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3 years ago

Find the distance between (-6,6.5), (-2,6.5)

Mathematics

1 answer:

natka813 [3]3 years ago

6 0

Answer:

Step-by-step explanation:

I did the math trust me, mark as brainliest please.

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Landscaping company planted 176 tress in 8 equal rows in the new park. Jow many tress did the company plant in each row

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22 plants in each row

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3 years ago

I'm so confused rn, please help a girl out

Svetach [21]

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3 years ago

Find the greatest common factor for pair 18,36

netineya [11]

The GCF of 18 and 36 is 18. To solve this, use the picture attached. After creating the boxes, find a number that both 18 and 36 can be divided by, for example, 9. Now that you have 9, divide both 18 and 36 by 9. That equals 2 and 4. Place the numbers underneath the boxes you made previously and find another number that both 2 and 4 are divisible by. 2 and 4 are both able to be divided into 2. Do 2 divided by 2 and 4 divided by 2. Now you have the numbers 1 and 2. There aren't any numbers that can be divided into 1 and 2, so now you are left with the numbers 9 and 2. Multiply the numbers together to get a GCF of 18. Hope this helped!

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4 years ago

Which equation has the steepest graph?

solmaris [256]

Answer:

D. y=5x+2

Step-by-step explanation:

The y int. is 2, and the slope is 5/1. If you plug that into a calculator, you can see the slope being steep.

8 0

3 years ago

two telephone calls come into a switchboard at times that are uniformly distributed in a fixed one-hour period. assume that the

AleksandrR [38]

We wil assume a variable x to be the total number of calls received by the switchboard.

The question also says to assume that the calls were made independently.

Given:

Calls are independent.

Calls are uniformly distributed over a 1 hour period.

Ans (a). The calls are distributed uniformly over 1 hour, hence: (0, 1).

So we have,

f1(x1) = 1

f2(x2) = 1

X1 and X2 are considered to be independent of each other. Hence,

f(x1,x2) = f1 (x1) f2 (x2)

f(x1,x2) = 1 (1)

f(x1,x2) = 1

Thus,

P(X1 <= 0.5; X2 <= 0.5) = ∫0.50 ∫0.50 f(x1,x2) dx2 dx1

= ∫^0.5 0 ∫^0.5 0 (1) dx2 dx1

= ∫^0.5 0 (x2^0.5 0) dx1

= ∫^0.5 0 (0.5 - 0) dx1

= 0.5 ∫^0.5 0 dx1

= 0.5 (x1^0.5 0)

= 0.5 (0.5 - 0)

= 0.25

Therefore, the probability that the calls were received within the first 30 minutes or first half hour is 0.25.

Ans (b). Steps 1 and 2 are the same as the above answer.

Probability = [∫^11/12 0 ∫^x1 + 1/12 x1 1dx2 dx1] + [∫^1 1/12 ∫^x1 x1-1/12 1dx2 dx1

= [∫^11/12 0 (x2 ^x1+1/12 x1 dx1] + [∫^1 1/12 (x2 ^x1 x1-1/12 dx1)]

= [∫^11/12 0 (x1 + 1/12 -x1) dx1] + [∫^1 1/12 (x1 - x1 + 1/12) dx1]

= [(1 + x1/12 - 1) ^11/12 0] + [( 1 - 1 + x1/12) ^1 1/12]

= 11/144 + 11/144

= 0.1528

Therefore, the probability that the calls were received within five minutes of each other is 0.15.

#SPJ4

7 0

1 year ago

Find the distance between (-6,6.5), (-2,6.5)​

Find the distance between (-6,6.5), (-2,6.5)