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erastova [34]
3 years ago
9

On a number line, what other number is the same distance from -2.8 as -7.2 is?​

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
3 0

Answer;

1.6

Explanation

What we shall do here is to calculate the distance between the two points on the number line

Mathematically, that would be ;

-2.8 -(-7.2) = -2.8 + 7.2 = 4.4

This means that -7.2 is 4.4 units away from -2.8

So to find other number which is this same distance;

we have -2.8 + 4.4 = 1.6

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Simplify (write without the absolute value sign): |x−(−12)|, if x<−12
VladimirAG [237]

Answer:

|x + 12|

Step-by-step explanation:

The question I took form you I assume looks like this...

|x - (-12)|

Therefore, if we simplify this, the answer is |x + 12|

This is the answer because you would multiply -1 by -12

3 0
3 years ago
KINNDAAAA need help. Will give Brainly! Need the full equation pls!
Alexxandr [17]
Hey, I’m pretty sure the answer is y=1/2x + 0.

To explain, we can plug the x value into the equation for all of the x values, and we will get our y value.

So the first one,

y=1/2(1) + 0 —> 1/2

The second one,

y=1/2(2) + 0 —> 1

And so forth, the rest of the equations will be correct. I hope that helps ! :D
4 0
3 years ago
Multiply this expression :: -2/7 • (-3 5/8)​
Helen [10]

Answer: 29/28

Step-by-step explanation:

-2/7 x (-3 5/8)

= -2/7 x -29/8

= 2/7 x 29/8 = 1/7 x 29/4

= 29/28

7 0
3 years ago
What are the missing blanks?
Ksenya-84 [330]

Answer:

1 - 3 (1 quarter is 25 and a dime is 10, so, 3.)

4 - 10

6 - 15

14 - 33

Step-by-step explanation:

7 0
3 years ago
A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for
tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of <em>t</em> are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29

s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11

s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}

  =\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}

  =-4.123

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}

   =\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}

   =26.55\\\approx 27

Compute the critical value for <em>α</em> = 0.05 as follows:

t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052

*Use a <em>t</em>-table for the values.

Thus, the critical values of <em>t</em> are ± 2.052.

3 0
3 years ago
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