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Sand is poured onto a surface at 13 cm3/sec, forming a conical pile whose base diameter is always equal to its altitude. How fast is the altitude of the pile increasing when the pile is 1 cm high? Note that the volume of a cone is 13πr2h where r is the radius of the base and h is the height of the cone.

Answers

Answer 1

Answer:

Answer:

Altitude of the pile will increase by 16.56 cm per second.

Step-by-step explanation:

Sand is poured onto a surface at the rate = 13 cm³ per second

Or $(dV)/(dt)=13$

It forms a conical pile with a diameter d cm and height of the pile = h cm

Here d = h

Volume of the pile $V=(1)/(3)* \pi r^(2)h$ cm³per sec.

Since h = d = 2r [r is the radius of the circular base]

r = $(h)/(2)$

$V=(1)/(3)\pi ((h)/(2))^(2)h$

$V=(1)/(3)\pi ((h^(2)))/(4)(h)$

$V=(1)/(12)\pi h^(3)$

$(dV)/(dt)=(1)/(12)\pi * 3(h)^(2)(dh)/(dt)$

$(dV)/(dt)=(1)/(4)\pi * h^(2)* (dh)/(dt)$

Since $(dV)/(dt)=13$ cm³per sec.

13 = $(1)/(4)\pi (1)^(2)(dh)/(dt)$ [For h = 1 cm]

$(dh)/(dt)=(13*4)/(\pi )$

$(dh)/(dt)=(52)/(3.14)$

$(dh)/(dt)=16.56$ cm per second.

Therefore, altitude of the pile will increase by 16.56 cm per second.

Answer 2

Answer:

Final answer:

To solve this problem, we first find the expression for the volume of the cone in terms of the height. We then differentiate this expression to get the relation between the rates of change of the volume and the height. By substituting the given values, we can find the rate of change of the height when the cone is 1 cm high.

Explanation:

The question is related to the application of calculus in Physics, specifically rates of change in the context of real-world problem involving a three dimensional geometric shape - a cone. The student asks how fast the altitude of a pile of sand is increasing at a given time if sand is being poured onto a surface at a constant rate and the pile forms a cone whose base diameter is always equal to its altitude.

We know that the volume of a cone is given by V = (1/3)πr²h, where r is the radius of the base and h is the altitude. Since in this problem the base diameter is always equal to its altitude, we have d = 2r = h, or r = h/2.

Replace r in the volume formula, yielding V = (1/3)π(h/2)²h = (1/12)πh³. Differentiate this expression with respect to time (t) to find the rate of change of V with respect to t, dV/dt = (1/4)πh² * dh/dt.

Given that sand is poured at a constant rate of 13 cm³/sec (that is, dV/dt = 13), we can solve for dh/dt when h = 1cm. Substituting the given values into the equation, 13 = (1/4)π(1)² * dh/dt, we find dh/dt = 13/(π/4) = 52/π cm/sec. Therefore, when the conical pile is 1 cm high, the altitude is increasing at a rate of 52/π cm/sec.

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Using the table of ordered pairs, which equation shows the linear relationship between the x and y values? ---- ----

x|y

-2|2

1|5

3|7

---- ----

A. y = x + 4

B. y = -x

C. y = x - 4

D. y = 4 - x

Answers

Answer:

y = x + 4

Step-by-step explanation:

y = mx + b (m = slope; b = y-intercept)

I used 2 points (-2 , 2) and (1 , 5) to calculate the slope of the line

m= (y₂ - y₁)/(x₂ - x₁)

(5-2)/(1--2) = 3/3= 1

y = 1x + b

using point (1, 5)

5 = 1(1) + b

5 = 1 + b

b = 5-1

b=4

y= 1x + 4 or simply y= x + 4

Answer:

A y = x + 4

yeh i did the test

Step-by-step explanation:

discrete random variable X has the following probability distribution: x 13 18 20 24 27 P ( x ) 0.22 0.25 0.20 0.17 0.16 Compute each of the following quantities. P ( 18 ) . P(X > 18). P(X ≤ 18). The mean μ of X. The variance σ 2 of X. The standard deviation σ of X.

Answers

Answer:

(a) P(X = 18) = 0.25

(b) P(X > 18) = 0.53

(c) P(X ≤ 18) = 0.47

(d) Mean = 19.76

(e) Variance = 22.2824

(f) Standard deviation = 4.7204

Step-by-step explanation:

We are given that discrete random variable X has the following probability distribution:

X P (x) X * P(x) $X^(2)$ $X^(2)$ * P(x)

13 0.22 2.86 169 37.18

18 0.25 4.5 324 81

20 0.20 4 400 80

24 0.17 4.08 576 97.92

27 0.16 4.32 729 116.64

(a) P ( X = 18) = P(x) corresponding to X = 18 i.e. 0.25

Therefore, P(X = 18) = 0.25

(b) P(X > 18) = 1 - P(X = 18) - P(X = 13) = 1 - 0.25 - 0.22 = 0.53

(d) Mean of X, $\mu$ = ∑X * P(x) ÷ ∑P(x) = (2.86 + 4.5 + 4 + 4.08 + 4.32) ÷ 1

= 19.76

(e) Variance of X, $\sigma^(2)$ = ∑ $X^(2)$ * P(x) - $(\sum X * P(x))^(2)$

= 412.74 - $19.76^(2)$ = 22.2824

(f) Standard deviation of X, $\sigma$ = $√(variance)$ = $√(22.2824)$ = 4.7204 .

Final answer:

The probabilities for the given X values are calculated by summing the relevant given probabilities. The mean of X is computed as a weighted average, and the variance and standard deviation are calculated using formula involving the mean and the individual probabilities.

Explanation:

The probability P(18) is given as 0.25 according to the distribution. The probability P(X > 18) is the sum of the probabilities for all x > 18, so we add the probabilities for x=20, x=24, and x=27, giving us 0.20 + 0.17 + 0.16 = 0.53. The probability P(X ≤ 18) includes x=18 and any values less than 18. As 18 is the lowest value given, P(X ≤ 18) is just P(18), or 0.25.

The mean μ of X is the expected value of X, computed as Σ(xP(x)). That gives us (13*0.22) + (18*0.25) + (20*0.20) + (24*0.17) + (27*0.16) = 2.86 + 4.5 + 4 + 4.08 + 4.32 = 19.76.

The variance σ 2 of X is computed as Σ [ (x - μ)^2 * P(x) ]. That gives us [(13-19.76)^2 * 0.22] + [(18-19.76)^2 * 0.25] + [(20-19.76)^2 * 0.20] + [(24-19.76)^2 * 0.17] + [(27-19.76)^2 * 0.16] = 21.61. The standard deviation σ of X is the sqrt(σ^2) = sqrt(21.61) = 4.65.

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The hypotenuse of the triangle shown below is 12 inches. What is the lengthof a side in inches?

Answers

Can you send a picture of the triangle ?

20% of the 35 vendors at a craft show sell jewelry. How many of the vendors sell jewelry? Please HURRYYYY

Answers

Answer:

what's the question tell please

We wish to study the quality of our production line. We take a random sample of 1000 widgets from our line. A quality rating was determined for each of the 1000 widgets, and the average quality rating in the sample was 4. The sample standard deviation was 0.5. Also, 6 of the 1000 widgets were found to be defective.Estimate the average quality rating for widgets from our production line, and include the uncertainty in this estimate in the form, a ± b.

Answers

Answer:

Average quality rating was 4.54+-0.00549

Final answer:

The estimate for the average quality rating from the production line, given this sample, is 4, with a degree of uncertainty expressed by a 95% confidence interval of 4 ± 0.031. The confidence interval represents a range whereby we can be 95% confident that the true mean lies within.

Explanation:

Since you have the average (mean) quality rating and standard deviation from a sample size of 1000 widgets, we can use these statistics to establish an estimate for the entire production line. The estimate of the average quality rating is given as 4. However, to account for the uncertainty of our estimate due to it being based upon a sample rather than the entire population, we use the concept of a confidence interval.

The formula for a confidence interval is mean ± z* (standard deviation/sqrt(n)), where z is a z-score corresponding to our desired level of confidence. For simplicity, we can use a z-score of 1.96 to represent a confidence level of 95%.

Therefore, the uncertainty in this estimate (at 95% confidence) is calculated as:1.96 * (0.5/sqrt(1000)), approximately equal to 0.031. So the confidence interval for the average quality of widgets is 4 ± 0.031.

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Part A How much voltage must be used to accelerate a proton (radius 1.2 ×10−15m) so that it has sufficient energy to just penetrate a silicon nucleus? A silicon nucleus has a charge of +14e, and its radius is about 3.6 ×10−15m. Assume the potential is that for point charges.