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3 years ago

Find the difference in the volume and total area of a cylinder with both a radius and height of 1.

Mathematics

2 answers:

Naya [18.7K]3 years ago

5 0

I think its 2pi hope it helps

diamong [38]3 years ago

3 0

Answer:

The difference in the volume of cylinder and total surface area of cylinder = $3\pi$ .

The number of sq units of total area exceeds the number of cu.units in the volume $3\pi$ .

Step-by-step explanation:

Given radius of cylinder =1 unit

Height of cylinder= 1 unit

We know that formula of volume of cylinder = $\pi r^{2} h$

Where r= Radius of the cylinder

h= Height of the cylinder

By using this formula we can find the volume of cylinder

volume of cylinder = $\pi \times 1\times 1$ = $\pi cubic units.Formula of total surface area of cylinder: Total surface area of cylinder = [tex]2\pi r(r+h)$

By using this formula we can find total surface area of cylinder

Total surface area of cylinder = $2\pi \times 1(1+1)[/tex} Total surface area of cylinder=[tex]4\pi$ sq units .

Difference between volume of cylinder and total surface area of cylinde= $4\pi -\pi$ = $3\pi$

<h3>Total surface area of cylinder - volume of cylinder= $3\pi$ </h3><h3>Total surface area of cylinder = $3\pi$ + volume of cylinder </h3><h3>Hence, the number of sq units of total surface area exceed the number of cu.units in the volume by $3\pi$ .</h3>

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When we toss a coin, there are two possible outcomes: a head or a tail. Suppose that we toss a coin 100 times. Estimate the appr

marin [14]

Answer:

96.42% probability that the number of tails is between 40 and 60.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

100 tosses, so $n = 100$

Two outcomes, both equally as likely. So $p = \frac{1}{2} = 0.5$

$E(X) = np = 100*0.5 = 50$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.5*0.5} = 5$

Estimate the approximate probability that the number of tails is between 40 and 60.

Using continuity correction.

$P(40 - 0.5 \leq X \leq 60 + 0.5) = P(39.5 \leq X \leq 60.5)$

This is the pvalue of Z when X = 60.5 subtracted by the pvalue of Z when X = 39.5. So

X = 60.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60.5 - 50}{5}$

$Z = 2.1$

$Z = 2.1$ has a pvalue of 0.9821

X = 39.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{39.5 - 50}{5}$

$Z = -2.1$

$Z = -2.1$ has a pvalue of 0.0179

0.9821 - 0.0179 = 0.9642

96.42% probability that the number of tails is between 40 and 60.

8 0

3 years ago

Please help!

Xelga [282]

Answer:

The volume of cone = Maximum holding of glass = 60 cube cm(cm^3)

Step-by-step explanation:

Cone formula:

$v = \frac{1}{3} \pi \: r {}^{2} h$

The radius is 3cm and Volume is 60cm^3.

if pi =~ 3 So the "h" is 6.66 cm.

"h" is distance from cone's head to its base.

The "H"( height of glass) = 6.66 + 7 = 13.6 =~ 14

3 0

3 years ago

Read 2 more answers

Need help on this too

andrezito [222]

Answer:

I think it is distributive

Step-by-step explanation:

When a single number is next to parentheses, it's called distributive property

I hope this is right :/

Please mark brainliest :/

6 0

3 years ago

Evaluate.<br> 1/2 (11)(15 + 17) =

Andrei [34K]

Answer: 176

Step-by-step explanation:

To evaluate the equation we have to remove the parentheses and add 15 + 17...

1/2 (11)(15 + 17) =

= 11 * 32 * 1/2

Now, we have to turn it into a fraction and multiply it...

$\frac{1*11*32 }{2}$

= $\frac{352}{2}$

352 ÷ 2 = 176

I hope this helps!

8 0

3 years ago

Read 2 more answers

Differentiate the function. y = (3x - 1)^5(4-x^4)^5

TiliK225 [7]

Answer:

$\displaystyle y' = -5(3x-1)^4(4 - x^4)^4(15x^4 - 4x^3 - 12)$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

Distributive Property

<u>Algebra I</u>

Terms/Coefficients
Factoring

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

y = (3x - 1)⁵(4 - x⁴)⁵

<u>Step 2: Differentiate</u>

Product Rule: $\displaystyle y' = \frac{d}{dx}[(3x - 1)^5](4 - x^4)^5 + (3x - 1)^5\frac{d}{dx}[(4 - x^4)^5]$
Chain Rule [Basic Power Rule]: $\displaystyle y' =[5(3x - 1)^{5-1} \cdot \frac{d}{dx}[3x - 1]](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^{5-1} \cdot \frac{d}{dx}[(4 - x^4)]]$
Simplify: $\displaystyle y' =[5(3x - 1)^4 \cdot \frac{d}{dx}[3x - 1]](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot \frac{d}{dx}[(4 - x^4)]]$
Basic Power Rule: $\displaystyle y' =[5(3x - 1)^4 \cdot 3x^{1 - 1}](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot -4x^{4-1}]$
Simplify: $\displaystyle y' =[5(3x - 1)^4 \cdot 3](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot -4x^3]$
Multiply: $\displaystyle y' = 15(3x - 1)^4(4 - x^4)^5 - 20x^3(3x - 1)^5(4 - x^4)^4$
Factor: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 3(4 - x^4) - 4x^3(3x - 1) \bigg]$
[Distributive Property] Distribute 3: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 12 - 3x^4 - 4x^3(3x - 1) \bigg]$
[Distributive Property] Distribute -4x³: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 12 - 3x^4 - 12x^4 + 4x^3 \bigg]$
[Brackets] Combine like terms: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4(-15x^4 + 4x^3 + 12)$
Factor: $\displaystyle y' = -5(3x-1)^4(4 - x^4)^4(15x^4 - 4x^3 - 12)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

6 0

3 years ago