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Oxana [17]
3 years ago
8

What is the median for 76,84,93,67,82,and 76

Mathematics
2 answers:
xxTIMURxx [149]3 years ago
5 0

Put the numbers in order:

<span>62, 76, 76, 82, 84, 93</span>

<span /><span>Which number is in the middle. Since two numbers are in the middle 76 and 82 you add them together and divide them by 2 to get the answer:</span>

<span /><span>\frac{76+82}{2} =  \frac{158}{2} = 79</span>

<span>The answer is 79</span>

Lapatulllka [165]3 years ago
3 0
First arrange the numbers from least to greatest:

67, 76, 76, 82, 84, 93

The median is the middle number. But since we have two numbers that are in the middle we have to find the average of them.

76 + 82 = 158
158/2 = 79

So your answer is 79
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There are 210210210 students in a twelfth grade high school class. 909090 of these students have at least one sister and 1051051
BlackZzzverrR [31]

Answer:

(1) The value of P (A) is 0.4286.

(2) The value of P (B) is 0.50.

(3) The value of P (A ∩ B) is 0.2143.

(4) The the value of P (B|A) is 0.50.

(5) The events <em>A</em> and <em>B</em> are independent.

Step-by-step explanation:

The events are defined as follows:

<em>A</em> = a student in the class has a sister

<em>B</em> = a student has a brother

The information provided is:

<em>N</em> = 210

n (A) = 90

n (B) = 105

n (A ∩ B) = 45

The probability of an event <em>E</em> is the ratio of the favorable number of outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

The conditional probability of an event <em>X</em> provided that another event <em>Y</em> has already occurred is:

P(X|Y)=\frac{P(A\cap Y)}{P(Y)}

If the events <em>X</em> and <em>Y</em> are independent then,

P(X|Y)=P(X)

(1)

Compute the probability of event <em>A</em> as follows:

P(A)=\frac{n(A)}{N}\\\\=\frac{90}{210}\\\\=0.4286

The value of P (A) is 0.4286.

(2)

Compute the probability of event <em>B</em> as follows:

P(B)=\frac{n(B)}{N}\\\\=\frac{105}{210}\\\\=0.50

The value of P (B) is 0.50.

(3)

Compute the probability of event <em>A</em> and <em>B</em> as follows:

P(A\cap B)=\frac{n(A\cap B)}{N}\\\\=\frac{45}{210}\\\\=0.2143

The value of P (A ∩ B) is 0.2143.

(4)

Compute the probability of <em>B</em> given <em>A</em> as follows:

P(B|A)=\frac{P(A\cap B)}{P(A)}\\\\=\frac{0.2143}{0.4286}\\\\=0.50

The the value of P (B|A) is 0.50.

(5)

The value of P (B|A) = 0.50 = P (B).

Thus, the events <em>A</em> and <em>B</em> are independent.

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