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2 years ago

What are the coordinates of the point X'if the image is rotated 90° counter clockwise

Mathematics

1 answer:

Wewaii [24]2 years ago

3 0

Step-by-step explanation:

When rotating a point 90 degrees counterclockwise about the origin our point A(x,y) becomes A'(-y,x). In other words, switch x and y and make y negative

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The length of a rectangle is 5 inches less than twice its width. The perimeter of the rectangle is 26 inches. What is the width

Yakvenalex [24]

Answer:

w= 6

Step-by-step explanation:

I just started by making an educated guess using the values already given. Then I inserted that into the problem to see if it worked.

L= 2w - 5

I used 6 as a random, educated guess for the value of w.

L = 2(6) - 5

L = 12-5

L = 7

Then, multiply L by 2 to account for both side lengths of the rectangle.

7(2)= 14

Subtract that value from the total perimeter to find what the width must equal.

26 - 14 = 12

Divide that answer by 2 since there are two sides for width.

12/2 = 6

I know this was kind of long, but I hope it helps! :)

3 0

2 years ago

Please answer please please answer need the answer

Korvikt [17]

A = 1/2 bh
A = 1/2 (21)(8)
A = 1/2 (168)
A = 84

5 0

2 years ago

Simplify: 3 . 3^2 + 8 ÷ 2 - (4 + 3) A: 32 B: 24 C: 23 D: 30

yaroslaw [1]

Answer:

its none of these answers ;-;

Step-by-step explanation:

3^2 + 8 ÷ 2 - (4 + 3)

we do what's in parentheses first. (pemdas)

so now we have, 3^2 + 8 ÷ 2 - 7

next exponents, 3^2=9

9 + 8 ÷ 2 - 7

next division, 8÷2=4

9 + 4 - 7

next addition, 9+4=13

13-7

finally subtraction, 13-7=6

final answer: 6

7 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:

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