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2 years ago

7

Michael is watching a movie that is 2 1/3 hours long. If he has already watched 1 2/9 hours of the movie, how much of the movie

is left?

1 answer:

MrMuchimi2 years ago

6 0

Answer:

1.1

Step-by-step explanation:

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5y+x2÷2 when x=4 and y= -3. when x=4 and y=-3, 5y + x2 ÷2=

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There is a new treatment to help smokers stop smoking in an experiment, 80% of 300 smokers quit after 10 days of treatment which

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Answer:

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1 year 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:

a

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

b

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

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

So

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

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

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

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

So

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

$P(A' n B) =0.0403$

7 0

3 years ago

Round 0.977 to the nearest hundredth

UkoKoshka [18]

.98 since the hundredth place is higher than a five it is rounded up

5 0

2 years ago