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ludmilkaskok [199]
3 years ago
5

David recorded his finishing time, in seconds, for each time he ran the 100 m race during one track season. What was the median

of his finishing times?
11. 9 11.9 12.0 12.1 12.1 12.2 12.3 12.4 12.4
Mathematics
1 answer:
Vikentia [17]3 years ago
5 0
The median is 12.1 because it is the center number in the sequence. This only applies if the numbers are in order from least to greatest, and also if there are a even number of numbers in the sequence you must take the average of the two center ones because there is no center one.
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Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili
lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:

P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}

Use the Bayes' theorem to compute the value of P (B | D) as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

                                                                                     =0.3333\times 0.3333\\=0.11108889\\\approx 0.1111

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0
3 years ago
How do i do this im very confused <br>9+(-2-8)=
Aliun [14]

Use PEMDAS form. (Parentheses, exponents, multiplication, division, addition, subtraction.) You will first solve -2-8 which is -10, then finish with 9 + (-10), or 9-10, which is -1. The answer is -1.

4 0
3 years ago
Parker marks sixths on number line. He writes 5/6 just before 1. What fraction does he write on the first mark to the right of 1
VashaNatasha [74]

Answer: 7/6

Step-by-step explanation:

If he is writing sixths, then we have multiples of 1/6.

this is:

0*(1/6) = 0

1*(1/6) = 1/6

.

.

.

5*(1/6) = 5/6 (the numer he wrote at the left of 1)

6*(1/6) =  6/6 = 1

7*(1/6) = 7/6

So the number next to 1, (at the right of 1) must be 7/6.

You also can find it by adding 1/6 to 1.

1/6 + 1 = 1/6 + 6/6 = 7/6.

4 0
3 years ago
Read 2 more answers
What is the domain of the function on the graph?
Dennis_Churaev [7]

Answer:

All real numbers greater than or equal to -3

Step-by-step explanation:

First look at graph where the line points to which direction of the graph

And look for any closed or open circles in the graph

Since in the graph has a close circle at (-3,-2) meaning it includes that x-value for its domain.

With the graph going to positive infinity it states that the domain is all real numbers.

So in conclusion it has a domain of all real numbers greater than or equal to -3

5 0
4 years ago
Seven times the difference of a number and 1
weeeeeb [17]
Describe that better
8 0
3 years ago
Read 2 more answers
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