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3 years ago

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An exam has two probability problems, 1 and 2. If 37% of the students solved problem 1 and 12% of the students solved both probl

ems 1 and 2, what percent of the students who solved problem 1 also solved problem 2?

1 answer:

olga55 [171]3 years ago

7 0

Answer:

about 32%

Step-by-step explanation:

The desired probability is the conditional probability that problem 2 was solved given that problem 1 was solved. That is ...

P(2|1) = P(1&2)/P(1) = 12%/37% ≈ .324 ≈ 32%

____

You can also think of it like this. If there were 100 students taking the exam, 37 solved the first problem. Of those 37, 12 solved the second one. 12 is about 32% of 37.

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You roll a six-sided die. Find the probability of each of the following scenarios

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Answer:

a) $1/2$

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c) $2/3$

Step-by-step explanation:

a)

Rolling a number greater than 3 means you can roll a 4, 5, or 6, which contains rolling a 6. So, the probability is $(1+1+1)/6=3/6=\boxed{1/2}$

b)

Rolling a number less than 5 means you can roll a 1, 2, 3, or 4, but because you could also roll an even number, so you could also roll a 6.

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Step-by-step explanation:

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4 years ago

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(Small sample confidence intervals for a population mean) suppose you are taking a sampling of 15 measurements. you find that x=

Luda [366]

Answer:

The 99% confidence interval is $71.67 < \mu < 78.33$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 15$

The sample mean is $\= x = 75$

The standard deviation is $s = 5$

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$\alpha = 100 - 99$

$\alpha = 1\%$

$\alpha = 0.01$

Next we obtain the critical values of $\frac{ \alpha }{2}$ from the normal distribution table

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$Z_{\frac{ \alpha }{2} } = 2.58$

Generally the margin for error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{ s}{ \sqrt{n} }$

=> $E = 2.58 * \frac{ 5}{ \sqrt{15} }$

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3 years ago

Drag the ordered pair that are 5.5 units apart.

Alexus [3.1K]

Answer:

(-1.5, -3.5) and (-1.5, 2)

Step-by-step explanation:

The x-coordinates in increasing order are ...

-3, -1.5, 5

The distances between pairs of these coordinates are ...

-1.5 -(-3) = 1.5

5 -(-3) = 8

5 -(-1.5) = 6.5

Clearly, any pair with x = 5 is too far away from any pair with a different x-value.

<u>∆x = 1.5</u>

If the distance between pairs is 5.5, then the distance formula tells us the y-coordinate differences for pairs with x-coordinates of -3 and -1.5 must be ...

d = √((x2 -x1)^2 +(y2 -y1)^2)

5.5 = √(1.5^2 +(y2 -y1)^2) . . . substitute known values

30.25 = 2.25 +(y2 -y1)^2 . . . . square both sides

√28 = (y2 -y1) . . . . . . . . . . . . . . subtract 2.25 and take the square root

This value is irrational. Clearly none of the y-coordinates is irrational, so there are no point pairs with x-coordinates -3 and -1.5 that are 5.5 units apart.

__

<u>∆x = 0</u>

If the x-coordinates are the same, then the y-coordinates must differ by 5.5 in order for the points to be 5.5 units apart. The coordinates in order are ...

for x = -1.5, y = -3.5, 2, 2.5 . . . . . differences of 5.5, 6, 0.25

for x = 5, y = -3.5, 1.5 . . . . . . . difference of 5

The only pair we can find that is 5.5 units apart is ...

(-1.5, -3.5) and (-1.5, 2).

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