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Damm [24]
2 years ago
13

SALE 40% OFF! If a badminton set has a sale price of $18, what was the original price?

Mathematics
2 answers:
harina [27]2 years ago
8 0
I believe that $10.80 is the answer :) hope this helps
Rina8888 [55]2 years ago
3 0
The answer is 25.2 Yurr
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( 2 pts) Find the mean, median and mode of the data below, if possible: The responses of a sample of 1022 people who were asked
Dmitry [639]

The data is categorical data

Since the data is categorical data so it's become impossible to calculate either mean or median.

Now mode:

The data with the highest frequency is the mode.

So watchful is the mode because it contained 672 peoples

5 0
3 years ago
There is a bag with 6 marbles. Three marbles are blue, 2 are red and 1 is green. What is the probability that you will randomly
Dominik [7]

Answer:

Mike is not correct.

Step-by-step explanation:

The probability is an "out of" statement. meaning out of all 6 marbles if 2 are red how what is the probability of choosing a red marble the answer is 2 out of 6 or 1/3 or 0.3333333etc. For green the probability is 1 out of 6 or 1/6 or 0.16666666etc

8 0
3 years ago
Math HELP PLEASEE URGENT NOWWWWWW MATH!!!!
Viefleur [7K]

Answer:

A

Step-by-step explanation:

Answer is the first choice. you can tell by the notation.

5 0
3 years ago
Read 2 more answers
Translate into an expression the sum of 4 and m
jek_recluse [69]

Answer:

4m

Step-by-step explanation:

8 0
3 years ago
In 1943​, an organization surveyed 1100 adults and​ asked, "Are you a total abstainer​ from, or do you on occasion​ consume, alc
Fynjy0 [20]

Answer:

We conclude that the proportion of adults who totally abstain from alcohol​ has changed.

Step-by-step explanation:

We are given that in 1943​, an organization surveyed 1100 adults and​ asked, "Are you a total abstainer​ from, or do you on occasion​ consume, alcoholic​ beverages?"

Of the 1100 adults​ surveyed, 429 indicated that they were total abstainers. In a recent​ survey, the same question was asked of 1100 adults and 352 in

<u><em>Let p = proportion of adults who totally abstain from alcohol.</em></u>

where, p = \frac{429}{1100} = 0.39

So, Null Hypothesis, H_0 : p = 39%      {means that the proportion of adults who totally abstain from alcohol​ has not changed}

Alternate Hypothesis, H_A : p \neq 39%      {means that the proportion of adults who totally abstain from alcohol​ has changed}

The test statistics that would be used here <u>One-sample z proportion statistics</u>;

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

where, \hat p = sample proportion of adults who totally abstain from alcohol = \frac{352}{1100} = 0.32

           n = sample of adults surveyed = 1100

So, <u><em>test statistics</em></u>  =  \frac{0.32-0.39}{\sqrt{\frac{0.32(1-0.32)}{1100} } }  

                              =  -4.976

The value of z test statistics is -4.976.

<em>Now, at 0.10 significance level the z table gives critical values of -1.645 and 1.645 for two-tailed test.</em><em> </em>

<em>Since our test statistics doesn't lie within the range of  critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the proportion of adults who totally abstain from alcohol​ has changed.

4 0
3 years ago
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