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3 years ago

The shoes that Sofia wanted to buy at the mall were marked down from $30 to $24. What is the percentage of discount?

Mathematics

1 answer:

Nataliya [291]3 years ago

8 0

30-24=6
6/30=1/5 aka 20%

The answer is 20%

Hope this helps.

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There are 5 different coloured cups in a row.

Marizza181 [45]

There are 12 possible arrangements.

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Considering that the blue cup has to be on one end, we consider the number of arrangements possible with the other 4 cups, then multiply by 2(considering the blue on each end).

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Considering green and yellow before red and purple, there are $2 \times 2 = 4$ outcomes(G-Y-R-P, G-Y-P-R, Y-G-R-P and Y-G-P-R).
Another two possible outcomes are yellow-purple and green-red, or vice-versa, thus 4 + 2 = 6 outcomes.
To account for the blue on the end, multiplying by 2. $2 \times 6 = 12$
There are 12 possible arrangements.

A similar problem is given at brainly.com/question/24617788

6 0

3 years ago

Please help and explain how to do this

Mnenie [13.5K]

Answer:

4√3

Step-by-step explanation:

The distance formula applies. It tells you ...

distance = √((x2 -x1)² +(y2 -y1)²)

Filling in the given values, you have ...

distance = √((-√32 -(-4√2))² +(2√3 -(-√12))²)

= √((-4√2+4√2)² +(2√3 +2√3)²)

= √(0 + (4√3)²)

distance = 4√3

___

We make use of the fact that ...

$a\sqrt{b}=\sqrt{a^2b}\\\\\sqrt{32}=\sqrt{16\cdot 2}=4\sqrt{2}\\\\2\sqrt{3}=\sqrt{2^2\cdot 3}=\sqrt{12}$

3 0

3 years ago

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. So we use the binomial probability distribution 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 all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

3 years ago

Sam has $\frac{5}{8}$ of a pound of chocolate. If he eats $\frac{1}{3}$ of the chocolate he has, how many pounds of chocolate do

givi [52]

1/3 of 5/8...." of " means multiply
1/3 * 5/8 = 5/24 lbs

3 0

3 years ago

Read 2 more answers

Rewrite the rational exponent as a radical by extending the properties of integer exponents. 2 to the 3 over 4 power, all over 2

Zepler [3.9K]

We are asked in this problem to determine the simplified expression of the statement given. The rules that apply in exponential functions is that when an exponential term is raised to the power of an integer, the simplified term has a degree that is equal to the product of the integers involved. The operations involved should be applicable to terms with the same base number only. In this problem, we thus write:

2^3/4 / 2^1/2 = 2^3/4 * 2^-1/2 = = 2^(3/4 - 1/2) = 2^ 1/4. hence the answer is 2^0.25 or simply equal to 1.1892 determined using a calculator.

6 0

3 years ago