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ELEN [110]
2 years ago
7

Here is questions for you because such thing as spring break is over ;)

Mathematics
2 answers:
stiv31 [10]2 years ago
7 0

Answer:yes

It I’ve it started on 29 and end on Monday

zvonat [6]2 years ago
6 0

Answer:

I don't now why but by seeing this photo my mouth is watery

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An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand
Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

b) We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

Step-by-step explanation:

Previous concepts

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

Part b

We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

4 0
3 years ago
The length of a rectangle is half the width. The area is 25 square meters
salantis [7]

Answer:

width= 5sqrt2 meters

length= 5/sqrt2 meters

Step-by-step explanation:

7 0
2 years ago
Kevin spends $11.25 on lunch every week during the school year. If there are 35.5 weeks during the school year, how much does Ke
ASHA 777 [7]

Answer:

399.37

Step-by-step explanation:

multiply the amount each week by the amount of weeks to get the answer :)

7 0
2 years ago
Solve each of the following inequalities for x. a. 3 + x > 8 b. x + 12 < 10 c. x – 23 > 15 d. x – 49 <6
krek1111 [17]

Answer:

a.

x > 5

b.

x < -2

c.

x > 38

d.

x < 55

8 0
3 years ago
Read 2 more answers
Which phrase represents 3p 6/7p-9
sammy [17]
(3p + 6) divided by (79 - 9)
5 0
2 years ago
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