All categories

3 years ago

Evaluate show the solution of 14C6 and 15C4

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

8 0

Answer:

Evaluate each expression for a = 4, b = 2, and c = 5.

c − a

You might be interested in

Select the correct answer.

Likurg_2 [28]

Answer:

Step-by-step explanation:

i did this test and question beforer si know what it iso

5 0

3 years ago

GIVING BRAINLIEST TO THE FIRST PERSON WHO CAN SOLVE THESE QUESTIONS. :)

Vanyuwa [196]

Answer:

1) y = 2(x+3)^2 -4

2) x+6/x-4

3) 2x^2 + 6x +20 = 0

Step-by-step explanation:

5 0

1 year ago

Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm

Vlad1618 [11]

Answer:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar x =512$ represent the sample mean

$\sigma=100$ represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

$Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

$25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}$

And we can find the critical value $z_{\alpha/2}$ like this:

$z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576$

And we know that on the right tail of the z score =2.576 we have $\alpha/2$ of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998= $\alpha/2$ , so then $\alpha=0.00498*2=0.009995 \approx 0.01$ , and then we can find the confidence like this: