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Lisa [10]
3 years ago
13

The cylinders are similar the volume of the larger cylinder is 1600 cubic centimeters what is the volume of the smaller cylinder

Mathematics
1 answer:
cestrela7 [59]3 years ago
8 0

We are given Volume of the larger cylinder is = 1600 cubic centimeters.

Height of the larger cylinder = 16 cm.

We know formula of volume of a cylinder V=\pi r^2h, where r is the radius of cylinder,  h is the height of the cylinder.

Plugging value of V and h of larger cylinder, we get

1600=\pi r^2(16).

Dividing both sides by 16, we get

\frac{1600}{16}=\frac{\pi r^2 (16)}{16}

r^2=100

Taking square root on both sides, we get

r=10.

Therefore, radius of the larger cylinder is 10 cm.

We are given cylinders are similar .

<u>Note: The radii and heights of similiar cylinders are in same ratio.</u>

Therefore, we can setup a proportion:

Let us take radius of small cylinder is x.

\frac{x}{10}=\frac{4}{16}

\frac{x}{10}=\frac{1}{4}

Multiplying both sides by 10, we get

10 \times \frac{x}{10}=10 \times\frac{1}{4}

x=2.50.

Therefore, radius of the small cylinder = 2.5 cm.

Now, plugging radius =2.5 and height  = 4 in the formula of volume the cylinder, we get

V=\pi (2.5)^2(4)=\pi (6.25)(4) =25 \pi \ cm^3.

Therefore, correct option is 25 pi cm^3.

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a coin will be tossed 10 times. Find the chance that there will be exactly 2 heads among the first five tosses and exactly 4 hea
777dan777 [17]

Answer:

The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.

Step-by-step explanation:

To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.

Both heads and tails have an individual probability p=0.5.

Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.

The probability that k heads are in the sample is:

P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{5}{k}\cdot0.5^k\cdot0.5^{5-k}

Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:

P(x=2)=\dbinom{5}{2}\cdot0.5^{2}\cdot0.5^{3}=10\cdot0.25\cdot0.125=0.3125\\\\\\

For the last five tosses, the probability that are exactly 4 heads is:

P(x=4)=\dbinom{5}{4}\cdot0.5^{4}\cdot0.5^{1}=5\cdot0.0625\cdot0.5=0.1563\\\\\\

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P(H_1=2;H_2=4)=P(H_1=2)\cdot P(H_2=4)=0.3125\cdot0.1563=0.0488

7 0
3 years ago
2/3 cup of oatmeal is needed to make 10 granola bars. how many such granola bars can be made with 20 cups of oatmeal? Kindly ple
aleksandrvk [35]
2/3 cup of oatmeal - 10 granola bars

When the number of cups of oatmeal increases, the number of granola bars also increases.
If you use for example twice as many cups of oatmeal, you'll make twice as many granola bars.
Therefore, the first step is to calculate how many times 20 cups of oatmeal is more than 2/3 cup of oatmeal. To do it, divide 20 by 2/3.

20 \div \frac{2}{3} = 20 \times \frac{3}{2}=\frac{60}{2}=30

You have 30 times as many cups of oatmeal, so you'll make 30 times as many granola bars.

10 \times 30= 300

300 granola bars can be made with 20 cups of oatmeal.
5 0
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5w = 23 - 3f and 4f = 12 - 2w
Kryger [21]

Answer:

f = 1, w = 4

Step-by-step explanation:

Given the 2 equations

5w = 23 - 3f → (1)

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Multiplying (1) by 4 and (2) by - 3 and adding the result will eliminate f

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