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sammy [17]
3 years ago
6

I need help please?!!!):

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
8 0

Please see attachment

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new type of cell phone is being released to the public. The function below is used to predict f(t), the total sales in thousands
MrMuchimi

Answer:

40,000 units on the first day then 18% per week

Step-by-step explanation:

Exponential function:

An exponential function has the following format:

f(t) = f(0)(1+r)^{t}

In which r is the rate of change, as a decimal.

In this question:

The number of phones sold in thousands of units per day, in t weeks since the date of release, is:

f(t) = 40(1.18)^{t}

Since f(0) = 40, it solds 40,000 on the first week.

The rate of percentage increase is:

1 + r = 1.18

r = 0.18

So an increase of 0.18 = 18% a week.

So the correct option is:

40,000 units on the first day then 18% per week

4 0
3 years ago
4×+3+(-3×)+7×-6×<br> I've been trying to figure ut out but I cant
kozerog [31]

kAnswer: There is a mistake into this i think

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
(b) how many measurements must be averaged to get a margin of error of ±0.0001 with 98% confidence? (round your answer up to the
goblinko [34]
Given that the scale readings of the laboratory scale is normally distributed with an unknown mean and a standard deviation of 0.0002 grams.

Part A:

If the weight is weighted 5 times and the mean weight is 10.0023 grams, the 98% confidence interval for μ, the true mean of the scale readings is given by:

98\% \ C. \ I.=\bar{x}\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right) \\  \\ =10.0023\pm2.33\left(\frac{0.0002}{\sqrt{5}}\right) \\  \\ =10.0023\pm2.33(0.000089) \\  \\ =10.0023\pm0.00021 \\  \\ =(10.0023-0.00021, \ 10.0023+0.00021) \\  \\ =(10.0021, \ 10.0025)

Thus, we are 98% confidence that the true mean of the scale readings is between 10.0021 and 10.0025.



Part B:

To get a margin of error of +/- 0.0001 with a 98% confidence, then

\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right)=\pm0.0001 \\ \\ \Rightarrow2.33\left(\frac{0.0002}{\sqrt{n}}\right)=0.0001 \\  \\ \Rightarrow\left(\frac{0.0002}{\sqrt{n}}\right)= \frac{0.0001}{2.33} =0.0000429 \\  \\ \Rightarrow\sqrt{n}= \frac{0.0002}{0.0000429} =4.66 \\  \\ \Rightarrow n=4.66^2=21.7156\approx22

Therefore, the number of <span>measurements that must be averaged to get a margin of error of ±0.0001 with 98% confidence is 22.</span>
8 0
3 years ago
I WILLL GIVE BRAINLIEST ANSWER ASAP
kipiarov [429]

Answer: 2ND ONE

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
HELP PLZ WILL GET BRAINLIEST
Nostrana [21]

for B I would get a large number of participants and see how much they walk in 1 day multiply it by 365 then by the average life span of human beings A might be similar. Hope that helped.

8 0
3 years ago
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