Please help test due in 1 hour and it’s a grade! Will give good review and brainliest!

Lindsay and Lorraine are trying to match the jump rope world record. Together, they need to jump 48 times in a row. Lindsay has

Sunny_sXe [5.5K]

Answer: 21

Step-by-step explanation:

Given : Lindsay and Lorraine are trying to match the jump rope world record.

Number of jumps they need to jump together in a row = 48

Lindsay has gotten 14 jumps in a row, and Lorraine has gotten 13.

Total jumps = 14+13=27

Let 'j' be the number of more jumps they need to match the record.

j= Total jumps needed - Total jumps they already gotten

⇒ j= 48-27

⇒ j= 21

Hence, they need 21 more jumps to match the record.

5 0

2 years ago

This need help for homework

Troyanec [42]

The answer is J because when you add all the scores you get 76.

7 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST IF CORRECT!!<br> HELP PLS!!

bulgar [2K]

Answer:

the answer should be B

Step-by-step explanation:

7 0

2 years ago

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

2 years ago

Please Help, 35 Points.

Andrei [34K]

To answer this question, we have the start-up costs of $ 52,000

A monthly inflation of $ 0 is assumed

Operating costs are $680

The daily gain is $960

For the Part A.

The inequality that this situation represents

$960d-680d>52000$

So:

$d(960-680)>52000$

Where d represents the number of days.

For the Part B.

To start earning, you must replace all the initial investment and cover the expenses per day. The time that must pass for this to happen is obtained by clearing "d" from the inequality.

$d>\frac{52000}{960-680}$

d> 185.71 days

Then, the sum of the net profits will be greater than the initial investment after 186 days of starting the business.

4 0

2 years ago