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zaharov [31]
3 years ago
7

Write the point-slope form of the line that passes through (6, 1) and is parallel to a line with a slope of -3. Include all of y

our work in your final answer. Type your answer in the box provided to submit your solution.
Please explain how to do this :)
Mathematics
1 answer:
denpristay [2]3 years ago
3 0

Answer:

(y - 1) = -3(x - 6)

Step-by-step explanation:

The equation of a line in point-slope form is

(y-y_{1}) = m (x-x_{1})

where y1 and x1 are 2 known points, and m is the slope of the line. So simply just fill in the blanks.

y is a variable so it is left alone (since there is no subscript).

y1 we are told is 1.

m we are told is -3 (the slope)

x is a variable so it is left alone (since there is no subscript).

x1 we are told is 6.

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A Local group claims that the police issue more than 60 speeding tickets a day in thire area. To prove their point, they randoml
MissTica

Answer:

The group's claim is not supported at 1% significance level.

Step-by-step explanation:

given that a local group claims that the police issue more than 60 speeding tickets a day in thire area

The data they collected for 2 weeks is given below:

70 48 41 68 69 55 70

57 60 83 32 60 72 58

\bar x = 60.2\\s = 12.9\\n=14

Std error of mean = \frac{12.9}{\sqrt{14} } \\=3.4477

H_0: \bar x=60\\H_a: \bar x >60

(Right tailed test at 1% significance level)

Mean difference = 0.10

Test statistic t with df 13 = mean difference/std error

= 0.029

p value = 0.4886

Since p >0.01, we accept null hypothesis

The group's claim is not supported at 1% significance level.

7 0
3 years ago
What is (6-6)+7x7+4= ?
laiz [17]
<h2>Anser</h2>

53

Step-by-step explanation:

(6-6)+7×7+4

=0+7×7+4

=7×7+4

=49+4

=53

8 0
4 years ago
Read 2 more answers
GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m
Harrizon [31]

Answer:

(a) Null Hypothesis, H_0 : \mu \leq 3.50 mg  

     Alternate Hypothesis, H_A : \mu > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average mg of mercury in compact fluorescent bulbs.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 3.50 mg     {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, H_A : \mu > 3.50 mg     {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

                          T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean mg of mercury = 3.59

            \sigma = population standard deviation = 0.18 mg

            n = sample of bulbs = 25

So, <em><u>test statistics</u></em>  =  \frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }

                               =  2.50

The value of z test statistics is 2.50.

<u>Now, P-value of the test statistics is given by;</u>

         P-value = P(Z > 2.50) = 1 - P(Z \leq 2.50)

                                             = 1 - 0.9938 = 0.0062

<em />

<em>Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

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Melissa drove her car 200 kilometers in 2.5 hours. At this rate
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