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NikAS [45]
3 years ago
9

I think I have the answer but, I really need help with the steps!

Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
6 0

Answer:

simplify to:

6-32768/-60

simplify more

-32762/-60

take away the negatives

32762/60

simplify the fraction

16381/30

keep going till you get the answer

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S÷6/7=3/4. How do I solve for s
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Answer:

do 6/7-3/4 so your answer is 1 1/7

Step-by-step explanation:

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3 years ago
What is the value of the ratio 12 1/2 14 3/4 feet
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1.4 hope ithelped!



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If 430 pounds of pebbles cover 1500 square feet, how many pounds of pebbles are necessary to cover 75,000 square feet?
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4 years ago
According to the University of Nevada Center for Logistics Management, 6% of all mer-
Fudgin [204]

Answer:

a) The point estimate of the proportion of items returned for the population of

sales transactions at the Houston store = 12/80 = 0.15

b) The 95% confidence interval for the proportion of returns at the Houston store = [0.0718 < p < 0.2282].

c) Yes.

We set an hypothesis and construct a test statistics. The test statistics result gives us:

Z calculated  = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.

Step-by-step explanation:

a) Point estimate of the proportion = number of returned items/ total items sold = 12/80 = 0.15.

b) By formula of confident interval:

CI(95%) = p ± Z*\sqrt{\frac{p*(1-p)}{n} }  =  0.15 \pm 1.96 *\sqrt{\frac{0.15*(1-0.15)}{80} },

CI(95%) = [0.0718 < p < 0.2282]

c) The hypothesis:

H_{0}: The proportion of returns at the Houston store is not significantly different from the returns  for the nation as a whole.

H_{a}: The proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.

The test statistics:

Z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p*(1-p)}{n} }}, where p_{0} is the proportion of nation returns.

Z calculated  = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns  for the nation as a whole.

6 0
3 years ago
Suppose that in a random selection of 100 colored​ candies, 26​% of them are blue. The candy company claims that the percentage
quester [9]

Answer:  a) -0.2252, b) 0.8219

Step-by-step explanation:

Since we have given that

Sample size n = 100

Probability that candies are blue = p= 0.26

Probability that company claims that it is blue candy = P = 0.27

So, Q = 1-P= 1-0.27 = 0.73

So, Null hypothesis : H_0:p=P

Alternate hypothesis : H_1:p\neq P

So, the test statistic would be

z=\dfrac{p-P}{\sqrt{\dfrac{P.Q}{n}}}\\\\z=\dfrac{0.26-0.27}{\sqrt{\dfrac{0.27\times 0.73}{100}}}\\\\z=-0.2252

Since α = 0.05

So, critical value of z = 1.96

p-value = P(Z>Z(calculated)

Using the excel function , we get that

P(z>0.2252)\\\\=2\times 0.410.911845\\\\=0.8219

Hence, a) -0.2252, b) 0.8219

6 0
3 years ago
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