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Ludmilka [50]
3 years ago
15

Here this is the new one.

Mathematics
1 answer:
Helga [31]3 years ago
5 0

Formula #1: 0.15x + 22

Formula #2: 0.19x

0.15x + 22 = 0.19x

15x+2200=19x\\15x=19x-2200\\-4x=-2200\\\frac{-4x}{-4}=\frac{-2200}{-4}\\X = 550

So x = 550.

I hope this helps! :)

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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
Evaluate 10/16 ÷ 5/16
VikaD [51]

Answer:

2

Step-by-step explanation:

10/16 ÷ 5/16

0.625 ÷ 0.3125

2

8 0
3 years ago
A total of 150 students have taken an Algebra 2 final exam. The scores are normally distributed with a mean of 71% and standard
WARRIOR [948]

Answer:

102 students

Step-by-step explanation:

Note that 65% and 71% are both 1 standard deviation from the mean (71%).  According to the empirical rule, 68% of scores lie within 1 std. dev. of the mean.

68% of 150 students would be 0.68(150 students) = 102 students

3 0
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MrRa [10]
Donkey, shrike style
6 0
3 years ago
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Shawna, Dexter, and Tilana are solving the equation -2.5(5-n)+2=15. Shawna says, “I can begin by dividing each side of the equat
laila [671]
Hi!
Shawna is correct because since it's multiplied to the  5-n you have to do the opposite which is division. 
Dexter is also correct because after distributing you can continue solving, there is nothing wrong with that. 
Hope this helps!
4 0
3 years ago
Read 2 more answers
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