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defon
2 years ago
6

BRAINLIEST VOLUME OF A SPHERE

Mathematics
1 answer:
Rina8888 [55]2 years ago
4 0

Answer:

\displaystyle V = 972 \pi \ ft^3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Geometry</u>

Volume of a Sphere Formula: \displaystyle V = \frac{4}{3} \pi r^3

  • V is volume
  • r is radius

Step-by-step explanation:

<u>Step 1: Define</u>

Radius <em>r</em> = 9 ft

<u>Step 2: Solve for V</u>

  1. Substitute in <em>r</em> [Volume of a Sphere Formula]:                                               \displaystyle V = \frac{4}{3} \pi (9 \ ft)^3
  2. [Volume] Evaluate exponents:                                                                         \displaystyle V = \frac{4}{3} \pi (729 \ ft^3)
  3. [Volume] Multiply:                                                                                             \displaystyle V = 972 \pi \ ft^3
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Answer:

Step-by-step explanation:

1.

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1 year ago
Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.
Vinvika [58]

Answer:

The point estimate for population mean is 8.129 Mpa.

Step-by-step explanation:

We are given the following in the question:

Data on flexural strength(MPa) for concrete beams of a certain type:

11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0

a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured

We use the sample mean, \bar{x} as the point estimate for population mean.

Formula:

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{219.5}{27} = 8.129

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Read 2 more answers
A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye
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Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{x}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

 \sigma_{x}=\sqrt{n}\sigma

The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

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Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

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Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

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