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2 years ago

Please help!!! asap!!!

Mathematics

1 answer:

quester [9]2 years ago

5 0

Answer:

x = 11 cos 22 = 11 * .927 = 10.2

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3 less than the product of 5 and x?

amm1812

Answer:

5x - 3

The product of 5 and x, minus 3

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3 years ago

What is an equivalent equation for each given equation?

Studentka2010 [4]

Answer:

Step-by-step explanation:

4x + 11 = 15

4x = 4

x = 1

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3 years ago

What is the volume of the cylinder?

mamaluj [8]

Answer/Step-by-step explanation:

Volume of a cylinder is given as V = πr²h

Where,

radius (r) = 6 cm

height (h) = 8 cm

Plug in the values

V = π*6²*8

V = π*36*8

V = π*288

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3 years ago

Lilly takes a train each day to work that averages 35 miles per hour. On her way home, her train ride follows the same path and

ivann1987 [24]

Lilly's home is 49.22 miles from her work.

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3 years ago

A doctor is interested in people who have experienced an unexplained episode of vitamin D intoxication. She took a SRS of size 1

Mila [183]

Answer:

(a) (2.573, 3.167) is the 99% confidence interval for $\mu$ the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication.

(b) There is a 99% probability that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.

Step-by-step explanation:

If we have a random sample of size n from a normal distribution with mean $\mu$ and standard deviation $\sigma$ , then we know that $\bar{X}$ is normally distributed with mean $\mu$ and standard deviation $\sigma/n$ . Therefore we can use $(\bar{X}-\mu)/\frac{\sigma}{\sqrt{n}}$ as a pivotal quantity.

We have a sample size of n = 12. The sample mean is $\bar{x}$ = 2.87 mmol/l and the standard deviation is $\sigma = 0.40$ . The confidence interval is given by $\bar{x}\pm z_{\alpha/2}(\frac{\sigma}{\sqrt{n}})$ where $z_{\alpha/2}$ is the $\alpha/2$ th quantile of the standard normal distribution. As we want the 99% confidence interval, we have that $\alpha = 0.01$ and the confidence interval is $2.87\pm z_{0.005}(\frac{0.40}{\sqrt{12}})$ where $z_{0.005}$ is the 0.5th quantile of the standard normal distribution, i.e., $z_{0.005} = -2.5758$ . Then, we have $2.87\pm (-2.5758)(\frac{0.40}{\sqrt{12}})$ and the 99% confidence interval is given by (2.573, 3.167)

(b) We found a 99% confidence interval for the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication. Therefore, there is a 99% probability that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.

8 0

4 years ago