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S_A_V [24]
4 years ago
15

Alfredo is buying a watch for $200. He will make an initial payment of $50 and then pay the remaining balance in payments for $1

5 per week. Which function can be used to find a, the amount he will still need to pay at the end of n weeks? *
Mathematics
1 answer:
Alexandra [31]4 years ago
6 0

Answer:multiplication

Step-by-step explanation:

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I need to know how to do this step by step
mariarad [96]
Multiply negative 0.5 times negative 8
The answer is 4
5 0
3 years ago
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Calculate the length of the line segment. round your answer to the nearest tenth.
Sunny_sXe [5.5K]

Answer:

c) 8.6

Step-by-step explanation:

Take the coordinates of the two points.

(0,8) and (7,3)

Formula:

√(x₁ - x₂)² + (y₁ - y₂)²

Substitute:

√(0 - 7)² + (8 - 3)²

√(-7)² + (5)²

√(49) + (25)

√(74) ≈ 8.6

8 0
3 years ago
In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0
4 years ago
A length of a rope is 8 meter long how many 2/5 meter pieces can be cut from the length of the rope
kozerog [31]
8 /  \frac{2}{5}  = 8 *  \frac{5}{2}  = 4*5 = 20
7 0
3 years ago
Find (f*f)(x) for the function f(x)=8x+9 and simplify
Alexus [3.1K]

Answer:

f(fx) = 64x + 81

Step-by-step explanation:

f(x) = 8x + 9

f(fx) = 8(8x+9) + 9

= 64x + 72 + 9

= 64x + 81

(Correct me if i am wrong)

5 0
3 years ago
Read 2 more answers
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