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STALIN [3.7K]
3 years ago
6

In order to estimate the mean amount of time computer users spend on the internet each month, how many computer users must be su

rveyed in order to be 95% confident that your sample mean is within 11 minutes of the population mean? Assume that the standard deviation of the population of monthly time spent on the internet is 202 min. The minimum sample size required is_______computer users. (Round UP to the nearest whole number.)
Mathematics
1 answer:
prisoha [69]3 years ago
6 0

Answer:

1568

Step-by-step explanation:

The computation of the minimum sample size is shown below:

Given that

Error E = 11

The population standard deviation \sigma =217

Confidence level 1-α = 95%

Now following formula should be used

n > (Zalpha ÷ 2 × sigma ÷ E)^2

As we know that

Zalpha ÷ 2= 1.96

Now the sample size is

n> (1.96 × 202  ÷ 10)^2

= 1568

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aniked [119]

I think its D I hope you do good on your test! Good luck!

6 0
3 years ago
Answer question one and show work.
likoan [24]

Answer:

x = 28

Step-by-step explanation:

angle 1 and angle 2 are complementary (their sum is 90°)

x + 5 + 2x + 1 = 90 add like terms

3x + 6 = 90 subtract 6 from both sides

3x = 84 divide both sides by 3

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3 0
3 years ago
Problem 81
netineya [11]

it hasn't been specified which ticket the question's is asking but I'm guessing the price of the adult ones so $30.00

altogether that's two adults and 3 children who all paid a total of $105.00

let's simplify that to algebra

children tickets = c adults tickets = a

3c + 2a = $105.00

we know that 1c is 0.5a because each child ticket is half of an adults one <em>as stated in the question</em>

so in terms of (adults) a: half of 3 meaning 3×0.5= $1.5 meaning 1.5a

so now <em>3c = 1.5a</em>

so we can substitute and have like terms

meaning 1.5a + 2a= $105

3.5a =$105.00

divide by 3.5 on both sides to get a by itself thus the value of a:

a = $30

so each adult tickets cost $30

8 0
3 years ago
Use the given minimum and maximum data​ entries, and the number of​ classes, to find the class​ width, the lower class​ limits,
Georgia [21]

Answer:

Step-by-step explanation:

Given that minimum is 8 and maximum equals 82

Range = 84-8 = 76

No of classes =6

Class width = 76/6 ~13

But not given whether variable is discrete or continuous.

If discrete, we have classes as

8-20, 21-33, 34-46, 47-59, 60-72, 73-85

If continuous, we have classes as

8 to <21

21 to <34

and ... ending 73-<86

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3 years ago
Help someone if you not sleep please
mart [117]

Hello from MrBillDoesMath!

Answer:

+\- 10i

Discussion:

Fortunately, I can do this one in my sleep...... );


+\- sqrt(-100)  =

+\- 10 i                         where i = sqrt(-1)


Thank you,

MrB

8 0
3 years ago
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