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Paraphin [41]
2 years ago
7

A scoop of ice cream in the shape of a whole sphere sits in a right cone. the radius of the ice cream scoop is 2.5 cm and the ra

dius of the cone is 2.5 cm. How tall must the height of the cone be to fit all the ice cream without spilling if it melts? SHOW ALL YOUR WORK!!!
Mathematics
1 answer:
Alexxandr [17]2 years ago
4 0

Answer:

(At least) 10 centimeters.

Step-by-step explanation:

The radius of the ice cream scoop is 2.5 cm and the radius of the cone is also 2.5 cm.

We want to determine the height of the cone such that it will fit all of the ice cream when it melts without any spilling.

First, we will find the volume of the ice cream scoop. The volume for a sphere is given by:

\displaystyle V=\frac{4}{3}\pi r^3

Since the radius is 2.5 cm, the volume of the full ice cream scoop is:

\displaystyle V=\frac{4}{3}\pi (2.5)^3

Use a calculator:

V\approx 65.4498\text{ cm}^3

The volume of a cone is given by:

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

The radius of the cone is 2.5 cm. Therefore:

\displaystyle V=\frac{1}{3}\pi(2.5)^2h=\frac{1}{3}\pi(6.25 h)=\frac{6.25\pi}{3}h

The volume of the cone should be equal to the volume of the scoop. So:

\displaystyle 65.4498=\frac{6.25\pi}{3}h

Solve for h:

\displaystyle h\approx 65.4498\Big(\frac{3}{6.25\pi}\Big)=10\text{ cm}

The height of the cone should be (at least) 10 cm.

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A baseball player hits a ball at an angle of 52 degrees and at a height of 4.5 ft. if the ball's initial velocity after being hi
Dafna1 [17]

Answer:

t\approx 5.865\,s

Step-by-step explanation:

The motion equations that describe the ball are, respectively:

x = \left[\left(152\,\frac{ft}{s} \right)\cdot \cos 52^{\circ} \right] \cdot t

y = 4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}

The time required for the ball to hit the ground is computed from the second equation. That is to say:

4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{m}{s^{2}} \right) \cdot t^{2} = 0

Given that formula is a second-order polynomial, the roots of the equation are described below:

t_{1}\approx 5.865\,s and t_{2} \approx -0.048\,s

Just the first root offers a realistic solution. Then, t\approx 5.865\,s.

5 0
3 years ago
Please solve it for me!! The answer is 97.3cm
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Answer:

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A man, 1.5m tall, is on top of a building. He observes a car on the road at an angle of 75°. If the building is 30m, how far is
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2 years ago
he owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on w
stellarik [79]

Answer:

90% confidence interval is ( -149.114, -62.666   )

Step-by-step explanation:

Given the data in the question;

Sample 1                                Sample 2

x"₁ = 259.23                            x"₂ = 365.12

s₁  = 34.713                              s₂ = 48.297

n₁ = 5                                       n₂ = 10

With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1

Since we are to assume that variance are equal and they are know, we will use pooled variance;

Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13

Now, pooled estimate of variance will be;

S_p^2 = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

S_p^2 = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

S_p^2 = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [  4 + 9 ]

S_p^2 = [ 4819.9692 + 20993.4 ] / [  13 ]

S_p^2 = 25813.3692 / 13

S_p^2 = 1985.64378

Now the Standard Error will be;

S_{x1-x2 = √[ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

S_{x1-x2 = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

S_{x1-x2 = √[ 397.128756 + 198.564378 ]

S_{x1-x2 = √595.693134

S_{x1-x2 = 24.4068

Critical Value = t_{\frac{\alpha }{2}, df = t_{0.05, df=13 = 1.771  { t-table }

So,

Margin of Error E =  t_{\frac{\alpha }{2}, df × [ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

Margin of Error E = 1.771 × 24.4068

Margin of Error E = 43.224

Point Estimate = x₁ - x₂ = 259.23 - 365.12 = -105.89

So, Limits of 90% CI will be; x₁ - x₂ ± E

Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114

Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

Therefore, 90% confidence interval is ( -149.114, -62.666   )

3 0
3 years ago
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