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2 years ago

What is the volume of the cone?

Mathematics

1 answer:

lina2011 [118]2 years ago

6 0

Answer:

The volume of the cone is 75.36 cm³

Step-by-step explanation:

The rule of the volume of a cone is V = $\frac{1}{3}$ π r² h, where

r is the length of the radius of its base

h is the height of the cone

In the given figure

∵ The length of the radius of the cone is 3 cm

∴ r = 3 cm

∵ The length of the height of the cone is 8 cm

∴ h = 8 cm

→ Substitute the values of r and h in the rule of the volume above

∵ π ≅ 3.14

∵ V = $\frac{1}{3}$ (3.14)(3)²(8)

∴ V = 75.36 cm³

∴ The volume of the cone is 75.36 cm³

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The data given below contain the state cigarette taxâ€‹ (in $) for all 20 regions of a country.

g100num [7]

Answer:

(a) $\bar x= 1.583$ -- Population Mean

(b) $s = 1.032$ --- Population standard deviation

Step-by-step explanation:

Given:

Cigarette tax for 20 regions

Solving (a): The population mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

$\sum x = 1.36 + 1.70 + 2.50 + 0.45 + 1.18 + 0.64 + 3.46 + 0.57 + 2.00 + 0.80 + 1.60 +0.98 + 0.36 + 2.24 + 4.35 + 0.62 + 2.70 + 1.78 + 1.53 + 0.84$

$\sum x = 31.66$

$n = 20$

So, we have:

$\bar x= \frac{31.66}{20}$

$\bar x= 1.583$

Solving (b): The population standard deviation

This is calculated as:

$s = \sqrt{\frac{\sum( x - \bar x)^2}{n}$

$\sum (x -\bar x)^2 = (1.36 - 1.583)^2 + (1.70 - 1.583)^2+ (2.50 - 1.583)^2+ (0.45 - 1.583)^2+ (1.18 - 1.583)^2+ (0.64 - 1.583)^2+ (3.46 - 1.583)^2+ (0.57 - 1.583)^2+ (2.00 - 1.583)^2+ (0.80 - 1.583)^2+ (1.60 - 1.583)^2+(0.98 - 1.583)^2+ (0.36 - 1.583)^2+ (2.24 - 1.583)^2+ (4.35 - 1.583)^2+ (0.62 - 1.583)^2+ (2.70 - 1.583)^2+ (1.78 - 1.583)^2+ (1.53 - 1.583)^2+ (0.84- 1.583)^2$

$\sum (x -\bar x)^2 = 21.29222$

So:

$s = \sqrt{\frac{21.2922}{20}$

$s = \sqrt{1.06461}$

$s = 1.032$

Solving (c):

Population mean tells the average amount while the standard deviation represents the spread from the calculated mean

Option (4) is correct

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2 years ago

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2 years ago

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3 years ago

In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n = 50

sample mean $\overline x$ = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

$X \sim N ( \mu \sigma)$

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})$

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})$

$P(X>140) = 1 - P( Z\leq0.42)$

From z tables,

$P(X>140) = 1 - 0.6628$

$P(X>140) = 0.3372$

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

$Y \sim Binomial (np)$

$Y \sim Binomial (50,0.3372)$

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

$P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}$

P(Y ≥ 15) = 0.763

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3 years ago

What is the volume of the cone?​

What is the volume of the cone?