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Citrus2011 [14]
4 years ago
6

Anyone feel like answering this page for me

Mathematics
1 answer:
LUCKY_DIMON [66]4 years ago
4 0
1.7x+35
2.5w-20
3.-5m+25
4.18-9a
5.2y+6
6.-2x-14
7.35m-21
8.6n+24
9.-12c-48
10.-8k-10
11.2-k
12.4-28p
13.18r-63
14.-5k-4

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Point Eis on line segment DF. Given DE = 8 and EF= 2, determine the length<br> DF.
julia-pushkina [17]

Step-by-step explanation:

if that's what you mean

6 0
3 years ago
Marcia talks on the phone to Gabrielle for 1/3 of an hour How many minutes did she spend on the phone with Gabrielle
viva [34]

Answer:

20 minutes

Step-by-step explanation:

Theres 60 minutes in a hour so a third of that would be 20

4 0
4 years ago
A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard d
bagirrra123 [75]

The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>

Suppose that we have:

  • Sample size n > 30
  • Sample mean = \overline{x}
  • Sample standard deviation = s
  • Population standard deviation = \sigma
  • Level of significance = \alpha

Then the confidence interval is obtained as

  • Case 1: Population standard deviation is known

\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}

  • Case 2: Population standard deviation is unknown.

\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}

For this case, we're given that:

  • Sample size n = 90 > 30
  • Sample mean = \overline{x} = 138
  • Sample standard deviation = s = 34
  • Level of significance = \alpha = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).

At this level of significance, the critical value of Z is: Z_{0.1/2} = ±1.645

Thus, we get:

CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]

Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

Learn more about confidence interval for population mean from large samples here:

brainly.com/question/13770164

3 0
2 years ago
Please help
MrRa [10]

Answer:

About 99.7% of births would be expected to occur within 51 days of the mean pregnancy​ length

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Standard deviation = 17.

About what percentage of births would be expected to occur within 51 days of the mean pregnancy​ length?

51/17 = 3.

So, within 3 standard deviations of the mean.

About 99.7% of births would be expected to occur within 51 days of the mean pregnancy​ length

4 0
3 years ago
Which of these sentences is always true for a parallelogram?
Murljashka [212]
The answer is d. opposite angles are congruent.
4 0
3 years ago
Read 2 more answers
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