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3 years ago

Write an equation for the line that passes through (4,-6) and (-7,-6). Is the line parallel or perpendicular to the y-axis?

Mathematics

1 answer:

alisha [4.7K]3 years ago

7 0

Answer:

y = -6, perpendicular

Step-by-step explanation:

You can use the points to find the slope of the line:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\frac{-6-(-6)}{-7-4} =\frac{-6+6}{-7-4}=\frac{0}{-11}=0$

Then, using point-slope form, choose one set of coordinates to use for y1 and x1:

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

$y-(-6)=0(x-4)$

$y+6=0$

$y=-6$

Since the slope of the line is zero, it's a horizontal line. The y-axis is a vertical line, which means that the line is perpendicular to the y-axis.

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I don’t know how to do this one.

serious [3.7K]

Answer:

(a) The unit circle is centered at (0,0) with a radius of 1.

(b) The equation of a circle of radius r, with a center located at (0,0):

x²+ y² = r².

(ii) P(0,1)

(iii) P(-1,0)

(iv) P(0,-1)

Step-by-step explanation:

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3 years ago

Read 2 more answers

(1 point) 1. The height (in feet) and volume of usable lumber (in cubic feet) of 32 cherry trees are measured by a researcher. T

Thepotemich [5.8K]

Answer:

C. volume

Step-by-step explanation:

The response variable in this scenario would be the volume of the usable lumber. That is because this variable depends completely on the height of the cherry trees that are being measured. The higher that the cherry trees are the more volume can be expected to get from cutting these trees down. The opposite goes for trees that are smaller, they would decrease the total expected volume that will be received from the usable lumber since there would be less amount of tree to cut down.

3 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

$\mu_s=np=500*0.90=450$

The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$