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Ainat [17]
2 years ago
6

7500 portions of strawberries have been ordered for the tournament, but because of bad weather only 5 6 of the order has been de

livered. 10% of those delivered are given to the players and staff. How many portions of strawberries can be sold?
Mathematics
1 answer:
Sati [7]2 years ago
5 0
Do 56 times 0.10. Subtract the answer from 56 then you'll know how many portions can be sold.
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Answer: C) Water in the river

Step-by-step explanation:

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Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27
irina [24]

Answer:

Step-by-step explanation:

Data D has the largest range.

Data A: 61-48=13

Data B: 29-14=15

Data C:27-12=15

Data D:54-31=23

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What is the number of chocolate-covered pretzels per ounce
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If this is your question, the answer is 4
<span>There are 64 pretzels in a 16 ounce bag of chocolate covered pretzels. what is the number of chocolate covered pretzels per ounce</span>
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3 years ago
Divide. <br><br> 8.844 ÷ 0.04 = <br><br><br> and <br> Divide. <br><br> 0.48 ÷ 120 =
miskamm [114]
221.1 and .0004 are the answers.
6 0
3 years ago
Adecco Workplace Insights Survey sampled men and women workers and asked if they expected to get a raise or promotion this year
fgiga [73]

Answer:

(a) <u>Null Hypothesis,</u> H_0 : p_1-p_2 \leq 0  or  p_1 \leq p_2  

    <u>Alternate Hypothesis</u>, H_a : p_1-p_2 > 0  or  p_1 > p_2

(b)  \hat p_1 = sample proportion of men = \frac{104}{200} = 0.52

     \hat p_2 = sample proportion of women = \frac{74}{200} = 0.37

(c) P-value is 0.0011.

Step-by-step explanation:

We are given that Adecco Workplace Insights Survey sampled men and women workers and asked if they expected to get a raise or promotion this year (USA Today, February 16, 2012).

Suppose the survey sampled 200 men and 200 women. If 104 of the men replied yes and 74 of the women replied yes.

Let p_1 = population proportion of men who replied yes

     p_2 = population proportion of women who replied yes

(a) <u>Null Hypothesis,</u> H_0 : p_1-p_2 \leq 0  or  p_1 \leq p_2   {means that the population proportion of men is less than or equal to the population proportion of women}

<u>Alternate Hypothesis</u>, H_a : p_1-p_2 > 0  or  p_1 > p_2   {means that the population proportion of men is more than the population proportion of women}

(b)  \hat p_1 = sample proportion of men = \frac{104}{200} = 0.52

     \hat p_2 = sample proportion of women = \frac{74}{200} = 0.37

The test statistics that will be used here is <u>Two-sample z proportion statistics</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 men who replied yes = 0.52

            \hat p_2 = sample proportion of women who replied yes = 0.37

            n_1 = sample of men = 200

            n_2 = sample of women = 200

So, <u><em>test statistics</em></u>  =  \frac{(0.52-0.37)-(0)}{\sqrt{\frac{0.52(1-0.52)}{200} +\frac{0.37(1-0.37)}{200} } }

                               =  3.05

Now at 0.01 significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the population proportion of men is greater than the population proportion of women.

<u>P-value is given by the following formula</u>;

     P(Z > 3.05) = 1 - P(Z \leq 3.05)

                        = 1 - 0.99886 = 0.0011

Hence, the p-value is 0.0011.

8 0
3 years ago
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