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Paladinen [302]
3 years ago
6

Graph the line with the equation y= - 1/3 x -4?

Mathematics
1 answer:
Tamiku [17]3 years ago
5 0

Use desmo.com

its really helpful

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John takes out a personal loan for $5,000 at an annual interest rate 7.5%, compounded monthly. He plans to pay off the loan (inc
myrzilka [38]

Answer:

155.53

Step-by-step explanation:

I know where this question is from hahaha

6 0
3 years ago
Jane bought a packet of 12 cards for $15.00 <br>What would be the average price of a card?
nignag [31]
12 cards = $15.00

1 card = $15.00 ÷ 12 = $1.25


Answer: $1.25
5 0
3 years ago
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Which sentence is correct?<br> A. -4 &gt; 3<br> B. -4&lt; -6<br> C. -4 &gt; -7<br> D. -5 &gt; -4
prisoha [69]

Answer:

C. -4 > -7

Step-by-step explanation:

answer= option C

3 0
3 years ago
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Identify the equations as parallel lines, perpendicular lines, or neither. ​
Sever21 [200]

Answer:

Perpendicular

Step-by-step explanation:

A perpendicular line has an opposite slope. one of these has 5x and one has -5x, these are opposite, so they are perpendicular.

6 0
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/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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