which is the side length, 1.825*1.825=20.2575 and then ypu divide 20.2575 by 3 because it is a pyramid and that gives you 6.7525

4 0

3 years ago

Answer to this please ?

stepan [7]

Answer:

(-a,b)

Step-by-step explanation:

You use midpoint formula which is (x1+x2/2),(y1+y2/2)

8 0

3 years ago

What are the answers?

madam [21]

Positive numbers is the last one

7 0

2 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.

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

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

$X - 3.6 = 0.4*1.75$

$X = 4.3$

They last at least 4.3 minutes

7 0

2 years ago