20 point for however can tell me what event in rodeo this is and a brainliest

When you are trying to find the mean in a dot plot do you want to use the number of dots of the actual number underneath of the

yuradex [85]

The number of dots tells us how many times we need to add the actual number underneath the dots. So you need to use both.

<h3>Which number should you use for the mean?</h3>

Let's suppose we have the next set:

{2, 2, 2, 4, 5, 5, 6}

If you graphed it in a dot plot, then you will have:

3 dots above the number 2.
1 dot above the number 4.
2 dots above the number 5.
1 dot above the number 6.

A total of 7 dots.

To find the mean, we just compute:

$M = \frac{2 + 2 + 2 + 4 + 5 + 5 + 6}{7} = 3.7$

So the number of dots tells us how many times we need to add the actual number underneath the dots. In conclusion, to find the mean you need to use both, the number of dots and the number underneath.

If you want to learn more about dot plots:

brainly.com/question/15853311

#SPJ1

5 0

2 years ago

Can someone help. Please

KonstantinChe [14]

Answer:

Answer below Please give brainliest!!

Step-by-step explanation:

Sue Black: 8x5=40 hours 40 x $6.75 = $270

Jerry Atrix: 7.25 + 3 + 6.75 + 8 + 5 = 30 hours

30 x $5.15 = $154.50

Sam Ting: 7.75 x 5 = 38.75 hours

38.75 x $8.75 = 339.0625 = $339.06

5 0

4 years ago

30 POINTS URGENT! When solving the following system of equations, which variable would be the easiest to solve for?

AnnyKZ [126]

Answer:

b

Step-by-step explanation:

8 0

3 years ago

Answer please as best as you can

mash [69]

The volume of a larger cone:

$V_l=\dfrac{1}{3}\pi\cdot4.5^2\cdot(3+6)=\dfrac{1}{\not3_1}\pi\cdot20.25\cdot\not9^3=(20.25)(3)\pi=60.25\pi\ cm^3$

The volume of a smaller cone:

$V_s=\dfrac{1}{3}\pi\cdot3^2\cdot6=\dfrac{1}{\not3_1}\pi\cdot\not3^1\cdot3\cdot6=18\pi\ cm^3$

The volume of the frustum:

$V=V_l-V_s\\\\V=60.25\pi-18\pi=42.75\pi\ cm^3$

<h3>Answer: 42.75π cm³.</h3>

4 0

3 years ago

A manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process,

Anastasy [175]

Step-by-step explanation:

A manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process, at most 7 chips each lot would be defective, but if something does go wrong, there could be far more defective chips. If something goes wrong with a given lot, they discard the entire lot. It would be prohibitively expensive to test every chip in every lot, so they want to make the decision of whether or not to discard a given lot on the basis of the number of defective chips in a simple random sample. They decide they can afford to test 100 chips from each lot. You are hired as their statistician.

There is a tradeoff between the cost of eroneously discarding a good lot, and the cost of warranty claims if a bad lot is sold. The next few problems refer to this scenario.

Problem 8. (Continues previous problem.) A type I error occurs if (Q12)

Problem 9. (Continues previous problem.) A type II error occurs if (Q13)

Problem 10. (Continues previous problem.) Under the null hypothesis, the number of defective chips in a simple random sample of size 100 has a (Q14) distribution, with parameters (Q15)

Problem 11. (Continues previous problem.) To have a chance of at most 2% of discarding a lot given that the lot is good, the test should reject if the number of defectives in the sample of size 100 is greater than or equal to (Q16)

Problem 12. (Continues previous problem.) In that case, the chance of rejecting the lot if it really has 50 defective chips is (Q17)

Problem 13. (Continues previous problem.) In the long run, the fraction of lots with 7 defectives that will get discarded erroneously by this test is (Q18)

Problem 14. (Continues previous problem.) The smallest number of defectives in the lot for which this test has at least a 98% chance of correctly detecting that the lot was bad is (Q19)

(Continues previous problem.) Suppose that whether or not a lot is good is random, that the long-run fraction of lots that are good is 95%, and that whether each lot is good is independent of whether any other lot or lots are good. Assume that the sample drawn from a lot is independent of whether the lot is good or bad. To simplify the problem even more, assume that good lots contain exactly 7 defective chips, and that bad lots contain exactly 50 defective chips.

Problem 15. (Continues previous problem.) The number of lots the manufacturer has to produce to get one good lot that is not rejected by the test has a (Q20) distribution, with parameters (Q21)

Problem 16. (Continues previous problem.) The expected number of lots the manufacturer must make to get one good lot that is not rejected by the test is (Q22)

Problem 17. (Continues previous problem.) With this test and this mix of good and bad lots, among the lots that pass the test, the long-run fraction of lots that are actually bad is (Q23)

8 0

3 years ago