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user100 [1]
3 years ago
15

What is the expression for "the difference of five times a number and twice that number"?

Mathematics
1 answer:
inna [77]3 years ago
6 0
"The difference of" means subtraction
"of 5 times a number" means 5N
"and twice that number" means 2N
So the statement is the subtraction of 5N and 2N

5N - 2N
The second option is correct.
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Need help with dilation with a scale factor of 1/2 centered at (−1, 3)
Studentka2010 [4]

A scale factor of 1/2 makes the triangle half the size of the original.


The original is 4 units tall, so 1/2 scale makes the new one 2 units tall.

It is 6 units long, so 1/2 would be 3 units long.


Now since the scale factor is centered at (-1,3) fine the midpoint between The top left corner of the triangle and the center point:


Midpoint ( -1-5/2 , 3+7/2)

Midpoint = (-6/2 , 10/2)

Midpoint = -3,5


This is where the top left corner of the scaled triangle would be.


See attached picture:


4 0
4 years ago
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Gala2k [10]
15 + 1 because you are using apsolute value so that negative 1 becomes a positive
8 0
3 years ago
1/4x + 8<br><br> ya help 14 points
Luden [163]

Answer:

ummm I think the answer is Carrot

7 0
3 years ago
What is the solution of this equation -12x-7=53
umka21 [38]

Answer:

<em>x </em>= -5

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You have to work backwards.

53 + 7 = 60

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8 0
2 years ago
Read 2 more answers
A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life
alisha [4.7K]

Answer:

a)

The null hypothesis is H_0: \mu \leq 10

The alternative hypothesis is H_1: \mu > 10

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

H_0: \mu \leq 10

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

H_1: \mu > 10

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that \mu = 10

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that n = 45, X = 10.5, s = 1.8

Then

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}

t = 1.86

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

5 0
3 years ago
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