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3 years ago

1. what is the setting of home sick by jean fritz

Mathematics

1 answer:

galben [10]3 years ago

8 0

Answer:

2. Jean a ten-year-old girl in the town of Hankow. Her mother and father work for the YMCA doing relief work. She attends a British school there and hates her teacher, Miss Williams, along with a couple of the other students. In general, Jean feels culturally conflicted, the product of two cultures but not belonging to either one, and she longs for a long-promised trip to America in a couple of years.

Step-by-step explanation:

Homesick is author Jean Fritz's story of her childhood growing up in China in a town called Hankow. While Jean adopts a first-person perspective and includes mostly real-life details from her life, some parts of Homesick are fictionalized in order to add dramatic interest.

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3 years ago

A, B, and C are collinear points, where B is between A and C. Find x if AB = x, BC = x + 6, and AC = 24. Then find the measures

Alekssandra [29.7K]

Answer:

AB = 9, BC = 15

Step-by-step explanation:

Since, A, B, and C are collinear points, where B is between A and C. i. e. A - B - C

$\therefore AC = AB + BC\\\therefore 24 = x + x + 6\\\therefore 24 = 2x + 6\\\therefore 24 - 6 = 2x \\\therefore 18 = 2x \\ \therefore \: \frac{18}{2} = x \\ \therefore \: x = 9$

$\therefore AC = AB + BC\\\therefore 24 = x + x + 6\\\therefore 24 = 2x + 6\\\therefore 24 - 6 = 2x \\\therefore 18 = 2x \\ \therefore \: \frac{18}{2} = x \\ \therefore \: x = 9 \\ \because \: AB = x \\ \huge \red{ \therefore AB = 9} \\ \\ \because \:BC = x + 6 \\ \therefore \: BC = 9 + 6 \\ \huge \purple{ \therefore \: BC = 15}$

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3 years ago

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2 years ago

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A diagnostic test for a disease is such that it (correctly) detects the disease in 90% of the individuals who actually have the

Ne4ueva [31]

Answer:

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Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Test positive

Event B: Has the disease

Probability of a positive test:

90% of 3%(has the disease).

1 - 0.9 = 0.1 = 10% of 97%(does not have the disease). So

$P(A) = 0.90*0.03 + 0.1*0.97 = 0.124$

Intersection of A and B:

Positive test and has the disease, so 90% of 3%

$P(A \cap B) = 0.9*0.03 = 0.027$

What is the conditional probability that she does, in fact, have the disease

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.027}{0.124} = 0.2177$

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

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Answer:

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Step-by-step explanation:

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