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NemiM [27]
3 years ago
7

A sphere and a cylinder have the same radius and height. The volume of the cylinder is 8 meters cubed. Yolanda found the volume

of the sphere.
A sphere with height h and radius r. A cylinder with height h and radius r.

Her work is shown below.

V = four-thirds (8) cubed. V = four-thirds (512). V = StartFraction 2,048 Over 3 EndFraction meters cubed.

What is Yolanda’s error?
Yolanda should have found the volume by multiplying 8 by Two-thirds.
Yolanda should have found the volume by multiplying 8 by Four-thirds.
Yolanda should have found the volume with the formula V = two-thirds pi (8) cubed.
Yolanda should have found the volume with the formula V = two-thirds (8) cubed.

Mathematics
2 answers:
LenKa [72]3 years ago
7 0

Answer:

Yolanda should have found the volume by multiplying 8 by 2/3.

Step-by-step explanation:

Yuki888 [10]3 years ago
6 0

we know that

the height of cylinder is equal to the diameter of sphere

so

h=2*r------> equation 1

volume of cylinder=pi*r²*h

volume=8 m³

substitute equation 1 in the formula of volume

8=pi*r²*(2*r)------> 8=2*pi*r³----->  equation 2

find the volume of a sphere

volume sphere=(4/3)*pi*r³-----> (2/3)*[2*pi*r³]

substitute the equation 2 in the formula of volume

volume sphere=(2/3)*[8]----> 16/3 m³

the answer is

Yolanda should have found the volume by multiplying 8 by 2/3.

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1/35

Step-by-step explanation:

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Find the value of the variable.
Zielflug [23.3K]

Answer:

x = 8, y = 8\sqrt{2}

Step-by-step explanation:

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3 0
3 years ago
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A manufacturer of processing chips knows that 2\%2%2, percent of its chips are defective in some way. Suppose an inspector rando
kipiarov [429]

The data in the question seems a bit erroneous. I am writing the correct question below:

A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective? Lets break this problem up into smaller pieces to understand the strategy behind solving it.

Answer:

The probability that at least one of the selected chips is defective is 0.0776.

Step-by-step explanation:

The question states that the probability of defective chips is 2% i.e. 0.02. Let p denote the probability of selecting a defective chip so, p = 0.02

An inspector selects 4 chips, which means n=4 and we need to compute the probability that at least one of the selected chips is defective. Let X be the number of defective chips selected. We need to compute P(X≥1) which means either 1, 2, 3 or 4 chips can be defective.

We will use the binomial distribution formula to solve this problem. The formula is:

<u>P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ</u>

where n = total no. of trials

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          x = no. of successful trials

          q = probability of failure = 1-p

we have n=4, p=0.02 and q=1-0.02=0.98.

We need to compute P(X≥1) which is equal to:

P(X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

A shorter method to do this is to use the total probability theorem:

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ (0.02)⁰(0.98)⁴⁻⁰

          = 1 - (0.98)⁴

          = 1 - 0.9224

P(X≥1) = 0.0776

4 0
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Answer:

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