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Mathematics

1 answer:

klasskru [66]3 years ago

8 0

Answer:

D. A translation 1 unit to the right followed by a 270-degree counterclockwise rotation about the origin

Step-by-step explanation:

See the picture for better visual

Take segments ST and S'T'. If we extend them they will intersect at right angle.

It is the indication that the rotation is 90° or 270° but not 180°, when the corresponding segments come parallel.

The QRST is in the quadrant IV and Q'R'S'T' is in the quadrant III, which mean the rotation is 90° clockwise or 270° counterclockwise.

This rotation rule is:

(x, y) → (y, -x)

We also see the points S and T have x-coordinate of 5 but their images have y-coordinates of -6. It means the translation to the right by 1 unit was the step before rotation.

We now can conclude the correct choice is D:

A translation 1 unit to the right followed by a 270-degree counterclockwise rotation about the origin

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Write a ratio for the following word phrase. Write fractions in lowest terms.42 in. to 6 ft

ollegr [7]

42 feet to 6 feet can be written as both ratio and a fraction:

42:6
42/6

If we look at the fraction, we can clearly see that we can simplify it:

42/6 = 7/1 = 7

So we can write the ratio in its lowest terms as 7:1.

6 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:

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