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kotegsom [21]
3 years ago
8

Suppose that you are taking a T/F exam and you have no idea at all about the answers to the last three questions. You choose ans

wers randomly and therefore have a 50/50 chance of being correct on any one question. 1) what is the probability that you guessed right on exactly one of the three questions? [a] 2) what is the probability that you guessed right on zero of the three questions? [b] 3) what is the probability that you guessed right on at least two of the questions? [c]
Mathematics
1 answer:
podryga [215]3 years ago
7 0

Answer:

0.375, 0.125, 0.5

Step-by-step explanation:

given that you are taking a T/F exam and you have no idea at all about the answers to the last three questions.

Probability for correct guess = 0.5 remains constant for each of the three questions

X - no of questions is binomial with n =3 and p = 0.5

1) the probability that you guessed right on exactly one of the three questions

= P(X=1)

= 3C1(0.5)^3 = 0.375

2) the probability that you guessed right on zero of the three questions

= P(X=0)\\= 0.5^3=0.125

3) the probability that you guessed right on at least two of the questions

=P(x=3)+P(2)\\= 1-0.5\\=0.5

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lisov135 [29]
<h3>Answer: B. 11</h3>

An exponential has the base at the bottom, or the lower portion. Think of "basement" to help remember this. The exponent is the number up top, so 12 is the exponent.

8 0
3 years ago
One sample has a mean of and a second sample has a mean of . The two samples are combined into a single set of scores. What is t
Murrr4er [49]

Answer:

a) For this case we can use the definition of weighted average given by:

M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}

And if we replace the values given we have:

M = \frac{8*4 + 16*4}{4+4}= 12

b) M = \frac{8*3 + 16*5}{3+5}= 13

c) M = \frac{8*5 + 16*3}{5+3}= 11

Step-by-step explanation:

Assuming the following question: "One sample has a mean of M=8 and a second sample has a mean of M=16 . The two samples are combined into a single set of scores.

a) What is the mean for the combined set if both of the original samples have n=4 scores "

For this case we can use the definition of weighted average given by:

M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}

And if we replace the values given we have:

M = \frac{8*4 + 16*4}{4+4}= 12

b) what is the mean for the combined set if the first sample has n=3 and the second sample has n=5

Using the definition we have:

M = \frac{8*3 + 16*5}{3+5}= 13

c) what is the mean for the combined set if the first sample has n=5 and the second sample has n=3

Using the definition we have:

M = \frac{8*5 + 16*3}{5+3}= 11

7 0
3 years ago
Mr. Williams is building a sand box for his children. It cost $228 for the sand if he builds a regular-sandbox with dimensions 9
erastovalidia [21]

Answer:

It will cost $512.73

Step-by-step explanation:

9 in by 6 in sandbox cost $228

i.e., 9*6 = 54 in²

54 in² cost $228

1 in² will cost $228/54

≈ $4.22

If the new dimension is 13 1/2 in by 9 in

then area will be 13 1/2*9

27*9/2 = 121.5 in²

The cost of 121.5 in² will be

121.5*4.22 = $512.73

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umka2103 [35]

Answer:

d

Step-by-step explanation:

mutipily

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Kerry bought 3 books that were all the same price he paid a total of $40.20 what is the cost of one book
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Dont forget the dollar sign - $13.4
Cost of one book is $13.4
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