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2 years ago

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:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

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

V is volume
r is radius

Step-by-step explanation:

Step 1: Define

Radius r = 9 ft

Step 2: Solve for V

Substitute in r [Volume of a Sphere Formula]: $\displaystyle V = \frac{4}{3} \pi (9 \ ft)^3$
[Volume] Evaluate exponents: $\displaystyle V = \frac{4}{3} \pi (729 \ ft^3)$
[Volume] Multiply: $\displaystyle V = 972 \pi \ ft^3$

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Answer:

Step-by-step explanation:

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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:

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

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

Thus, the point estimate for population mean is 8.129 Mpa.

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3 years ago

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3 years ago

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

Katena32 [7]

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 μ and standard deviation σ and we select appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, 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:

μ = $970

σ = $129

n = 102

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

Then,

$\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)$