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Igoryamba
1 year ago
12

A spear has a diameter of 14 ft which equation finds the volume of the sphere

Mathematics
1 answer:
fiasKO [112]1 year ago
6 0

The volume of the sphere is 1437.33 ft³

<h3></h3><h3>What is volume of sphere?</h3>

A sphere's volume V is four-thirds of pi times the radius cubed.   A hemisphere's volume is half of the volume of the associated sphere.

The volume of a spherical.

V= 4/3 πr³

radius= 7ft

So,

V= 4/3*22/7*7*7*7

V= 30184/21

V = 1437.33  ft³

Learn more about this concept here:

brainly.com/question/8549007

#SPJ1

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Computing the rate we get:

\frac{300}{10}\frac{cameras}{\text{hour}}=30\frac{cameras}{\text{hour}}

Then, when the assembly line runs at the same rate for 24 hours they will produce:

24\text{hour}\cdot30\frac{cameras}{\text{hour}}=720\text{ cameras}

Answer: Option C) 720.

7 0
1 year ago
Derek buys 4 packages of cheese. Each package weighs 5/8 pounds.
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5/8 x 4 = 5/2 = 2 1/2

answer is <span>D. 2 1/2</span>
7 0
2 years ago
The angles shown are supplementary. What is the value of x ?
almond37 [142]

Answer:

D

Step-by-step explanation:

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6 0
3 years ago
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.

5 0
2 years ago
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