Can somebody help me please with percent increases

Please please please answer this I have the answer I just need the work for it!!!!!

MrRa [10]

Answer:

x=80

Step-by-step explanation:

First of all, you must find the length of the side to add with 36 to equal 100. This would give you 64. So once you do that, you can do pythagorean theorem to find the length of x. This would give you 64^2 + 48^2 = x^2. Once you have this equation, you square your terms to get; 4,096 + 2,304 = x^2. After you add them you will now have 6,400 = x^2. Once you have this, you will need to do the square root of both sides of the equation; sqrt(6400) = x. This gives you x = 80.

Hope this helps!

Brainliest?

3 0

3 years ago

The numbers 1 through 50 are written with red marker on slips of paper while the numbers 51 through 100 are written with blue ma

Marizza181 [45]

Given:
1 - 50 written in red marker
51 - 100 written in blue marker

Probability of selecting a number greater than 81; 100 - 81 = 19 possible numbers. 1 draw. 1/19

Probability of selecting a number written in red: 1/50

Probability of selecting a number written in blue: 1/50

Probability of selecting a number that is a multiple of 10. There are 10 instances; 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 ; 1/10

5 0

3 years ago

Jimmy has a diecast metal car that is a scale model of an actual race car. If the actual length of the car is 10 feet 6 inches a

pickupchik [31]

10 feet 6 inches = 126 inches. Length of the car is 126 inches and the model length is 7 inches.
126 : 7 = 18
The scale factor of the model to the actual car is 1 : 18.

8 0

3 years ago

Write an expression for the calculation divide 15 into thirdsand subtract from 14 divided in half

Neporo4naja [7]

15/3 - 14/2

This is the expression you are looking for

( I'm not 100% sure but I believe this is correct ! )

6 0

3 years ago

According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se

belka [17]

Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

50% of adult smokers started smoking before 21 years old.

This means that $p = 0.5$

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that $n = 5$ .

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875$

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

$P(X \leq 4) = 1 - P(X = 5)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875$

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125$

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

4 0

2 years ago