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blondinia [14]
3 years ago
5

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

Mathematics
1 answer:
astraxan [27]3 years ago
8 0

Answer:

x=40

Step-by-step explanation:

What the problem is telling you is that the unlabeled angle is the same as the 70 degree angle. Since all the angles of a triangle add up to 180 degrees. First take 70+70, which equals 140. Now take the 180-140=40 ,so 40 would be the missing angle.

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What is the slope of the line through (3, 2) and (-3, 4)?​
Angelina_Jolie [31]

Answer:

Slope =-1/3

Step-by-step explanation:

(3,2)..... x1 =3, y1=2

(-3,4).... x2=-3, y2 =4

Slope=(y2-y1) /(x2-x1)

Slope=(4-2) /(-3-3)

Slope =2/(-6)

Slope=-1/3

7 0
3 years ago
A rhombus with congruent sides is shown. Diagonals are drawn. The length of one diagonal is 6, and the length of the other diago
dezoksy [38]

Answer:

its 24

Step-by-step explanation:

i got it right on ed 2020

8 0
3 years ago
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Pls help plsssssssssssssssssssssssssssssssssss
lozanna [386]

Answer:

i belive its 1

Step-by-step explanation:

Because its SUM so its adding and the sum is 22 and 20+2=22.

6 0
2 years ago
In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0
3 years ago
4(a + 2) = -4<br> whats A
OLga [1]
The answer is -3 because -3+2=-1 then -1(4)=-4
4 0
3 years ago
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