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Mashcka [7]
3 years ago
6

Which of the following sets of numbers could not represent the three sides of a right triangle?

Mathematics
1 answer:
umka21 [38]3 years ago
7 0

Answer:

{48, 64, 81}

Step-by-step explanation:

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WILL MARK BRAINLIEST!!!
patriot [66]

Answer:

option D.

Step-by-step explanation:

average rate of change in steel production over the eight days=-0.15 tons per day

Step-by-step explanation:

The production rate of a steel factory is declining due to old machinery. Over the last eight days production has dropped by 1.2 tons.

Production is dropped by 1.2 tons, so its negative

Average rate of change = Change in production / number of days

Average rate of change = -1.2/ 8= -0.15

so sorry -0.15

6 0
3 years ago
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
2 years ago
You are on your way to work. You drive 3.9 miles to a gas station and buy gas. Your workplace is 1.39 miles from the gas station
Whitepunk [10]

Answer:

Approx. 5.3 miles

Step-by-step explanation:

4 0
3 years ago
Any suggestions or answers
Bad White [126]

Answer:

3m - 4

Step-by-step explanation:

Divide.

(15mn)/(5n) = 3m

(-20n)/(5n) = -4

3m - 4 is your answer.

~

7 0
3 years ago
An air transport association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maxim
Flura [38]

Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population mean, when the population standard deviation is not provided is:

CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}

The sample selected is of size, <em>n</em> = 50.

The critical value of <em>t</em> for 95% confidence level and (<em>n</em> - 1) = 49 degrees of freedom is:

t_{\alpha/2, (n-1)}=t_{0.05/2, 49}\approx t_{0.025, 60}=2.000

*Use a <em>t</em>-table.

Compute the sample mean and sample standard deviation as follows:

\bar x=\frac{1}{n}\sum X=\frac{1}{50}\times [1+5+6+...+10]=6.76\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{49}\times 31.12}=2.552

Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:

CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}

     =6.76\pm 2.000\times \frac{2.552}{\sqrt{50}}\\\\=6.76\pm 0.722\\\\=(6.038, 7.482)\\\\\approx (6.0, 7.5)

Thus, the 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

7 0
2 years ago
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