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3 years ago

6

Solve the problem. For a standard normal distribution, find the percentage of data that are between 3 standard deviations below

the mean and 1 standard deviation above the mean.

1 answer:

solong [7]3 years ago

7 0

Answer:

83.85%

Step-by-step explanation:

We will use the empirical rule. For a standard normal distribution the mean is equal to 0 and the standard deviation is equal to 1. Then, between 3 standard deviations below the mean, we have 2.35% + 13.5% + 34% = 49.85% of data, and 1 standard deviation above the mean we have 34% of data. Therefore, the percentage of data that are between 3 standard deviations below the mean and 1 standard deviation above the mean is 49.85%+34% = 83.85%.

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Two angles supplementary. the first angle measures 60 what's the second

IgorLugansk [536]

choices? i cant tell without choices

4 0

3 years ago

15 POINTS!<br> What is the correct answer for Dean's problem? Explain where he went wrong.

prohojiy [21]

Answer:

The slope is 5/-2.

Step-by-step explanation:

Slope is y1-y2 over m1-m2 (rise over run). The first ordered pair is -2 (m1) and 11 (y1). We then subtract the second ordered pair (4 (m2) and -4 (y2)) from the first.

11 - (-4) = 11 + 4 = 15

-2 - 4 = -6

Remember, slope is rise over run (y over x), so the slope is 15/-6. Now, we must simplify. 15/-6 = 5/-2

Dean went wrong because he thought that slope was run over rise (x over y). If he had switched the two numbers, his answer would have been correct.

8 0

3 years ago

5(n+1) = 2n + 20 can someone tell me the answer and explain

finlep [7]

Answer:

n=15

Step-by-step explanation:

5(n+1)=2n+20

U start by expanding the first term which will give you 5n+5

Then after that u collect like terms giving u

5n-2n=20-5

Finally u evaluate it giving u

n=15

8 0

3 years ago

Read 2 more answers

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

3 years ago

Recall that m(t) = (1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500g sample of phosphorus-32 decays

katrin2010 [14]

The question is incomplete, here is the complete question:

Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

<u>Answer:</u> The half life of the sample of phosphorus-32 is $14.28days^{-1}$

<u>Step-by-step explanation:</u>

The equation used to calculate the half life of the sample is given as:

$m(t)=m_o(1/2)^{t/h}$

where,

m(t) = amount of sample after time 't' = 356 g

$m_o$ = initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

$356=500\times (\frac{1}{2})^{7/h}\\\\h=14.28days^{-1}$

Hence, the half life of the sample of phosphorus-32 is $14.28days^{-1}$

7 0

3 years ago