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Dvinal [7]
4 years ago
8

PLease help me out. I will mark brainliest! promise.

Mathematics
1 answer:
m_a_m_a [10]4 years ago
3 0

Answer:

Option D

Hope It helps

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Shenelle has 100100100 meters of fencing to build a rectangular garden. the garden's area (in square meters) as a function of th
cluponka [151]
Area as a function of width, w is:
a(w)=(w-25)^2+625
for maximum area, the width will be:
a'(w)=2(w-25)+0=0
solving for w we egt
2w-50=0
2w=50
w=25 m
given that the perimeter is 100, the length will be:
100=2(L+W)
solving for L we get:
L=50-W
but W=25m
hence
L=50-25=25 m
thus the maximum area will be:
A=L*W=25*25=625m^2

8 0
3 years ago
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Imagine you have five total nieces and nephews and their ages are 4, 10, 8, 9, and 5. What is the average age of your nieces and
Eddi Din [679]

Answer:

The mean is 7.2 and the median ( average) is 8

Step-by-step explanation:

4 + 10 + 8 + 9 + 5 = 36 / 5

The mean is 7.2 and the median ( average) is 8

7 0
3 years ago
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In ΔMNO, n = 78 cm, o = 14 cm and ∠M=12°. Find the length of m, to the nearest centimeter
STALIN [3.7K]

Answer:

64

Step-by-step explanation:

i calculated using https://www.calculator.net/triangle-calculator.html :/

5 0
3 years ago
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Help ASAP please !!
nignag [31]

Option 4

Answered by Gauthmath must click thanks and mark brainliest

7 0
4 years ago
A survey organization has used the methods of our class to construct an approximate 95% confidence interval for the mean annual
lawyer [7]

Answer:

\bar X = \frac{66000+70000}{2}= 68000

We can estimate the margin of error with this formula:

ME= \frac{Upper -Lower}{2}= \frac{70000-66000}{2}= 2000

And the margin of error is given by:

ME = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

And we can rewrite the margin of error like this:

ME =z_{\alpha/2}*SE

Where SE= \frac{\sigma}{\sqrt{n}}

For 95% of confidence the critical value is z_{\alpha/2}= \pm 1.96

The Standard error would be:

SE= \frac{ME}{z_{\alpha/2}}= \frac{2000}{1.96}= 1020.408

For 99% of confidence the critical value is z_{\alpha/2}= \pm 2.58

And the new margin of error would be:

ME = 2.58* 1020.408 = 2632.653

And then the interval would be given by:

Lower = 68000- 2632.653 = 65367.347

Upper = 68000+ 2632.653 = 70632.653

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 for the sample  

\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 z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The 95% confidence interval is given by (66000 , 70000)

We can estimate the mean with this formula:

\bar X = \frac{66000+70000}{2}= 68000

We can estimate the margin of error with this formula:

ME= \frac{Upper -Lower}{2}= \frac{70000-66000}{2}= 2000

And the margin of error is given by:

ME = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

And we can rewrite the margin of error like this:

ME =z_{\alpha/2}*SE

Where SE= \frac{\sigma}{\sqrt{n}}

For 95% of confidence the critical value is z_{\alpha/2}= \pm 1.96

The Standard error would be:

SE= \frac{ME}{z_{\alpha/2}}= \frac{2000}{1.96}= 1020.408

For 99% of confidence the critical value is z_{\alpha/2}= \pm 2.58

And the new margin of error would be:

ME = 2.58* 1020.408 = 2632.653

And then the interval would be given by:

Lower = 68000- 2632.653 = 65367.347

Upper = 68000+ 2632.653 = 70632.653

7 0
4 years ago
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