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zubka84 [21]
3 years ago
14

Which of the following is NOT true about “yield”? a. the value of the actual yield must be given in order for the percent yield

to be calculated b. the actual yield may be different from the theoretical yield because reactions do no always go to completion c. the percent yield is the ratio of the actual yield to the theoretical yield d. the actual yield may be...
Biology
1 answer:
Law Incorporation [45]3 years ago
8 0
The right answer for the question that is being asked and shown above is that: "b. the act<span>ual yield may be different from the theoretical yield because reactions do no always go to completion." This is the statement that is not true about "yield."
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Gregor Mendel

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He was the first to describe how characteristics are passed down from generation to generation.

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Deer ticks can carry both Lyme disease and human granulocytic ehrlichiosis (HGE). In a study of ticks in the Midwest, it was fou
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(a) 0.16

(b) 1

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Let Probability that ticks in the Midwest carried Lyme disease, P(L) = 0.16

Probability that ticks in the Midwest carried HGE disease, P(H) = 0.10

Probability that ticks in the Midwest carried either Lyme disease or HGE disease, P(L \bigcup H) = 0.10

(a) Probability that a tick carries both Lyme disease (L) and HE (H) is given by

    P(L \bigcap H);

As we know that P(A \bigcup B) = P(A) + P(B) - P(A \bigcap B)

So, in our question;

       P(L \bigcup H) = P(L) + P(H) - P(L \bigcap H)

        0.10 = 0.16 + 0.10 - P(L \bigcap H)

       P(L \bigcap H) = 0.16 + 0.10 - 0.10 = 0.16

Therefore, the probability that a tick carries both Lyme disease (L) and HE (H) is 0.16 or 16% .

(b) <em>Conditional Probability P(A/B) is given by</em> = \frac{P(A\bigcap B)}{P(B)}

So, the conditional probability that a tick has HE given that it has Lyme disease is given by = P(H/L)

    P(H/L) = \frac{P(H\bigcap L)}{P(L)} = \frac{0.16}{0.16} = 1 .

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Please see attachment

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