How many ordered quadruples (w, x, y, z) of positive odd integers are there

kirill115 [55]

We are going to need more contex to awnser this

5 0

3 years ago

$10,788 per dozen, find for EACH computer (a) dollar markup, (b) percent markup on cost, and (c) percent markup on selling price

swat32

It’s A dollar markup

5 0

3 years ago

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

5 0

3 years ago

Find the second derivative of f(x)=sec(x).

marysya [2.9K]

The derivative of sec x is equal to sec x tan x. The derivative of the first derivative can be determined using the rule of products. The derivative is equal to sec x sec^2 x + tan x * sec x tan x. The simplified answer is sec^3 x + sec^2 x tan x equal to sec^2 x ( sec x + tanx )

8 0

3 years ago

Read 2 more answers

Need help with this math question, keep getting the first part correct but the other 2 parts of the function are incorrect. Not

Akimi4 [234]

Step-by-step explanation:

Here are four steps to help solve any math problems easily:

Read carefully, understand, and identify the type of problem. ...

Draw and review your problem. ...

Develop the plan to solve it. ...

Solve the problem.

4 0

2 years ago