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erastovalidia [21]
3 years ago
14

Alison wants to find out how much time people spend reading books. She is going to use a questionnaire.

Mathematics
1 answer:
neonofarm [45]3 years ago
5 0
Well, for her questionnaire she could use and create questions or queries that are obviously related to her hypothesis or study. These could be done in a likert type of scale. <span><span>
1.       </span>I read most often.
</span> <span><span>a.       </span>Strongly Agree </span> <span><span>b.      </span>Agree</span> <span><span>c.       </span>Disagree </span> <span><span>d.      </span>Strongly Disagree</span> <span><span>

2.       </span>When I read my books its takes me 24 hours a day</span> <span><span>
a.       </span>Strongly Agree </span> <span><span>b.      </span>Agree</span> <span><span>c.       </span>Disagree </span> <span><span>d.      </span>Strongly Disagree</span> <span><span>

3.       </span>When I start reading I can’t stop</span> <span><span>
a.       </span>Strongly Agree </span> <span><span>b.      </span>Agree</span> <span><span>c.       </span>Disagree </span> <span><span>d.      </span>Strongly Disagree</span>



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Gabby is a makeup artist and carries a lot of makeup around with her. she has 20 cosmetics in her bag, including 4 eye shadows.
myrzilka [38]

Answer:

1/5

Step-by-step explanation:

this is because 4/20 simplifies to 1/5

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7 0
2 years ago
Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to
RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75

And for the margin of error we have the following estimation:

ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62

5 0
3 years ago
What is the prime factorization of 121?
Diano4ka-milaya [45]
121|11
11|11
1

121=11^2
3 0
3 years ago
Whats 926 divided by 30
balandron [24]

30.87 ✅

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

\frac{926}{30}  \\  = 30 \frac{26}{30}  \\ ( \: or \: ) \\  = 30.8666 \\  = 30.87

\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}

4 0
2 years ago
Can someone please help with this ?
olga nikolaevna [1]

Answer:1/32

Step-by-step explanation:

8 0
3 years ago
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