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Vika [28.1K]
3 years ago
13

Try your best to answer it​

Mathematics
2 answers:
Ahat [919]3 years ago
8 0

Answer:

y=-4(x,y)=(0,-4)

y=-3(x,y)=(2,-3)

y=-2(x,y)=(4,-2)

y=-1(x,y)=(6,-1)

y=0(x,y)=(8,0)

y=1(x,y)=(10,1)

Step-by-step explanation:

inn [45]3 years ago
8 0

Answer:

Answer attached

Step-by-step explanation:

I think you can plot it in the graph.

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What is the standard equation for a circle with the center (2,1) and passing through (0, 1)
Volgvan
The radius is the distance from the center to any point.  The distance from (2,1) to (0,1) is of course 2.  The general formula for a circle with radius r and center (a,b) is

(x-a)^2 + (y-b)^2 = r^2

In our case we get

(x-2)^2 + (y-1)^2 = 4

4 0
3 years ago
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Enguun earns $17 per hour tutoring student-athletes at Brooklyn University.
Juliette [100K]

Answer:

a. Earnings of Enguun for 12 hours this month (G)

G = 204 dollars

b. Earnings of Enguun for  19.5 hours last month (G)

G = 331.5 dollars

Step-by-step explanation:

Nomenclature

G: Total earnings of Enguun in dollars

E: Enguun's earnings for 1 hour of tutoring (dollars/hour)

n: number of hours that Enguun tutored per month (hour/month)

Formula

G = E*n

E = $17 per hour = 17 dollars/hour

Problem development

a. Earnings of Enguun for 12 hours this month

n = 12 hours

G = E*n

G= 17\frac{dollars}{hour} * 12 hour  : We eliminate hours

G = 204 dollars

b. Earnings of Enguun for 19.5 hours last month

G = E*n

G = 17 * 19.5

G = 331.5 dollars

7 0
3 years ago
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Can somebody help me with this? Thank you
melomori [17]

Answer:

x≥1 or x≤−3

Step-by-step explanation:

x+1≥2  possibility 1

x+1−1≥2−1  Subtract 1 from both sides

x≥1

x+1≤−2  Possibility 2

x+1−1≤−2−1  Subtract 1 from both sides

x≤−3

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If they had 2 bikes, each person will ride for 8 miles, 4 miles on each bike, and then walk 4 miles.

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3 years ago
A fair die is cast four times. Calculate
svetlana [45]

Step-by-step explanation:

<h2><em><u>You can solve this using the binomial probability formula.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
8 0
3 years ago
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