there are three choices of jellybeans - grape ,cherry and orange. if the probability of getting a grape is 3/a0

g The total national health expenditures per capita (in dollars) for 2006 and projected to 2021 can be modeled with the equation

Alex73 [517]

Answer:

a) The predicted per capita health expenditures for 2018 is $11943.5.

b) After 2018, it will take 16 years for the 2018 expenditures to double

Step-by-step explanation:

The total national health expenditures per capita, in dollars, is given by the following equation:

$H(t) = 6791e^{0.04343t}$

In which t is the number of years after 2005.

a. Find the predicted per capita health expenditures for 2018.

2018 is 13 years after 2005, so this is H(13).

$H(t) = 6791e^{0.04343t}$

$H(13) = 6791e^{0.04343*13}$

$H(13) = 11943.5$

The predicted per capita health expenditures for 2018 is $11943.5.

b. According to the model, how long will it take the 2018 expenditures to double?

First we find how many years after 2005, in which t is found when H(t) = 2*11943.5.

So

$H(t) = 6791e^{0.04343t}$

$2*11943.5 = 6791e^{0.04343t}$

$e^{0.04343t} = \frac{2*11943.5}{6791}$

$e^{0.04343t} = 3.5174$

Applying ln to both sides

$\ln{e^{0.04343t}} = \ln{3.5174}$

$0.04343t = \ln{3.5174}$

$t = \frac{\ln{3.5174}}{0.04343}$

$t = 29$

It will take 29 years after 2005 = 29-13 = 16 years after 2018.

After 2018, it will take 16 years for the 2018 expenditures to double

5 0

3 years ago

Car A went 60 km in 3/4 hour while a car B went 80 km in 4/5 hour. Which car was faster? How many times faster?

laila [671]

Answer:

Car B would go faster because it goes 20 more km and that would mean that it drove that fast in 48 minutes, for 1/5 of 60 minutes (1 hour) is 12 minutes. As for Car A, that goes 60 km is 45 minutes. The only time difference between the two is 3 minutes, which is 3 minutes longer for Car B, although it goes faster. Car B also goes 20 more km faster.

**Also any Brainliests from the questioner would help :)

*I do hope that this helps!

-EarthGirl :3

4 0

4 years ago

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago

Which angles are alternate interior angles?

GuDViN [60]

Answer:

<RSQ and <WVX

Step-by-step explanation:

They are just alternate interior angles based on how they are placed

4 0

3 years ago

In the equation 3x ÷ 4 = 15 solve for x

Nutka1998 [239]

Answer:

20

Step-by-step explanation:

you will get it

4 0

3 years ago

Read 2 more answers