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stealth61 [152]
4 years ago
14

A candy shop sells a box of chocolates for $30. It has $29 worth of chocolates plus $1 for the box. The box includes two kinds o

f candy; caramels and truffles. Lita knows how much the different types of candies cost per pound and how many pounds are in the box. She said if x is the number of caramels included in the box and y is the number of pounds of truffles in the box then I can write the following equations based on what I know about one of these boxes:
X+y=3
8x+12y+1=30


How many pounds of candy are in the box?

What is the price per pound of the caramels?

What does the term 12y in the second equation represent?

What does 8x+12y+1 in the second equation represent?
Mathematics
1 answer:
ANEK [815]4 years ago
5 0
Dm for the answer can’t do it right now
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A corporation randomly selects 150 salespeople and finds that 66% who have never taken a self- improvement course would like to
Citrus2011 [14]

Answer:

We conclude that there is no difference in the two population proportions using α = 0.05.

Step-by-step explanation:

We are given that a corporation randomly selects 150 salespeople and finds that 66% who have never taken a self- improvement course would like to do so.

The firm did a similar study 10 years ago and found that 70% of a random sample of 160 salespeople wanted a self-improvement course.

Let p_1 (\pi_1) = <u><em>true proportion of workers who would like to attend a self-improvement course in the recent study.</em></u>

p_2 (\pi_2) = <u><em>true proportion of workers who would like to attend a self-improvement course in the past study.</em></u>

SO, Null Hypothesis, H_0 : p_1=p_2      {means that there is no difference in the two population proportions}

Alternate Hypothesis, H_A : p_1\neq p_2      {means that there is a difference in the two population proportions}

The test statistics that would be used here <u>Two-sample z test for</u> <u>proportions</u>;

                       T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of salespeople who would like to attend a self-improvement course in recent study = 66%

\hat p_2 = sample proportion of salespeople who would like to attend a self-improvement course in past study = 70%

n_1 = sample of salespeople in recent study = 150

n_2 = sample of salespeople in past study = 160

So, <u><em>the test statistics</em></u>  =  \frac{(0.66-0.70)-(0)}{\sqrt{\frac{0.66(1-0.66)}{150}+\frac{0.70(1-0.70)}{160} } }

                                      =  -0.755

The value of z test statistics is -0.755.

<u>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that there is no difference in the two population proportions.

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BaLLatris [955]

Answer:

i dont get the question. can you be more especific.

Step-by-step explanation:

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