I need help please?!!!):

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