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9 months ago

Mr. Parker spent $96.30 on 3 hats and 6 key chains. if each hat cost $15.50 how much did each key chain cost?

Mathematics

1 answer:

Vladimir [108]9 months ago

6 0

1) Given that Mr. Parker spent that we can write

3h +6k = 96.30

h=15.5

2) So now let's solve system system

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2 years ago

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2 years ago

Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the nu

olga_2 [115]

Answer:

The probability that the selected joint was judged to be defective by neither of the two inspectors is $P(A' n B' ) = 0.8855$

The probability that the selected joint was judged to be defective by inspector B but not by inspector A is $P(A' n B) =0.0403$

Step-by-step explanation:

From the question we are told that

The sample size is $n_s = 10000$

The number of outcome for inspector A is $n__{A}} = 742$

The number of outcome for inspector B is $n__{B}} = 745$

The number of joints judged defective by both inspector is $n(A u B) = 1145$

The the probability that the selected joint was judged to be defective by neither of the two inspectors is mathematically represented as

$P(A' n B' ) = 1 - P(A u B)$

Now

$P(A\ u \ B) = \frac{n(Au B)}{n_s }$

substituting values

$P(A\ u \ B) = \frac{1145}{ 10 000 }$

$P(A' n B' ) = 1 - \frac{1145}{10 000}$

$P(A' n B' ) = 0.8855$

the probability that the selected joint was judged to be defective by inspector B but not by inspector A is mathematically represented as

$P(A' n B) = P(A \ u \ B) -P(A)$

Now

$P(A) = \frac{n__{A}}{n_s}$

substituting values

$P(A) = \frac{742}{10 000}$

$P(A' n B) = \frac{1145}{10 000} - \frac{742}{10 000}$

$P(A' n B) =0.0403$

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3 years ago