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cricket20 [7]
3 years ago
10

If we flip an unfair coin, suppose the probability to get a 'Head' is 0.6 each time. In a random sample of 75 tosses, let p deno

te the proportion of getting a 'Head' in the 75 tosses. What is the standard error of the sample proportion p .
Mathematics
1 answer:
Lerok [7]3 years ago
3 0

Answer:

Step-by-step explanation:

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Fractions and decimals can be written as a percent
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F(x)=3/x+2-√x-3 f(19)=​
77julia77 [94]

Answer:

  • - 27/7

Step-by-step explanation:

Given

  • f(x)=3/(x+2) - √(x-3)

To find

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Solution

  • f(19) =​ 3/(19 + 2) - √(19 - 3) = 3/21 - √16 = 1/7 - 4 = - 27/7
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Well, we see that it is a straight line
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Christopher had 3 1/2 pies left over from the party. He qually divided the 3 and 1/2 pies between 4 of his friends. How much did
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The answer is 3/8. Christopher with have to share 3/8 of pie with each of his friends.
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About Exercise 2.3.1: Proving conditional statements by contrapositive Prove each statement by contrapositive
Sladkaya [172]

Answer:

See proofs below

Step-by-step explanation:

A proof by counterpositive consists on assuming the negation of the conclusion and proving the negation of the hypothesis.

a) Assume that n is not odd. Then n is even, that is, n=2k for some integer k. Hence n²=4k²=2(2k²)=2t for some integer t=2k². Then n² is even, therefore n² is not odd. We have proved the counterpositive of this statement.

b) Assume that n is not even, then n is odd. Thus, n=2k+1 for some integer k. Now, n³=(2k+1)³=8k³+6k²+6k+1=2(4k³+3k²+3k)+1=2t+1 for the integer t=4k³+3k²+3k. Thus n³ is odd, that is, n³ is not even.

c) Suppose that n is not odd, that is, n is even. Now, n=2k for some integer k. Then 5n+3=10k+3=2(5k+1)+1, thus 5n+3 is odd, then 5n+3 is not even.

d) Suppose that n is not odd, then n is even. Now, n=2k for some integer k. Then n²-2n+7=4k²-4k+7=2(2k²-2k+3)+1. Hence n²-2n+7 is odd, that is, n²-2n+7 is not even.

e) Assume that -r is not irrational, then -r is rational. Since -1 is rational, then (-1)(-r)=r is rational. Thus r is not irrational.

f) Assume that 1/z is not irrational. Then 1/z is rational. Multiplucative inverses of rational numbers are rational, hence z is rational, that is, z is not irrational.

g) Suppose that z>y. We will prove that z³+zy²≤z²y+y³ is false, that is, we will prove that z³+zy²>z²y+y³. Multiply by the nonnegative number z² in the inequality z>y to get z³>z²y (here we assume z and y nonzero, in this case either z³>0=y³ is true or z³=0>y³ is true). On the other hand, multiply by z² (positive number) to get zy²>y³. Add both inequalities to obtain z³+zy²>z²y+y³ as required.

h) Suppose than n is even. Then n=2k, and n²=4k² is divisible by 4.

i) Assume that "z is irrational or y is irrational" is false. Then z is rational and y is rational. Rational numbers are closed under sum, then z+y is rational, that is, z+y is not irrational.

3 0
2 years ago
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