All categories

2 years ago

A number increased by 5 then divided by 2

Mathematics

1 answer:

alex41 [277]2 years ago

4 0

Answer:

(x + 5) /2

Step-by-step explanation:

"a number" (this will represent as "x")

"increased" another way to say add or "+"

then divided by 2

If you wrote it out like this "x+5/2"it would be wrong because of PEMDAS making you want to divide first instead of after.

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

Find (f•g)(2) given f(x)=x-3 and g(x)=x+3 Please show your work

marissa [1.9K]

Answer:

$f.g(x) = f(g(x))\\f(g(2)) = ?\\\\g(x) = x+3 \\g(2) = 5\\\\f(x) = x-3\\f(g(2)) = f(5) = 2=f.g(2)$

Step-by-step explanation:

7 0

2 years ago

Another wooden plank is leaning against an outside wall of a building. The bottom of the plank is 9 ft from the wall. The length

Vladimir [108]

Given:
bottom of the plank or ground is 9 feet from the wall
length of the plant is 41 ft
height of the wall is unknown.

Let us use the Pythagorean theorem.

a² + b² = c²
a² + (9ft)² = (41ft)²
a² + 81 ft² = 1,681 ft²
a² = 1,681 ft² - 81 ft²
a² = 1,600 ft²
√a² = √1,600 ft²
a = 40 ft

The height of the wall is 40 ft.

3 0

3 years ago

I need help with number 3 please and thank you

murzikaleks [220]

So here we're dealing with equivalent fractions.
It's really simple to find the answer, so I'll try to explain the best I can.

2 dogs / 5 cats is really just 2/5.
If we want to find an equivalent fraction, we have to multiply the numerator and the denominator by the same number.

Currently the number of cats is 5, and we need it to be 140. What we need to do is find the number it has to be multiplied by to equal 140, which is 140 divided by 5. 140 divided by 5 is 28, so 5 x 28 = 140!

We need to multiply the denominator (5) by 28, so that we can get 140. What we have now is ?/140.

Like I said, to find an equivalent fraction, we need to multiply the numerator by the same number as we did the denominator, which is 28!

2 x 28 = 56.

So 2/5 is the same as 56/140.

The answer is D) 56 Dogs/140 Cats.

Hope this helps!
If you're confused about anything leave me a reply and I'll try to explain the best that I can!

3 0

2 years ago

Evaluate the functions

Anna11 [10]

Answer:

(btw the first question answer is -4, not 0)

Step-by-step explanation:

(btw the first question answer is -4, not 0)

So to solve the 2nd equation you substitute t with 30

so 30-2/3*30

30-20=10

6 0

2 years ago