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denpristay [2]
3 years ago
13

An ice cream shop sells cones at the volume of 94.2 cubic meters they want to double the volume of the cones without changing th

e diameter of the cone so that ice cream scoop will stay on the top of the cold what should the dimensions of the new cone be at the old cone had a height of 10 centimeters
Mathematics
1 answer:
marin [14]3 years ago
5 0
The volume of a cone is V=\pi r^2\frac{h}3 where r = radius and h = height. If the cone has a volume of 94.2 cm³ (I assume you didn't mean m³ because that would be ridiculously huge) and a height of 10 cm, we can plug these values into the formula to find the radius. Don't do any rounding.

94.2 = \pi r^2\frac{10}3 \\ 282.6 = \pi r^2 *10 \\ 28.26 = \pi r^2 \\ 8.99543738355 = r^2 \\ 2.99923946752=r

Now we know that's going to be the radius of our <em>new </em>cone as well since we're keeping the diameter the same. The volume is going to be double 94.2 which is 188.4. Let's solve for the height.

188.4 = \pi (2.99923946752)^2\frac{h}3 \\ 188.4 = \pi(8.99543738355)\frac{h}3 \\ 188.4 = 28.26\frac{h}3 \\ 565.2 = 28.26h \\\\ \boxed{h = 20, r\approx 3}
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Answer:

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Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the average using the finite correction factor is:

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The critical value of <em>z</em> for 95% confidence level is,

<em>z</em> = 1.96

Compute the 95% confidence interval for the average number of years until the first major repair as follows:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}

     =3.3\pm 1.96\times\frac{1.47}{\sqrt{183}}\times\sqrt{\frac{1500-183}{1500-1}}\\\\=3.3\pm 0.19964\\\\=(3.10036, 3.49964)\\\\\approx (3.1, 3.5)

Thus, the 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

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8 0
2 years ago
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[|] Answer [|]

\boxed{5 \ 1/9}

[|] Explanation [|]

Rewrite the equation with parts separated:

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7/9 - 6/9 = 1/9

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\boxed{[|] \ Eclipsed \ [|]}

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Answer:

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Step-by-step explanation:

Hope this helps! :)

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