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Shalnov [3]
3 years ago
6

Fis the midpoint of EG E has the

Mathematics
2 answers:
Gelneren [198K]3 years ago
5 0

Answer:

21

Step-by-step explanation:

Vadim26 [7]3 years ago
4 0

Answer:

21

Step-by-step explanation:

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alexira [117]
Volume of the carton = 1000 cubic inches. Dimensions of a carton are 10"×10"×10"
4 0
3 years ago
Do) 4. Claudia spends $2.15 a day for lunch. Her cafeteria lunch account has a
Shkiper50 [21]

Answer:

d=48/2.15

Step-by-step explanation:

hope it helps

easy peazy

8 0
2 years ago
For this question, f(x) = 4kx2 + (4k + 2)x + 1, where k is a real constant.
snow_tiger [21]

Answer:

fghhjhhfgkus+($+#($!?$)

6 0
2 years ago
Using the simple random sample of weights of women from a data​ set, we obtain these sample​ statistics: nequals45 and x overbar
Tomtit [17]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : n=45

\overline{x}=148.79\text{ lb}

\sigma=31.37\text{ lb}

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = \mu=148.79\text{ lb}

b) The confidence interval for the population mean is given by :-

\mu\ \pm E, where E is the margin of error.

Formula for Margin of error :-

z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given : Significance level : \alpha=1-0.90=0.1

Critical value : z_{\alpha/2}=z_{0.05}=\pm1.645

Margin of error : E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69

Now, the 90% confidence interval for the population mean will be :-

148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)

Hence, the 90​% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

3 0
3 years ago
Approximately 5% of calculators coming out of the production lines have a defect. Fifty calculators are randomly selected from t
baherus [9]

Answer:

0.2611 = 26.11% probability that exactly 2 calculators are defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. The probability of a calculator being defective is independent of any other calculator, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

5% of calculators coming out of the production lines have a defect.

This means that p = 0.05

Fifty calculators are randomly selected from the production line and tested for defects.

This means that n = 50

What is the probability that exactly 2 calculators are defective?

This is P(X = 2). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{50,2}.(0.05)^{2}.(0.95)^{48} = 0.2611

0.2611 = 26.11% probability that exactly 2 calculators are defective.

3 0
2 years ago
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