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2 years ago

Surface Area of Cylinder=2πrh+2πr2

Mathematics

1 answer:

Alex2 years ago

8 0

Find the surface area when r is 8 inches and h is 8 inches.

$\qquad$ A. 160π in²

$\qquad$ B. 154π in²

$\qquad$ C. 288π in²

$\qquad$ D. 256π in² ☑

We are given –

$\qquad$ ⇢Radius of cylinder , r = 8 inches

$\qquad$ ⇢ Height of cylinder, h = 8 inches.

We are asked to find surface area of the given cylinder.

Formula to find the surface cylinder given by –

$\bf \star \pink{ Surface\: Area_{(Cylinder)} = 2\pi rh +2\pi r^2 }$

Now, Substitute given values –

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2\pi rh +2\pi r^2$

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2 \pi \times 8 \times 8 + 2\pi \times 8^2$

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2 \pi \times 8^2 + 2\pi \times 8^2$

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 2\pi \times 64 +2\pi \times 64$

$\sf \twoheadrightarrow Surface\: Area_{(Cylinder)} =128 \pi + 128\pi$

$\purple{\bf \twoheadrightarrow Surface\: Area_{(Cylinder)} = 256\pi \: in^2 }$

Henceforth,Option D is the correct answer.

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8 cm 4 cm 3cm. 6cm What is the area of the figure in square centimeters?

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Answer:

576 cm sqyared

Step-by-step explanation:

8 x 4 x 3 x 6 = 576

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2 years ago

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a $\frac{1}{4} = 0.25$ probability of getting ir right. So $\pi = 0.25$ . There are five questions, so $n = 5$ .

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

$P = 0.75(0.25) = 0.1875$

There is a 18.75% probability that the first question that she gets right is the second question.

(b) What is the probability that she gets exactly 1 or exactly 2 questions right?

This is: $P = P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P = P(X = 1) + P(X = 2) = 0.3955 + 0.2637 = 0.6592$

There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

(c) What is the probability that she gets the majority of the questions right?

That is the probability that she gets 3, 4 or 5 questions right.

$P = P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 3) = C_{5,3}.(0.25)^{3}.(0.75)^{2} = 0.0879$

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P = P(X = 3) + P(X = 4) + P(X = 5) = 0.0879 + 0.0146 + 0.001 = 0.1035$

There is a 10.35% probability that she gets the majority of the questions right.

6 0

3 years ago

Scores on a certain IQ test are knownto have a mean of 100. A random sample of 43 students attend a series of coaching classes b

NeX [460]

Answer:

Part 1

Type II error

Part 2

No ; is not ; true

Step-by-step explanation:

Data provided in the question

Mean = 100

The Random sample is taken = 43 students

Based on the given information, the conclusion is as follows

Part 1

Since it is mentioned that the classes are successful which is same treated as a null rejection and at the same time it also accepts the alternate hypothesis

Based on this, it is a failure to deny or reject the false null that represents type II error

Part 2

And if the classes are not successful so we can make successful by making type I error and at the same time type II error is not possible

Therefore no type II error is not possible and when the null hypothesis is true the classes are not successful

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3 years ago

If the exponential parent function,

laila [671]

Answer: $f(x) =3^{-x}-4$