Is either x = 5 or x = 15 a solution to 25

What is the place value of 1 in 5,678.321

alukav5142 [94]

<span>The place value of 1 in 5,678.321 is one-one thousands. The one digit is the third place behind the decimal point. The first place behind the decimal point is three one-tenths. The second place behind the decimal is two one-hundredths. A decimal number is a division of one. In other words, it will take a thousand one-one thousands to equal one.</span>

6 0

4 years ago

Consider the circle with a radius of 30 cm.

Molodets [167]

Answer:

You are close it is (D)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A)

andriy [413]

Answer:

a) 1/10

b) 110

Step-by-step explanation:

A) we start by getting the total number of rolls

That would be;

11 + 22 + 7 + 8 + 13 + 19 = 80

The frequency of getting a 4 is 8

The probability that the next roll will be a 4 is;

8/80 = 1/10

b) Firstly, we get the probability of getting a 2

We have this as:

22/80

So for 400 rolls, the number of 2’s will be;

22/80 * 400

= 22 * 5 = 110

8 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago

MUST ANSWER ALL PARTS!!!!

dsp73

Hello there,

The outlier of the data set is 92. Lets first find everything with the outlier.
Mean: 60.08
Median: 57
Mode: 52
Range:46

Now let's take out the outlier of the set and see the difference.
Mean: 54.27
Median:56
Mode: 52
Range: 15

So as we can see without the outlier everything is closer together. The outlier shouldn't be included because that makes the entire set screwed up. The data is more accurate without the outlier.

Hope this helps!

6 0

3 years ago

Is either x = 5 or x = 15 a solution to 25 - x =5?