How do i change my username...they gave me a random gross one

1 answer:

7 0

Im not sureif this would work, but I suggest you to go into your account, push your profile, and then push edit profile, then go into your prefences and edit the first name section and to save it push change to save

Step-by-step explanation:

You might be interested in

Suppose that a random sample of 10 adults has a mean score of 62 on a standardized personality test, with a standard deviation o

aniked [119]

Answer:

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645*\frac{4}{\sqrt{10}} = 2.08$

The lower end of the interval is the sample mean subtracted by M. So it is 62 - 2.08 = 59.92

The upper end of the interval is the sample mean added to M. So it is 62 + 2.08 = 64.08.

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

5 0

2 years ago

Please answer ASAP. Suppose we write down the smallest (positive) 2-digit, 3-digit, and 4-digit multiples of 8.

GaryK [48]

Answer:

16, 104, 1000

Step-by-step explanation:

8*2

8*13

8*125

7 0

1 year ago