Help please will give brainly! Calculate the mean and median below and answer the questions.

I need the RIGHT answer to this please. :)

KATRIN_1 [288]

A is the answer. It has no repeating x points buy it repeats on the y coordinate, which is okay. The answer is A just to say it again.

3 0

3 years ago

Read 2 more answers

Which equation represents the function represented by the table? 17 POINTS!!!

dybincka [34]

Answer:

f(x) = -x -1

Step-by-step explanation:

You can make the correct choice by seeing which equation works for the first line of the table.

f(-1) = -(-1) -1 = 0 . . . . . the first equation works

f(-1) = -(-1) +1 = 2 . . . not zero

f(-1) = -1 -1 = -2 . . . not zero

f(-1) = 1 -(-1) = 2 . . . not zero (same as second equation)

6 0

2 years ago

The local theater sells tickets to its plays at a rate of $38 each. Patrons who attend multiple shows are encouraged to buy an a

Mekhanik [1.2K]

Answer:

Roger

Step-by-step explanation:

Since he is attending more shows, he will benefit. He will save a net amount of $40

7 0

2 years ago

While researching the cost of school lunches per week across the state, you use a sample size of 45 weekly lunch prices. The sta

Drupady [299]

We assume the lunch prices we observe are drawn from a normal distribution with true mean $\mu$ and standard deviation 0.68 in dollars.

We average $n=45$ samples to get $\bar{x}$ .

The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write

$\sigma = 0.68 / \sqrt{45} = 0.101$

Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains $\mu$ .

Our interval takes the form of $( \bar{x} - z \sigma, \bar{x} + z \sigma )$ as $\bar{x}$ is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".

Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.

Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.

With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.

We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is

$( \bar{x} - 1.65 (.101), \bar{x} + 1.65 (.101) )$