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PilotLPTM [1.2K]
3 years ago
9

I need to know this ASAP

Mathematics
1 answer:
sashaice [31]3 years ago
3 0

Answer:


Problem 1

x=2

y=3

Problem 2

x=-3

y=8


Goodluck :)

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Factor this polynomial completely.<br> x^2-8x+12
Sphinxa [80]

Answer:

(x - 2) (x - 6)

Step-by-step explanation:

Factor the following:

x^2 - 8 x + 12

The factors of 12 that sum to -8 are -2 and -6. So, x^2 - 8 x + 12 = (x - 2) (x - 6):

Answer:  (x - 2) (x - 6)

7 0
3 years ago
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PLEASE HELP! If you solve this I would be so grateful!
stealth61 [152]

Answer:

f(h)=2d+4.3

= 2 x 1.5+4.3

  h=7.3ft

4 0
2 years ago
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Help me please! BRAINLIEST
masya89 [10]

Answer:

1. 53 by 58 , 222 ft

2. 321000

3.1080

4.  3.5

Step-by-step explanation:

1. 23+15+15 = 53

28+15+15= 58

53+53+58+58= 222

6 0
3 years ago
Prove that &lt; AOD and &lt; BOD are supplementary
Karolina [17]

Angle AOD and < BOD are supplementary since they are both right angles which equal 180°.

<h3>What is a supplementary angle?</h3>

It should be noted that a supplementary angle simply means an angle that adds up to 180°.

In this case, the angles given are both rights angles. This will be:

= 90° + 90°

= 180°

In this case, Angle AOD and < BOD are supplementary since they are both rights angles which equal 180°.

Learn more about supplementary angles on:

brainly.com/question/404323

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4 0
2 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.

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7 0
1 year ago
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