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3 years ago

How does the volume of an oblique cone change if the height is reduced to 2/3 of its original size and the radius is doubled

Mathematics

1 answer:

spayn [35]3 years ago

3 0

The volume of oblique cone = $\frac{1}{3} \pi r^{2} h$

Now radius is doubled is now it is 2r

And height is reduced to 2/3 of its original size that is 2/3 h

So plugging the values we get volume

New Volume = $\frac{1}{3} \pi (2r)^2 (\frac{2}{3} h)$

= $\frac{1}{3} \pi 4r^2 (\frac{2}{3} h) = \frac{8}{6} \pi r^{2} h =\frac{8}{2} (\frac{1}{3} \pi r^{2} h )$ $=4*(\frac{1}{3} \pi r^{2} h )$

It means the volume becomes 4 times

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Parallelogram ABCD has the angle measures shown. Can you conclude that it is a rhombus, a rectangle, or a square? Explain.

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3 years ago

Read 2 more answers

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Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

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2 years ago

You throw a die twice. What is the probability of throwing a number less than four and then a six?

gogolik [260]

I like the answer is c because it doubles

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2 years ago

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second

Molodets [167]

Given:

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second.

To find:

The equation for the given situation if the sum of the numbers is 77.

Solution:

a. Let the first number in the set is x.

The second is 3 more than the first. So, the second number is $(x+3)$ .

The 3rd number is a square of the second. So, the third number is $(x+3)^2$ .

Therefore, the first, second and third numbers are $x,(x+3),(x+3)^2$ respectively.

b. The sum of the numbers is 77.

First number + Second number + Third number = 77

c. So, the equation in terms of x is:

$x+(x+3)+(x+3)^2=77$

Therefore, the required equation is $x+(x+3)+(x+3)^2=77$ .

d. On simplification, we get

$x+x+3+x^2+6x+9=77$ $[\because (a+b)^2=a^2+2ab+b^2]$

$x^2+(6x+x+x)+(3+9)=77$

$x^2+8x+12=77$

Subtract 77 from both sides.

$x^2+8x+12-77=77-77$

$x^2+8x-65=0$

Therefore, the simplified form of the required equation is $x^2+8x-65=0$ .

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2 years ago

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We know that:
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x=0.08 m/min
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(0.08+1/120)=0.088 m/min

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2 years ago