All categories

3 years ago

Find the percent increase. Round to the nearest percent.

Mathematics

2 answers:

noname [10]3 years ago

8 0

Answer:

I think it's a 72.36% increase

Step-by-step explanation:

I'm not completely sure but that's what the calculator I used said..

yan [13]3 years ago

8 0

Answer:

Step-by-step explanation:

Find the amount of change.

Amount of change = Greater Value − Lesser Value

= 131 − 127 Substitute values.

= 4 Subtract.

Find the percent increase. Round to the nearest percent.

Percent Change =

Amount of Change

Original Amount

131

Substitute values.

≈ 0.03 Divide and round to the nearest hundredth.

= 3% Write as a percent.

The percent decrease is 3%.

pls press the heart if this was helpful

You might be interested in

A set of silverware contains 30 pieces. The set contains only spoons, forks, and knives. It contains the same number of spoons a

Tema [17]

Let the number of knife be x.
Knife = x
Spoon = 2x
forks = 2x

Total = 30
x + 2x + 2x = 30
5x = 30
x = 6

Knife = x = 6
forks = 2x = 12
Spoon = 2x = 12

Answer: There are 12 forks.

5 0

3 years ago

Read 2 more answers

In a multiple choice quiz, there are 5 questions each with 4 answer choices (a, b, c, and d). Robin has not studied for the quiz

k0ka [10]

Answer:

A. 0.1035

B. 0.1406

C. 0.1025

Step-by-step explanation:

Given that:

the number of sample questions (n) = 5

The probability of choosing the correct choice (p) = 1/4 = 0.25

Suppose X represents the number of question that are guessed correctly.

Then, the required probability that she gets the majority of her question correctly is:

P(X>2) = P(X=3) + P(X =4) + P(X = 5)

$P(X>2) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4} + (^{5}C_{5}) \times (0.25)^5 (1-0.25)^{5-5}$

$P(X>2) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} +\dfrac{5!}{5!(5-5)!} \times (0.25)^5 (1-0.25)^{0} \Bigg ]$

P(X>2) = [ 0.0879 + 0.0146 + 0.001 ]

P(X>2) = 0.1035

Recall that

n = 5 and p = 0.25

The probability that the first Q. she gets right is the third question can be computed as:

$P(X=x) = 0.25 ( 1- 025) ^{x-1}$

Since, x = 3

$P(X = 3) = 0.25 ( 1- 0.25 ) ^{3-1}$

$P(X =3) = 0.25 (0.75)^{3-1}$

$P(X =3) = 0.25 (0.75)^{2}$

P(X=3) = 0.1406

The probability she gets exactly 3 or exactly 4 questions right is as follows:

P(X. 3 or 4) = P(X =3) + P(X =4)

$P(X=3 \ or \ 4) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4}]$

$P(X=3 \ or \ 4) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} \Bigg ]$

P(X = 3 or 4) = [ 0.0879 + 0.0146 ]

P(X=3 or 4) = 0.1025

3 0

3 years ago

Which exspression is equivalint to -9 - (-4 1/3

Naya [18.7K]

Hey there!

-9 - (-4 1/3)

= -9 - (-13/3)

= -9 + 13/3

= -14/3

= -4 2/3

Therefore, your answer is: -14/3 or -4 2/3 either should work because both of the numbers are equivalent to each other

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

8 0

3 years ago

Joel made some muffins. He gave 1\4 of the muffins to a neighbor.He took 3\8 of the muffins to school.What fraction of muffins i

forsale [732]

3/8 because we add what he gave off then subtract that by 8/8

6 0

3 years ago

Read 2 more answers

4 people are to be chosen from 10 men and 12 women to form a committee which contains at leat two women. How many different ways

Natali5045456 [20]

Answer with Step-by-step explanation:

The condition that at least 2 women are included is satisfied in the below cases:

Case 1) Exactly 2 women included

Thus the total number of ways to select the committee is

$N_1=\binom{12}{2}\times \binom{10}{2}=2970$

Case 2) Exactly 3 women included

Thus the total number of ways to select the committee is

$N_2=\binom{12}{3}\times \binom{10}{1}=2200$

Case 3) Exactly 4 women are included

Thus the total number of ways to select the committee is

$N_3=\binom{12}{4}\times \binom{10}{0}=495$

Thus the total number of commitees possible are

$N_1+N_2+N_3=2970+2200+495=5665$

Part 2)

If Mr and Mrs Simith are not to be both included then in that case the number of ways are the sum of

1) All cases of Mr Smith included and Mrs smith excluded

$N_4=\binom{11}{2}\binom{10}{2}+\binom{11}{3}\binom{10}{1}+\binom{11}{4}\binom{10}{0}\\\\N_4=4455$

2) Mrs smith included and Mr Smith Included

$N_5=\binom{11}{1}\binom{9}{2}+\binom{11}{2}\binom{9}{1}+\binom{11}{3}\binom{9}{0}\\\\N_5=1056$

Thus the cases are $N_4+N_5=5511$

6 0

4 years ago

Read 2 more answers