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Kryger [21]
3 years ago
13

How can the statement be rewritten as a conditional statement in if-then form? A rectangle with 4 congruent sides is a square.

Mathematics
2 answers:
Nady [450]3 years ago
7 0

Answer:

it is a

Step-by-step explanation:

larisa86 [58]3 years ago
3 0

Answer:

the answer is a

Step-by-step explanation:

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When the decimal is 5 or more, the number is rounded up by converting it into the next whole number.

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Please help! My teacher didn't go over this in the notes D:, as long as it gets answered before midnight im fine though
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15*9=135

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Would the answer to this be 5 or did I do something wrong
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Simplify the expression

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2 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/2th 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
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