Explain how to use the pattern in the table to rewrite a division problem involving fractions as a multiplication problem

Line m passes through the points (-2, 12) and (5, – 3). Which is the slope of a

schepotkina [342]

Answer:

the dog

Step-by-step explanation:

dfg

8 0

3 years ago

Hector is competing in a 42 mile bicycle race, he has already completed 18 miles of the race ad is traveling at a constant speed

kkurt [141]

Time it took Hector to complete 18 miles is given by 18 / 12 = 1.5 hours.

Let the time when Wanda will catch up to Hector be t, then, the distance covered by Hector at the time they meet is 18 + 12t and the distance covered by Wanda at the time they meet is 16t.

Then, 18 + 12t = 16t
16t - 12t = 18
4t = 18
t = 18 / 4 = 4.5

Therefore, Wanda will catch up with Hector after 4.5 hours.

7 0

3 years ago

What property is illustrated by the statement? 7+(4+4)=(7+4)+4

sergij07 [2.7K]

C cumulative property of multiplication

4 0

3 years ago

Read 2 more answers

How to find surface area of a triangular prism

Ugo [173]

Answer:if there is numbers then you would multiply then or divide them to find it

Step-by-step explanation:

8 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.

So

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

7 0

3 years ago