All categories

3 years ago

6500 is what percent of 32,500?

2 answers:

7 0

It is 20%. 20% can also be written as 0.20. 0.20 multiplied by 32500 is 6500.

$3200x = 6500$
$\frac{32500}{32500} x = \frac{6500}{32500}$

blagie [28]3 years ago

3 0

i belive its 6500 is 20% of 32500

You might be interested in

How did I get a 2 here (simplifying ratios)

PilotLPTM [1.2K]

Answer:

$\frac{2x {}^{2} + 14x + 24 }{4 {x}^{2} + 4x - 48 } = \frac{x + 4}{2(x - 2)}$

Step-by-step explanation:

Please see attached picture for full solution.

4 0

3 years ago

Pls help no links if u can pls explain the steps I’ll be grateful

valkas [14]

Answer:

WX = 9

Step-by-step explanation:

WY is the entire line - this tells you that the length of the entire line is 11.

XY is a section of the line, and WX is the section you need to find the length for. Since it is given that XY is of length 2, we can find the length of WX by subtracting the length of XY from WY.

11 (WY) - 2 (XY) = 9.

8 0

2 years ago

Brainliest If Correct

natta225 [31]

Answer:

Step-by-step explanation:

Messed up, 292+52=240, I hope it’s correct for you, I have Alzheimer, I hope is correct for you!

8 0

2 years ago

Which pair shows equivalent expressions?

viva [34]

Answer:

Step-by-step explanation:I think

3 0

2 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

2 years ago