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kari74 [83]
3 years ago
15

The data represents the body mass index​ (BMI) values for 20 females. Construct a frequency distribution beginning with a lower

class limit of 15.0 and use a class width of 6.0. 17.7. 33.5. 26.6 24.7
28.8. 23.8. 18.3. 22.1
19.2. 22.6. 22.7. 38.9
27.3. 44.9. 32.2. 23.9
Mathematics
1 answer:
Wittaler [7]3 years ago
4 0

Kindly note that the data supplied is not up to 20. Hence, the answer will be based on the number of data given.

Answer:

Step-by-step explanation:

Given the following data:

BMI of 20 females :

17.7. 33.5. 26.6 24.7

28.8. 23.8. 18.3. 22.1

19.2. 22.6. 22.7. 38.9

27.3. 44.9. 32.2. 23.9

Lower class limit (LCL) = 15.0

Class width = 6.0

Class_interval_____frequency

15.0 - 20.9 _________3

21.0 - 26.9 _________7

27.0 - 32.9 _________3

33.0 - 38.9 _________2

39.0 - 44.9 _________1

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Answer:

The answer is "96.864 ml".

Step-by-step explanation:

In this question, the formula of \bold{CI = X \pm t \times s}.

( where X is the mean, t is the coefficient, and s is the mean difference error)

As a result, only 2.5% of containers might include less than 100 ml of volume, its trust coefficient could indeed be used in accordance with 95%, which is t=1.96.  

And it can take \pm \ 1.6 \ ml to have been the full value the standard infinite:  \to CI = 100 \pm (1.96 \times 1.6) \\\\

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5 0
3 years ago
Which expression is equivalent to5(h+9)
Fantom [35]

Answer:

The answer would be 5h+45

Step-by-step explanation

5 (h+9)

You need to convert this equation into the equation

you multiply 5 by h and get 5h

Then multiply the same number outside the parenthesis, 5, by 9 which will give you the answer 45

Once you have 5h and 45 just put it together with the addition symbol which will will equal:

5 (h+9) = 5h + 45

Your Welcome! (hopefully this answers your question)

4 0
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The owners of an eyeglass store claim that they fill customers' orders, on average, in 60 minutes with a standard deviation of 1
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By applying z-score formula we got that if a reporter determines a mean order time of 63 minutes then value o Z is 1.19

<h3>What is standard deviation?</h3>

Standard deviation is a number used to tell how measurements for a group are spread out from the average (mean), or expected value

Here given that

The owners of an eyeglass store claim that they fill customers' orders, on average, in 60 minutes with a standard deviation of 15. 9 minutes. Based on a random sample of 40 orders, a reporter determines a mean order time of 63 minutes

mean=μ=60

standard deviation=σ=15.9

number of sample=n=40

\bar{x}=63 \\

Now we can calculate z by z-score formula as

z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}  \\\\&=\frac{63-60}{\frac{15 \cdot 9}{40}} \\ \\=\frac{3}{2.5140} \\ \\=1.19\end{aligned}

By applying z-score formula we got that if a reporter determines a mean order time of 63 minutes then value o Z is 1.19

To learn more about standard deviation visit :brainly.com/question/12402189

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