Mathematics, 13.09.2019 01:30, miaaa0

Acomputer assembling company receives 24% of parts from supplier x, 36% of parts from supplier y, and the remaining 40% of parts from supplier z. five percent of parts supplied by x, ten percent of parts supplied by y, and six percent of parts supplied by z are defective. if an assembled computer has a defective part in it, what is the probability that this part was received from supplier z?

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answered: senituliii

There is a 33% probability that this party was received from supplier Z.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of a defective part being supplied. For this probability, we have:

$P(A) = P_{1} + P_{2} + P_{3}$

In which $P_{1}$ is the probability that the defective product was chosen from supplier X(we have to consider the probability of supplier X being chosen). So:

$P_{1} = 0.24*0.05 = 0.012$

$P_{2}$ is the probability that the defective product was chosen from supplier Y(we have to consider the probability of supplier Y being chosen). So:

$P_{2} = 0.36*0.10 = 0.036$

$P_{3}$ is the probability that the defective product was chosen from supplier Z(we have to consider the probability of supplier Z being chosen). So:

$P_{2} = 0.40*0.06 = 0.024$

$P(A) = P_{1} + P_{2} + P_{3} = 0.012 + 0.036 + 0.024 = 0.072$

P(B) is the probability of the supplier chosen being Z, so P(B) = 0.4

P(A/B) is the probability of the part supplied being defective, knowing that the supplier chosen was Z. So P(A/B) = 0.06.

So, the probability that this part was received from supplier Z is:

$P = \frac{0.4*0.06}{0.072} = 0.33$

There is a 33% probability that this party was received from supplier Z.

answered: Guest

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

answered: Guest

22, 28, 34, 40, 46

b) f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c) the 14th row will have 100 seats

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100