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nekit [7.7K]
2 years ago
5

What is the equation of a horizontal line passing through the point (-3,8)

Mathematics
1 answer:
Tresset [83]2 years ago
5 0

Answer:

y=8

Step-by-step explanation:

Horizontal equations are of the form

y =

The y coordinate is 8

y=8

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The graph depicts the growth and costs using of line segment use the midpoint formula to approximate the cost during the year 19
marysya [2.9K]

Answer:

The approximate cost during the year 1987 was $7,533

Step-by-step explanation:

The complete question in the attached figure

we have

y -----> is the cost in dollars

x -----> is the year

Looking at the graph we have the points

(1983,5,047) and (1991, 10,019)

To determine the cost in the year 1987, find out the midpoint between the two given points

The formula to calculate the midpoint between two points is equal to

M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

substitute the values

M=(\frac{1983+1991}{2},\frac{5,047+10,019}{2})

M=(1987,7,533)

therefore

The approximate cost during the year 1987 was $7,533

7 0
3 years ago
Can anyone help me on this I would appreciate it
kaheart [24]

Answer:

14

Step-by-step explanation:

Triangle=180

180-90 (RA) = 90

90-41= 39

(2x+11)=39

2x=28

x=14

3 0
2 years ago
Read 2 more answers
Solve the inequality
Mazyrski [523]
The answer would be m ≥ 1  or m ≤ 1/3
7 0
3 years ago
According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are re
Luda [366]

Answer:

The questions asked are

If you randomly select 4 peanuts

1. Compute the probability that exactly three of the four M&M’s are brown

2. Compute the probability that two or three of the four M&M’s are brown.

3. Compute the probability that at most three of the four M&M’s are brown.

4. Compute the probability that at least three of the four M&M’s are brown.

Step-by-step explanation:

Given the following information

Brown=12%. P(B)=0.12

Yellow=15%. P(Y)=0.15

Red=12%. P(R), =0.12

Blue=23%. P(B) =0.23

Orange, =23%. P(O) =0.23

Green=15%. P(G)=0.15

Question 1.

They are independent events

If there are exactly three brown and the last is not brown

P(B n B n B n B')

P(B)×P(B)×P(B)×P(B')

0.12×0.12×0.12×(1-P(B))

0.001728×(1-0.12)

0.001728×0.88

0.00152.

0.152%

2. If two or three are brown

I.e we are going to two brown and two none brown or three brown and one not brown. (P(B)×P(B)×P(B')×P(B'))+ (P(B)×P(B)×P(B'))

(0.12×0.12×0.88×0.88)+(0.12×0.12×0.12×0.88)

0.0112+0.00152

0.0127

1.27%

3. At most three brown out of four then we are going to have

BBBB', BBB'B', BB'B'B', B'B'B'B'

These are the cases of at most three brown.

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B')×P(B') + P(B)×P(B')×P(B')×PB')+ P(B')×P(B')×P(B')×P(B')=

0.12×0.12×0.12×0.88+ 0.12×0.12×0.88×0.88+ 0.12×0.88×0.88×0.88+ 0.88×0.88×0.88×0.88=0.694

0.694

69.4%

4. At least 3 brown out of four selection

I.e BBBB', BBBB

These are the two options

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B)×P(B)=

0.12×0.12×0.12×0.88 + 0.12×0.12×0.12×0.12

0.001728

0.173%

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