(X2+6x-10)-(4x3+7x-8)

A school debate team has 4 girls and 6 boys. A total of 3 of the team members will be chosen to participate in the district deba

NeTakaya

The number of all possible 3-person groups
${10 \choose 3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{6}=120$

The number of all possible groups consisting of 1 girl and 2 boys
$4\cdot{6\choose 2}=4\cdot\dfrac{6!}{2!4!}=4\cdot\dfrac{5\cdot6}{2}=60$

The probability is $\dfrac{60}{120}=\dfrac{1}{2}$

6 0

3 years ago

Read 2 more answers

Evaluate 4 square root 9 square root 9 over 4 square root 9^5

Sedbober [7]

$\frac{\sqrt[4]{9}.\sqrt{9}}{\sqrt[4]{9^5}}$

$\frac{9^\frac{1}{4}.9^\frac{1}{2}}{9^\frac{5}{4}}$

$\frac{9^\frac{6}{8}}{9^\frac{5}{4}}$

$9^{\frac{6}{8}-\frac{5}{4}}$

$9^\frac{-1}{2}$

8 0

3 years ago

Suppose Miss Roxanne Davenport is 25 years old right now and puts away $1,800 per quarter in an account that returns 6% interest

Lena [83]

We have been given that miss Roxanne is 25 years old and she puts 1800 dollars per quarter that returns 6% interest.

(a) We need to figure out how much will be in the account when she turns 65 years old. When she turns 65 years old, the number of years during which she made deposits would be 40. Since she made quarterly deposits. She made a total of 160 deposits. We can now figure out the final amount in the account using future value of annuity formula.

$A=P\frac{(1+r)^{n}-1}{r}$

We have the values P=1800, r=6/4% = 1.5% = 0.015 and n=160.

Therefore, the amount in the account would be:

$A=1800\frac{(1+0.015)^{160}-1}{0.015}=1179415.39$

Therefore, miss Roxanne will be 1179415.39 dollars in her account when she turns 65 years old.

(b) In this part we need to figure out the total amount she deposited.

The total amount she deposited would be $1800*160=\$288000$ .

(c) We can find the interest earned by subtracting her contribution from the answer of part (a).

Interest earned = $1179415.39-288000=\$891415.39$

7 0

4 years ago

Read 2 more answers

Please answer #6 for me!

sdas [7]

Ok s we know the total cost is $320 so if the cost for the supplies is $70 the n we will subtract 70 from 320 leaving us with $250. The jobs is 5 hours so we will divide 250 by 5 and get 50 so the total cost per hour is $50
hope this helps!

7 0

3 years ago

In 2002, the mean age of an inmate on death row was 40.7 years with a standard deviation of 9.6 years according to the U.S. Depa

marissa [1.9K]

Answer:

The 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Step-by-step explanation:

The confidence interval of the mean is given by the next formula:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$ [1]

We already know (according to the U.S. Department of Justice):

The (population) standard deviation for this case (mean age of an inmate on death row) has a standard deviation of 9.6 years ( $\\ \sigma = 9.6$ years).
The number of observations for the sample taken is $\\ n = 32$ .
The sample mean, $\\ \overline{x} = 38.9$ years.

For $\\ z_{1-\frac{\alpha}{2}}$ , we have that $\\ \alpha = 0.05$ . That is, the level of significance $\\ \alpha$ is 1 - 0.95 = 0.05. In this case, then, we have that the z-score corresponding to this case is:

$\\ z_{1-\frac{\alpha}{2}} = z_{1-\frac{0.05}{2}} = z_{1-0.025} = z_{0.975}$

Consulting a cumulative standard normal table, available on the Internet or in Statistics books, to find the z-score associated to the probability of, $\\ P(z$ , we have that $\\ z = 1.96$ .

Notice that we supposed that the sample is from a population that follows a normal distribution. However, we also have a value for n > 30, and we already know that for this result the sampling distribution for the sample means follows, approximately, a normal distribution with mean, $\\ \mu$ , and standard deviation, $\\ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$ .

Having all this information, we can proceed to answer the question.

Constructing the 95% confidence interval for the current mean age of death-row inmates

To construct the 95% confidence interval, we already know that this interval is given by [1]:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

That is, we have:

$\\ \overline{x} = 38.9$ years.

$\\ z_{1-\frac{\alpha}{2}} = 1.96$

$\\ \sigma = 9.6$ years.

$\\ n = 32$

Then

$\\ 38.9 \pm 1.96*\frac{9.6}{\sqrt{32}}$

$\\ 38.9 \pm 1.96*\frac{9.6}{5.656854}$

$\\ 38.9 \pm 1.96*1.697056$

$\\ 38.9 \pm 3.326229$

Therefore, the Upper and Lower limits of the interval are:

Upper limit:

$\\ 38.9 + 3.326229$

$\\ 42.226229 \approx 42.23$ years.

Lower limit:

$\\ 38.9 - 3.326229$

$\\ 35.573771 \approx 35.57$ years.

In sum, the 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Notice that the "mean age of an inmate on death row was 40.7 years in 2002", and this value is between the limits of the 95% confidence interval obtained. So, according to the random sample under study, it seems that this mean age has not changed.

7 0

4 years ago