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faltersainse [42]

3 years ago

8

Can u help me with this and I don't wanna rush u but can we hurry because my homework has a timer and it will close I've been wo

rking on this all day

1 answer:

deff fn [24]3 years ago

3 0

Answer:

Step-by-step explanation:

C , looks like the right answer, and probably this is too late to help... :/ sorry

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What is 8,000 *20? Please help me, thanks!

Rainbow [258]

It is 160000 because 8*2 is 16 just add 0000 after the 16

6 0

3 years ago

Read 2 more answers

Solve the following compound inequality.<br><br> 5x+7<=-3 or 3x-4>=11

Kobotan [32]

Answer:

$x\leq -2$ or $x\geq 5$

Step-by-step explanation:

$5x+7\leq -3$ or $3x-4\geq 11\\$

$5x+7\leq -3\\5x\leq -10\\x\leq -2$

$3x-4\geq 11\\3x\geq 15\\ x\geq 5$

4 0

3 years ago

Write a model for the statement V varies jointly as h and the fifth power of r. Group of answer choices LaTeX: V=khr^5 V = k h r

hjlf

Answer:

Option A)

$V = khr^5$

Step-by-step explanation:

We are given the following in the question:

V varies jointly as h and the fifth power of r.

$V\propto h\\V\propto r^5$

Combing the two equations, we get,

$V\propto hr^5$

Now, removing the sign of proportionality and adding the constant of proportionality, we get,

$V = khr^5$

where k is the constant of proportionality.

Thus, the correct answer is

Option A)

$V = khr^5$

7 0

3 years ago

When 258 college students are randomly selected and surveyed, it is found that 106 own a car. Find a 99% confidence interval for

juin [17]

Answer: 0.332 < p < 0.490

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where n= sample size

$\hat{p}$ = sample proportion

z* = critical z-value.

As per given , we have

n= 258

Sample proportion of college students who own a car = $\hat{p}=\dfrac{106}{258}\approx0.411$

Critical z-value for 99% confidence interval is 2.576. (By z-table)

Therefore , the 99% confidence interval for the true proportion(p) of all college students who own a car will be : $0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)$

Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490

3 0

3 years ago

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago