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Ksju [112]
3 years ago
7

A line PQ passes through the points (-8, -1) and (-2, 3). The equation of line RS is -2y -3x =13. Is the line PQ parallel to RS?

Justify
Mathematics
1 answer:
Andrei [34K]3 years ago
6 0
First find the slope
Y-y
----
X-x

2/3 is the slope
Next find the slope of the other line
-2y -3x=13
-2y=3x+ 13
Y=-2/3+-6.5
No the lines are perpendicular not parallel
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Answer:

Correct option: (B)

Step-by-step explanation:

A one-sample <em>t</em>-test can be performed to determine whether the mean weight of female college students is still 59.4 kg.

The hypothesis can be defined as:

<em>H₀</em>: The mean weight of female college students has changed.

<em>Hₐ</em>: The mean weight of female college students has not changed.

The information provided is:

\bar x=61.65\\s=6.39\\n=20\\\alpha =0.10

As the there is no information about the population standard deviation we will use a <em>t</em>-test for the mean.

The test statistic is:

t=\frac{\bar x-\mu}{\s/\sqrt{n}}=\frac{61.65-59.4}{6.39/\sqrt{20}}=1.57

Decision rule:

If the if the <em>p</em>-value of the test is less than the significance level of the test <em>α</em> then the null hypothesis will be rejected and vice versa.

The degrees of freedom of the test is:

df=n-1=20-1=19

The test is two-tailed.

Compute the <em>p</em>-value of the test as follows:

p-value=2\times P(t_{0.10/2, 19}

*Use a <em>t</em>-table for the probability.

The <em>p</em>-value = 0.13 > <em>α</em> = 0.10

The null hypothesis was failed to be rejected.

As the null hypothesis was rejected it can be concluded that there is not sufficient evidence that the mean weight of female students has changed.

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Step-by-step explanation:

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8 less than the product of a number n and 1/7 is no more than 95 .
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Answer:

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95 ≤ 8 < 1/7n

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Example 1: Calculation of Normal Probabilities Using ????????-Scores and Tables of Standard Normal Areas The U.S. Department of
Molodets [167]

Answer:

i) 0.872

ii) 0.300

iii) 0.76

iv) 0.704

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  $261.50 per month

Standard Deviation, σ = $16.25

We are given that the distribution of monthly food  cost for a 14- to 18-year-old male is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

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P( x < 280) = P( z < \displaystyle\frac{280 - 261.50}{16.25}) = P(z< 1.138)

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b) P(More than $270)

P(x > 270)

P( x > 270) = P( z > \displaystyle\frac{270 - 261.50}{16.25}) = P(z > 0.523)

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Calculation the value from standard normal z table, we have,  

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c) P(More than $250)

P(x > 250)

P( x > 250) = P( z > \displaystyle\frac{250 - 261.50}{16.25}) = P(z > -0.707)

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Calculation the value from standard normal z table, we have,  

P(x > 250) = 1 - 0.240 = 0.76 = 76.0\%

d) P(Between $240 and $275)

P(240 \leq x \leq 275) = P(\displaystyle\frac{240 - 261.50}{16.25} \leq z \leq \displaystyle\frac{275-261.50}{16.25}) = P(-1.323 \leq z \leq 0.8307)\\\\= P(z \leq 0.8307) - P(z < -1.323)\\= 0.797 - 0.093 = 0.704 = 70.4\%

P(240 \leq x \leq 275) = 70.4\%

e) Thus, 0.704 is the probability  that the monthly food cost for a randomly selected 14- to 18-year-old male is between $240 and $275.

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3 years ago
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