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2 years ago

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Javier has a model of a cone with a radius of 3 inches and a height of 10 inches. Javier increased his model’s volume by doublin

g its radius and height. What is the volume of the larger cone in cubic inches?

2 answers:

yan [13]2 years ago

8 0

Answer:

240 pi in. ^2

Step-by-step explanation:

Just took the test. Correct me if I'm wrong!

Thank you

Tisa B Sora

statuscvo [17]2 years ago

3 0

Answer:

$240 \pi\ in^{3}$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

z------> the scale factor

x-----> the volume of the larger cone

y-----> the volume of the original cone

$z^{3}=\frac{x}{y}$

In this problem we have

$z=2$ -----> scale factor

substitute

$2^{3}=\frac{x}{y}$

$8=\frac{x}{y}$

$x=8y$

That means-----> The volume of the larger cone is 8 times the volume of the original cone

Find the volume of the original cone

$V=\frac{1}{3}\pi r^{2} h$

we have

$r=3\ in$

$h=10\ in$

substitute

$V=\frac{1}{3}\pi (3^{2})(10)=30 \pi\ in^{3}$

therefore

The volume of the larger cone is equal to

$8*30 \pi\ in^{3}=240 \pi\ in^{3}$

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

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Matt is showing his work in simplifying (6.2 − 1.6) − 4.4 + 7.8. Identify any error in his work or reasoning. Step 1: (6.2 − 1.6

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Answer:

The correct option is (d)-He wrote commutative instead of associative in Step 2.

Step-by-step explanation:

Commutative property: $a+b=b+a$

Associative property: $(a+b)+c=a+(b+c)$

Now consider the provided expression.

(6.2 − 1.6) − 4.4 + 7.8

Step 1

(6.2 − 1.6) − 4.4 + 7.8

Step 2 Use Associative property

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Step 3: Simplify

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Step 4: Use commutative property

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Hence, the wrong step is step 2. He wrote commutative instead of associative in Step 2.

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2 years ago

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Answer:

$3.13$

Step-by-step explanation:

We are given the following in the question:

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Degree of freedom = 67 - 1 = 66

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Chi square critical value for lower tail =

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80% confidence interval:

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Putting values, we get,

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Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

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