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3 years ago

Find the distance from P to l. Line l contains points (-4,2) and (3,-5).Point P has coordinates (1,2).

Mathematics

1 answer:

Annette [7]3 years ago

7 0

Hello,

A=(-4,2)
B=(3,-5)
AB≡y-2=(x+4)(-7)/7==>y=-x-2

Slope of the perpendicular: 1

y-2=(x-1)1==>y=x+1 is the equation of the perpendicular.

Intersection: y=x+1 and y=-x-2 ==>(-3/2,-1/2)=Q

|PQ|²=(-1/2-2)²+(1+3/2)²=25/2
|PQ|=5√2/2≈3.5355339059....

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Read 2 more answers

Which of the following is an equation of the line that passes through the point (−2, 3) and is perpendicular to the graph of the

Ludmilka [50]

The equation of the line is $y=-\frac{1}{3}+\frac{7}{3}$

Explanation:

The given equation is $y=3 x-2$

<u>Slope:</u>

Since, the equation of the line is perpendicular to the equation $y=3 x-2$ , then, the slope is given by

$m=-\frac{1}{3}$

Hence, the slope is $m=-\frac{1}{3}$

<u>Equation of the line:</u>

The equation of the line can be determined using the formula,

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3 years ago

An average computer mouse inspector can inspect 50 mice per hour. The 48 computer mice inspectors at a particular factory can on

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Answer: C. Yes, because -2.77 falls in the critical region .

Step-by-step explanation:

Let $\mu$ be the population mean .

As per given , we have

$H_0:\mu=50\\\\ H_a: \mu$

Since the alternative hypothesis is left-tailed and population standard deviation is not given , so we need to perform a left-tailed t-test.

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

Also, it is given that ,

n= 48

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s= 10

$t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}=\dfrac{-4}{\dfrac{10}{6.93}}\\\\=\dfrac{-4}{1.443}\approx-2.77$

Degree of freedom = df = n-1= 47

Using t-distribution , we have

Critical value = $t_{\alpha,df}=t_{0.025,47}=2.0117$

Since, the absolute t-value (|-2.77|=2.77) is greater than the critical value.

So , we reject the null hypothesis.

i.e. -2.77 falls in the critical region.

[Critical region is the region of values that associates with the rejection of the null hypothesis at a given probability level.]

Conclusion : We have sufficient evidence to support the claim that these inspectors are slower than average.

Hence, the correct answer is C. Yes, because -2.77 falls in the critical region

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