Given the figure below is a special type of trapezoid and WX || YZ, which angle pairs can be proven supplementary by the given i
nformation? Select all that apply.
∠W and ∠Z
∠W and ∠Y
∠X and ∠Y
∠X and ∠Z
∠W and ∠X
∠W and ∠Z
∠W and ∠Y
∠X and ∠Y
∠X and ∠Z
∠W and ∠X
2 answers:
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Answer:
The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.
Then, the mean of the distribution of the sum of values of X is given by,
And the standard deviation of the distribution of the sum of values of X is given by,
The information provided is:
<em>μ</em> = $970
<em>σ</em> = $129
<em>n</em> = 102
Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.
Then,
Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:
*Use a <em>z</em>-table for the probability.
Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.
Answer: 0.82
Step-by-step explanation:
The probability of the computer not containing neither a virus nor a worm is expressed as P(∩
) , where P(
) is the probability that the event V doesn't happen and P(
) is the probability that the event W doesn't happen.
P()= 1-P(V) = 1-0.17 = 0.83
P()=1-P(W) = 1-0.05 = 0.95
Since and
aren't mutually exclusive events, then:
P(∪
) = P(
) + P(
) - P(
∩
)
Isolating the probability that interests us:
P(∩
)= P(
) + P(
) - P(
∪
)
Where P(∪
) = 1 - 0.04 = 0.96
Finally:
P(∩
) = 0.83 + 0.95 - 0.96 = 0.82