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meriva
3 years ago
11

Which figures demonstrate a single reflection? Select each correct answer.

Mathematics
1 answer:
AysviL [449]3 years ago
5 0

Answer:

Please see the attached image below, to find more information about the graph

The figures that are obtained by a single reflection are shown in the image inside a red rectangle.

The axis of reflection is shown with a black line.

- The figure from the left shows horizontal reflection

- The figure from the right shows vertical reflection

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2 3/4 divided by 1/2 ?
Vladimir [108]

Answer:

5.5

Step-by-step explanation:

7 0
2 years ago
Please look at the attached image. Really need Help!! Who ever find answer first will get brainily.
nadya68 [22]

Answer:

27°

Step-by-step explanation:

XWY and VWY are supplementary

∴ XWY + VWY = 180°

∴ XWY = 180° - VWY

XYW and ZYW are supplementary

∴ XYW + ZYW = 180°

∴ XYW = 180° - ZYW

ΔXWY is a right triangle at X

∴ XWY + XYW = 90°

∴ 180° - VWY + 180° - ZYW = 90°

∴ 180° - 6x + 180° - 4x = 90°

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7 0
3 years ago
Read 2 more answers
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

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

P(X=280)=P(279.5

8 0
3 years ago
Which of the following is equivalent to 5^-9
VLD [36.1K]
0.0000000005 9 zeros
4 0
3 years ago
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Write a problem that can be solved using a flowchart and working backward. Then draw the flowchart and solve the problem.
Len [333]
I'm sorry but do you ha e a picture of a flowchart
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4 years ago
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