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3 years ago

1. Which of the following shows the correct use of the Distributive Property when solving 1/3( 33

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

5 0

Answer:

The option $(\frac{1}{3}\times 33)-\frac{1}{3}y=\frac{1}{3}135.2$ is correct

Step-by-step explanation:

Given equation is $\frac{1}{3}(33-y)=135.2$

The option shows correct usage of distributive property is

$(\frac{1}{3}\times 33)-\frac{1}{3}y=\frac{1}{3}135.2$

( by distributive property $a(x+y)=ax+ay$ here $a=\frac{1}{3}$ , x=33 and y=-y )

Therefore the option $(\frac{1}{3}\times 33)-\frac{1}{3}y=\frac{1}{3}135.2$ is correct.

You might be interested in

Which is a solution to the equation 6 X Plus y equals 4

artcher [175]

6x+y = 4
y = -6x + 4

0 = -6x + 4
6x = 4
x = 2/3

(2/3,0)

I Hope I helped

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7 0

3 years ago

A veterinarian’s assistant made a table that shows the animals seen in the office in one week. What is the probability that the

horsena [70]

Answer:

The probability that the pet seen was sick is 53.57%.

Step-by-step explanation:

The probability of an event E is the ration of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

Here,

n (E) = number of favorable outcomes

N = total number of outcomes

Denote the events as follows:

C = the pet is a cat

D = the pet is a dog

O = the pet is some other animal

S = the pet is sick.

The data provided is summarized as follows:

n (C) = 28

n (C ∩ S) = 3

n (D) = 42

n (B ∩ S) = 4

n (O) = 24

n (O ∩ S) = 8

Compute the probability that a cat was sick as follows:

$P(C\cap S)=\frac{n(C\cap S)}{n(C)}=\frac{3}{28}$

Compute the probability that a dog was sick as follows:

$P(D\cap S)=\frac{n(D\cap S)}{n(D)}=\frac{4}{42}=\frac{2}{21}$

Compute the probability that another animal was sick as follows:

$P(O\cap S)=\frac{n(O\cap S)}{n(O)}=\frac{8}{24}=\frac{1}{3}$

Compute the probability that the pet seen was sick as follows:

P (S) = P (C ∩ S) + P (D ∩ S) + P (O ∩ S)

$=\frac{3}{28}+\frac{2}{21}+\frac{1}{3}\\=\frac{9+8+28}{84}\\=\frac{45}{84}\\=0.5357$

Thus, the probability that the pet seen was sick is 53.57%.

8 0

3 years ago

Read 2 more answers

A line for tickets to a Broadway show had a mean waiting time of 20 minutes with a standard deviation of 5 minutes.

andrezito [222]

Answer:

5.48% of the people in line waited for more than 28 minutes

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean waiting time of 20 minutes with a standard deviation of 5 minutes.

This means that $\mu = 20, \sigma = 5$

What percentage of the people in line waited for more than 28 minutes?

The proportion is 1 subtracted by the p-value of Z when X = 28. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{28 - 20}{5}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452.

1 - 0.9452 = 0.0548.

As a percentage:

0.0548*100% = 5.48%

5.48% of the people in line waited for more than 28 minutes

6 0

3 years ago

Segment AC is a diagonal of a regular polygon ABCDEF. What does Angle ACD equal.

Marina86 [1]

Answer: the answer is D.108

4 0

3 years ago

Billy has 2/3 the amount of Instagram followers that is sister Stacy has if they have 1500 followers altogether how many does ea

olganol [36]

Answer:

Stacy has 900 followers. Billy has 600 followers.

Step-by-step explanation:

900+600=1500. And 2/3 of 900 is 600.

5 0

4 years ago