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Pavel [41]
3 years ago
7

Using the Formula n=1/(ME)2 to Determine Sample Size When we want confidence and use the conservative estimate of , we can use t

he simple formula to roughly determine the sample size needed for a given margin of error . Use this formula to determine the sample size needed for a margin of error of 0.005 . Enter the exact answer.
Mathematics
1 answer:
PIT_PIT [208]3 years ago
4 0

Answer:

The sample size needed is 40000.

Step-by-step explanation:

The sample size needed is:

n = \frac{1}{M^{2}}

Use this formula to determine the sample size needed for a margin of error of 0.005

This is n when M = 0.005. So

n = \frac{1}{M^{2}}

n = \frac{1}{0.005^{2}}

n = 40000

The sample size needed is 40000.

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Vikentia [17]

Answer:

-4(8x-4)+3x = -29x+16

Step-by-step explanation:

1. Simplify -4(8x-4) by using the distribution property:

<u>-4(8x) = -32x and -4(-4) = 16 so -4(8x-4) = -32x+16</u>

2. Add 3x to the remaining equation:

<u>-32x+16+3x = -29x+16 </u>

3. The answer is:

<u> -29x+16 </u>

6 0
3 years ago
Solve each equation. If exact roots cannot be found, state the consecutive integers between which the roots are located.
vodomira [7]

Answer:

x=2

Step-by-step explanation:

x^2+10x+24=0

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My math is rusty so this may not be the right answer.

3 0
3 years ago
4.
denis23 [38]
The answer is none. There are no solutions
6 0
3 years ago
Help... please.I don't understand TT​
pickupchik [31]

Answer:

13. The more passengers, the more Subway cars are needed; 15. There's more weight with the more packages?

Step-by-step explanation:

4 0
3 years ago
Hey so im also looking for the first answer and the choices were “ is not “ and “ is “
dexar [7]

so the investigator found the skid marks were 75 feet long hmmm what speed will that be?

s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}

nope, the analysis shows that Charlie was going faster than 35 m/h.

now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been

s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d

4 0
2 years ago
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