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Ket [755]
4 years ago
5

Hi, i need help with this question. Thank you so much if you decide to help me :3

Mathematics
2 answers:
irina1246 [14]4 years ago
7 0

Answer:

We can split this into ∛8 * ∛n⁹. ∛8 = 2 and ∛n⁹ = n³ so the answer is 2n³.

poizon [28]4 years ago
5 0

Answer:

2n^3

Step-by-step explanation:

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.......................look at the pic if u can......
Nataliya [291]

Answer:

A. 8x10^7 is 4000 times larger tham 2x10^4

Step-by-step explanation:


3 0
4 years ago
Questions on file. Thx.<br>Pls number questions to answer
ZanzabumX [31]
Second one
1/16
3/8 divided by 1/6=1/16
3 0
3 years ago
Read 2 more answers
Sari sold 300 boxes of greeting cards in November. In december she sold 500. What was the percentage increase in her sales
grigory [225]
There was a 66 \frac{2}{3}% increase from Sari's sales from November to December. 

To find the percentage increase in her sales, we must first find the amount her sales increased. We can do that by subtracting the greeting cards sold in November from the number of greeting cards sold in December. Substituting values and simplifying, we get that she sold 200 more greeting cards in December.

To find what percent increase this is, we must find what percent 200 is of the number of greeting cards sold in November. We can do this by dividing 200 by the number of cards sold in November (300), and then multiplying by 100 (to make it a percent). Doing this, we get \frac{2}{3} * 100 which gives us our answer of 66 \frac{2}{3}%.
3 0
3 years ago
A marketing researcher wants to find a 96% confidence interval for the mean amount those visitors spend per person per day while
ki77a [65]

Answer:

n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

Notation

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=12 represent the population standard deviation

n represent the sample size  

ME = 4 the margin of error desired

Solution to the problem

When we create a confidence interval for the mean the margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 96% of confidence interval now can be founded using the normal distribution. The significance is \alpha=1-0.96 =0.04. And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got z_{\alpha/2}=2.054, replacing into formula (b) we got:

n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38

So then the minimum sample to ensure the condition given is n= 38

5 0
3 years ago
Jessica has saved $50. She will add $25 to her savings each week. Ron has saved $40 and will add $25 to his savings each week.
MAVERICK [17]

Answer:

Jessica will have saved $175, and Ron will have saved $165.

Step-by-step explanation:

Jessica starts out with $50. She adds $25 each week, and it's asking how much money she would have in 5 weeks. So, you multiply $25 by 5 to get $125. Then, you add $50 to that to get $175. Same thing for Ron. Multiply $25 by 5 to get $125, but this time add $40 to it to get $165.

6 0
3 years ago
Read 2 more answers
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