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3 years ago

It takes a while for new factory workers to master a complex assembly process. During the first month that the new

Mathematics

1 answer:

Citrus2011 [14]3 years ago

8 0

Answer:

Will I think it will be A

Step-by-step explanation:

because it say it was a positive

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Find the center and radius of the circle with the equation : (x+5)^2 + (y-2)^2 =25

murzikaleks [220]

Answer:

Center (-5, 2) and radius 5

Step-by-step explanation:

The equation for a circle is $(x-h)^{2} +(y-k)^{2} =r^{2}$

Where (h, k) is the center and r is the radius.

6 0

3 years ago

Select the correct answer. Find the value of x. Please answer!!!

olga nikolaevna [1]

C. 64

Rewrite the logarithm:

8=x^.5

Solve by raising both sides of the equation to the power of the inverse (in this case, the inverse of .5, which is 2):

8^2=x^(.5*2)

64=x

6 0

3 years ago

2(x2 + 5) − 3(x + 1) + 3

mestny [16]

Answer:

2x^2 -3x +10

Step-by-step explanation:

2(x^2 + 5) − 3(x + 1) + 3

Distribute

2x^2 +10 -3x -3 +3

Combine like terms

2x^2 -3x +10 -3+3

2x^2 -3x +10

7 0

3 years ago

A(9,2) B(1,6) if (2,k) is equidistant from A and B, find the value of k.

Pani-rosa [81]

P is the point (2,k)
PA = PB
PA = √(49 + (2-k)²) and PB = √(1 + (6 - k)²)
√(49 + (2-k)²) = √(1 + (6 - k)²) => (49 + (2-k)² = (1 + (6 - k)²
=> 49 + 4 - 4k + k² = 1 + 36 - 12k + k² => 8k = 37 - 53 = -16 => k = -2

5 0

3 years ago

Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they

Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

$n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276$

Number of ways of choosing 3 students From all males

$n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220$

The probability that all are male of choosing '3' students

$P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }$

$P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}$

P(E) = 0.067 = 6.71%

Final answer:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0

4 years ago