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3 years ago

Choose the equation that represents this situation. Use n to represent the

Mathematics

2 answers:

Flauer [41]3 years ago

5 0

Answer:

Step-by-step explanation:

I know is C let em have brainliest

Alenkasestr [34]3 years ago

4 0

Answer:

I THINK ITS C

Step-by-step explanation:

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If r is ten more than the square root of l , then r = A) l2 + 10 B) (l + 10)2 C) l + 10 D) l + 10 E) l + 10

stiv31 [10]

Answer:

$r=\sqrt {I}+10$

Step-by-step explanation:

We first need to know the square root of I then add 10 on that square root since r is 10 more than the square root of I

Therefore,

$r=\sqrt {I}+10$

4 0

3 years ago

At the beach 29% of people have red towels and 18% have white towels if the rest have blue ,what percentage of the people at the

Sonbull [250]

The anwser is 53%....

8 0

3 years ago

Read 2 more answers

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:

B = bugs are present in a computer program.

D = 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 E given that another event X 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

A cube has a surface area of 864 square inches.

slamgirl [31]

The answer is (D) or 1728 In³

6 0

3 years ago

Read 2 more answers

Can some one help me ??? Anyone ?

Leya [2.2K]

$((-2)^2-4)^3+4\cdot(-5)=\\
(4-4)^3+(-20)=\\
0^3-20=\\
0-20=\\-20$

6 0

3 years ago