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Alchen [17]
3 years ago
5

Which expression would be easier to simplify if you used commutative property to change the grouping?

Mathematics
2 answers:
Brrunno [24]3 years ago
8 0

Answer:

I think c

Sory if it is wrong

Step-by-step explanation:

just olya [345]3 years ago
4 0
A because it’s just pluss what makes it easier
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HELP ME ASAP !!!!! ITS DUE TODAY
denis23 [38]

Answer a:

Figure 4 = 5 blocks up, 5 blocks right

Figure 5 = 6 blocks up, 5 blocks right

Answer b: Grows 2 blocks each time, 1 blocks at the top and 1 blocks on the right.

Answer c: Since you add 2 blocks each time, you do the opposite so you subtract 2 blocks. The answer will be 1 block.

Step-by-step explanation:

7 0
3 years ago
What is the anwser pls and thanks
Alja [10]

Answer:

400 atheletes (answer 2)

Step-by-step explanation:

35:21:14

x2

70:42:28

x2

140:84:56

x2

280:168:112

x1.5

420:252:168

420 + 252 + 168 = 840

5 0
3 years ago
Read 2 more answers
Based on information from a large insurance company, 67% of all damage liability claims are made by single people under the age
svetoff [14.1K]

Answer:

We conclude that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company.

Step-by-step explanation:

We are given that based on information from a large insurance company, 67% of all damage liability claims are made by single people under the age of 25.

A random sample of 52 claims showed that 43 were made by single people under the age of 25.

Let p = <u><em>population proportion of claims made by single people</em></u>

So, Null Hypothesis, H_0 : p \leq 67%     {means that the insurance claims of single people under the age of 25 is smaller than or equal to the national percent reported by the large insurance company}

Alternate Hypothesis, H_A : p > 67%      {means that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company}

The test statistics that will be used here is <u>One-sample z-test for</u> <u>proportions</u>;

                          T.S.  =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of claims made by single people under the age of 25 = \frac{43}{52} = 0.83

           n = sample of claims = 52

So, <u><em>the test statistics</em></u> =  \frac{0.83-0.67}{\sqrt{\frac{0.67(1-0.67)}{52} } }  

                                    =  2.454

The value of z-test statistics is 2.454.

Since in the question, we are not given the level of significance so we assume it to be 5%. Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 2.454 > 1.645, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company.

6 0
4 years ago
A restaurant has an electronic system that randomly selects customers when they pay for their meal to
polet [3.4K]

Using the binomial distribution, it is found that there is a 0.81 = 81% probability that NEITHER customer is selected to receive a coupon.

For each customer, there are only two possible outcomes, either they receive the coupon, or they do not. The probability of a customer receiving the coupon is independent of any other customer, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • For each customer, 10% probability of receiving a coupon, thus p = 0.1.
  • 2 customers are selected, thus n = 2

The probability that <u>neither receives a coupon is P(X = 0)</u>, thus:

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

P(X = 0) = C_{2,0}.(0.1)^{0}.(0.9)^{2} = 0.81

0.81 = 81% probability that NEITHER customer is selected to receive a coupon.

A similar problem is given at brainly.com/question/25326823

7 0
3 years ago
How many six digit number can be made from the digits 1, 1, 1, 2, 2, 3?
yulyashka [42]
The permutation of all digits is 6!, but we don't differentiate the same digits, so we divide it by the number of permutations of 1's (3!) and 2's (2!)

\dfrac{6!}{3!2!}=\dfrac{4\cdot5\cdot6}{2}=60

3 0
4 years ago
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