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3 years ago

I Write the explicit formula for the sequence 43, 46, 49, 52, .... *

Mathematics

1 answer:

yarga [219]3 years ago

6 0

Answer:55

Step-by-step explanation:in order it goes 43,46,49,52,55

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Credit unions are not subject to federal regulations. A. True B. False

Stolb23 [73]

Answer:

The correct answer is (FALSE)

Step-by-step explanation:

(APEX)

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2 years ago

After learning the theoretical probability of a two sided coin landing on any one side, students work in groups to flip the coin

Bumek [7]

The answer c, because they got 9 out of 15 flips being heads. This probability is 0.6. The other answers are not accurate.

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2 years ago

True or false (picture provided)

Eddi Din [679]

False. That does not satify the equation

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2 years ago

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Is the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed?

sp2606 [1]

Answer:

D. No, because the sample size is large enough.

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

If the sample size is higher than 30, on this case the answer would be:

D. No, because the sample size is large enough.

And the reason is given by The Central Limit Theorem since states if the individual distribution is normal then the sampling distribution for the sample mean is also normal.

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

If the sample size it's not large enough n<30, on that case the distribution would be not normal.

7 0

2 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

2 years ago