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2 years ago

6

These two cylinders are congruent. Cylinder A has a radius of 4 centimeters. Cylinder B has a volume of 176π cubic centimeters.

What is the height of cylinder B?

2 answers:

schepotkina [342]2 years ago

8 0

The correct answer is:

11

h = 11 cm

$|Huntrw6|$

Oduvanchick [21]2 years ago

4 0

It is given that both Cylinders A and B are congruent, thus, Cylinder B also has a radius of 4 centimeters. We now have the volume and the radius, which we will use to get the height of Cylinder B.

r= 4cm
V= 176π cubic cm

From the formula Volume of a cylinder, we can derive a formula to get the height.
From V = πr^2
to h= V/πr^2

h= 176π cubic cm / π(4cm)^2

We can just eliminate the π since it is both in the denominator and numerator.
h= 176 cubic cm / (4cm)^2
h= 176 cubic cm / 16cm^2
h= 11cm

The height of Cylinder B is 11cm.

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Answer:

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Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.

Step-by-step explanation:

Data given

Campus Sample size Mean Population deviation

1 330 33 8

2 310 31 7

$\bar X_{1}=33$ represent the mean for sample 1

$\bar X_{2}=31$ represent the mean for sample 2

$\sigma_{1}=8$ represent the population standard deviation for 1

$\sigma_{2}=7$ represent the population standard deviation for 2

$n_{1}=330$ sample size for the group 1

$n_{2}=310$ sample size for the group 2

$\alpha$ Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$

P value

Since is a one right tailed test the p value would be:

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VikaD [51]

The correct question is:

Consider the initial value problem

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(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

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c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

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a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

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Implies

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2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

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1 year ago