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3 years ago

Study the exercises in which both addends are negative. What do you notice about the sum?

1 answer:

7 0

This question would be a lot better if you had actually included the exercises. When a question talks about other questions that you already did, you need to include those too. Otherwise we don't know what the question is talking about. Fortunately I was able to figure it out without seeing the exercises, but just for future reference, you should include that extra information.

When two addends(things you add together) are negative, adding them is a lot like adding positive numbers; you just have a negative sign on the sum. For example,
4 + 5 = 9
-4 + -5 = -9

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Step-by-step explanation:

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3 years ago

Simplify: (45.22 – 73.91) – |65.85 – 79.15| A. -47.41 B. -41.99 C. 40.18 D. 45.45

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3 years ago

Read 2 more answers

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

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