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hammer [34]
1 year ago
6

(word sentence) Pooky eats three cans of cat food each day. How long will 27 cans of food last?

Mathematics
2 answers:
kolbaska11 [484]1 year ago
7 0

Answer: The 27 cans of food will last 9 days since 27 ÷ 3 = 9.

Step-by-step explanation:

pshichka [43]1 year ago
5 0

ANSWER:

9 days

Solution:

\frac{27\text{ cans}}{3\text{ cans per day}}\text{ = 9 days}

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The city of Omaha decided to build a new parking garage in hopes of luring more people to the Old Market district.The city will
dimulka [17.4K]

Answer:

Given:

Mean, u = 135

Sample size, n = 42

Sample mean, x' = 126

Standard deviation = 16

Significance level = 0.05

1) The null and alternative hypotheses:

H0 : u = 135 (the revenue per day is $135)

H1 : u < 135 (the revenue per day is less than $135)

2) In this case, we have a left tailed test.

Let's use the formula:

= \frac{x' - u}{s/ \sqrt{n}} = \frac{126 - 135}{16 / \sqrt{42}} = -3.645

t = - 3.645

Critical value: at a significance level of 0.05, left tailed test,

tcritical = -1.683

We reject null hypothesis, H0, since t calculated, -3.645 is less than tcritical, -1.683.

3) Given, a = 0.05, df = 41, left tailed test,

t-value = 2.02

For the corresponding confidence interval, we have:

CI = (x' - \frac{t * \sigma}{\sqrt{n}}, x' + \frac{t * \sigma}{\sqrt{n}})

CI = (126 - \frac{2.02 * 16}{\sqrt{42}}, 126 + \frac{2.02 * 16}{\sqrt{42}})

CI = (121.014, 130.986)

4) Conclusion:

We reject the estimated potential revenue advised by the consultant, H0, as it is too high, because the t-calculated falls in rejection region(i.e, less than critical value), also the upper limit of the confidence interval is less than $135.

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3 years ago
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Vladimir79 [104]

X is 5

...................

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3 years ago
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