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3 years ago

If the surface area of a cylinder is 288π cm2 with radius of 8 cm, what is the volume of the cylinder?

Mathematics

1 answer:

Furkat [3]3 years ago

7 0

Answer:

The volume of the cylinder is 640π cm³

Step-by-step explanation:

The formula of the surface area of a cylinder is SA = 2πrh + 2πr², where r is the radius of its base and h is its height

The formula of the volume of a cylinder is V = πr²h

Let us use the formula of the surface area to find the height of the cylinder, then use the formula of the volume to find its volume

∵ The surface area of a cylinder is 288π cm²

∵ Its radius is 8 cm

∵ SA = 2πrh + 2πr²

- Substitute the SA by 288π and r by 8 in the formula

∴ 288π = 2π(8)h + 2π(8)²

∴ 288π = 16πh + 128π

- Subtract 128π from both sides

∴ 160π = 16πh

- Divide both sides by 16π

∴ 10 = h

∴ The height of the cylinder is 10 cm

Now let us find the volume of the cylinder

∵ V = πr²h

- Substitute r by 8 and h by 10

∴ V = π(8)²(10)

∴ V = π(64)(10)

∴ V = 640π

The volume of the cylinder is 640π cm³

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In ten-pin bowling, the highest possible score in a single game is 300.

Natali [406]

Answer:

a) Is it possible for Fred to accomplish this? Yes

b) If it is possible, what score does he need in his next game? 293

Step-by-step explanation:

Step 1

Let us represent the number of games Fred bowled to get an average of 177 = X

His total points scored in that game would be

177 × X

= 177X

To achieve an average of 178 for his next game,

The total number of points he scored was 199, hence that is represented as:

178 = 177X + 199/ X + 1

178(X + 1) = 177X + 199

178X + 178 = 177X + 199

178X - 177X = 199 - 178

X = 21

Hence, Fred bowled 21 games to achieve an average of 178

Therefore, the score he needs to have on the 23rd game is obtained as:

= (23 × 183) - (22 × 178) = 4209 - 3916 = 293

Therefore, it is possible for Fred to raise his average from 178 to 183 in a single game, but he must bowl 293 in his next game to do this.

5 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .