All categories

3 years ago

Need answer Now!! 618 is 75% what is 100%?

Mathematics

2 answers:

aliya0001 [1]3 years ago

6 0

Answer:

824 I think?

Step-by-step explanation:

614 divided by 75 is 8.24.

8.24 x 100 is 824

ch4aika [34]3 years ago

4 0

824, 100% of 824 is 824, therefore 75 percent of 824 equals 618.

You might be interested in

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 radio station installs a vhf radio tower that stands 1500 ft

aleksley [76]

Whats the question u just told us some of the information

4 0

3 years ago

In AABC, MZA = 16°, mZB = 49°, and a = 4. Find c to the nearest tenth.

zhannawk [14.2K]

Answer:

Step-by-step explanation:

The sum of angles in a triangle is 180°, so angle C is ...

C = 180° -A -B

C = 180° -16° -49° = 115°

The measure of side c can be found from the Law of Sines:

c/sin(C) = a/sin(A)

c = a·sin(C)/sin(A) = 4·sin(115°)/sin(16°)

c ≈ 13.2

6 0

2 years ago

One-eighth times three-elevenths

wolverine [178]

Answer:

3/88

Step-by-step explanation:

8 0

3 years ago

Find the mean for the following group of data items. 4.1, 8.9, 3.2, 1.9, 7.3, 6.3, 6.7, 8.6, 3.2, 2.3, 5.9 (Round to 3 decimal p

Alex787 [66]

Answer:

The mean is 5.309.

Step-by-step explanation:

Given group of data,

4.1, 8.9, 3.2, 1.9, 7.3, 6.3, 6.7, 8.6, 3.2, 2.3, 5.9,

Sum = 4.1+ 8.9 + 3.2 + 1.9 + 7.3 + 6.3 + 6.7 + 8.6 + 3.2 + 2.3 + 5.9 = 58.4,

Also, number of observations in the data = 11,

We know that,

$Mean=\frac{\text{Sum of all observation}}{\text{Total observations}}$

Hence, the mean of given data = $\frac{58.4}{11}=5.30909\approx 5.309$

4 0

3 years ago