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love history [14]
3 years ago
14

2

Mathematics
1 answer:
Anestetic [448]3 years ago
6 0

Answer:

price after discount and how much you saved...

Step-by-step explanation:

price after discount 9.60

you saved 2.40

please mark brainliest if correct :)

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Confidence Interval Mistakes and Misunderstandings—Suppose that 500 randomly selected recent graduates of a university were as
kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

n=500\\\bar x=8.6\\\sigma=2.2

(a)

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)

(b)

The (1 - <em>α</em>)% confidence interval for the parameter implies that there is (1 - <em>α</em>)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - <em>α</em>)% confidence interval for population parameter implies that there is a (1 - <em>α</em>) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

<em>n </em>= 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

7 0
3 years ago
Josie's Bagel Shop recently sold 3 poppy seed bagels and 3 other bagels. What is the experimental probability that the next bage
yuradex [85]

Answer:

ffbhjdhjdfahdsfsdfakfawlefgwaheghfgawlygfailwegfalawegfuwlgweilfygweiyaglifwafawegwegerg

Step-by-step explanation:

3 0
3 years ago
What's the solution to this inequality?
dezoksy [38]
P< -9 IS THE ANSWER
  _
I'm trying to write, less than and equal to
8 0
3 years ago
Read 2 more answers
The starting salary for a particular job is 1.2 million per annum. The salary increases each year by 75000 to a maximum of 1.5mi
MrRa [10]

The maximum salary reached in 5th month.

<h3>What is Arithmetic Progression?</h3>

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant

For the first year, salary  = 1,200,000

For the second year, salary

= 1,200,000 + 75,000 = 1,275,000

For the last year, salary  = 1,500,000

We get the AP as

1,200,000,  1,275,000, .... ,  1,500,000

a= 1,200,000  and d = 75000

Now using

L = a + (n - 1)d

1,500,000 = 1,200,000 + (n-1) 75,000

300,000 = 75,000(n-1)

300000/75000= n-1

4= n-1

n= 5

Hence, in the 5th year the maximum salary will be reached.

Learn more about this concept here:

brainly.com/question/11553395

#SPJ1

4 0
2 years ago
Please help. Answer only if u can show your work or explain
Elenna [48]
Gropu like terms
(-1/2)x+(3/2)x-4x+3+5=
(2/2)x-4x+8=
1x-4x+8=
-3x+8
3 0
3 years ago
Read 2 more answers
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