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3 years ago

I NEED HELP PLEASE!! PLEASE IM BEGGING YOU!!

Mathematics

1 answer:

Gennadij [26K]3 years ago

7 0

A.) r=C/2(pi)
b.) 6.0

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Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

According to a polling organization, 24% of adults in a large region consider themselves to be liberal. A survey asked 200 resp

Varvara68 [4.7K]

Answer:

$z=\frac{0.375 -0.24}{\sqrt{\frac{0.24(1-0.24)}{200}}}=4.47$

Now we can calculate the p value based on the alternative hypothesis with this probability:

$p_v =P(z>4.47)=0.00000391$

The p value is very low compared to the significance level of $\alpha=0.05$ then we can reject the null hypothesis and we can conclude that the true proportion of people liberal is higher than 0.24

Step-by-step explanation:

Information given

n=200 represent the random sample taken

X=75 represent the number of people Liberal

$\hat p=\frac{75}{200}=0.375$ estimated proportion of people liberal

$p_o=0.24$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true proportion of adults liberal is higher than 0.24:

Null hypothesis: $p \leq 0.24$

Alternative hypothesis: $p > 0.24$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.375 -0.24}{\sqrt{\frac{0.24(1-0.24)}{200}}}=4.47$

Now we can calculate the p value based on the alternative hypothesis with this probability:

$p_v =P(z>4.47)=0.00000391$

The p value is very low compared to the significance level of $\alpha=0.05$ then we can reject the null hypothesis and we can conclude that the true proportion of people liberal is higher than 0.24

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3 years ago

A school replaced 20% of it computers with new ones . what the total number of computers in the school if 55 computers were repl

sladkih [1.3K]

The total is 275 because 100% divided by 20% is 5 and if 20%= 55 then that means that 55 goes into 100% 5 times so you multiply 55 by 5 to get 275. Hope I helped! Please rate me as the brainliest!

7 0

3 years ago

During the women's basketball championship game, team a scored 33 more points than team

Olin [163]

Team A scored a total of 80,597 points I believe

161,161 + 33 = 161,194
161,194 ÷2 = 80,597
80,597- 33 = 80,564
80,564 + 80,597 = 161,161

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3 years ago

Dean and Tara are photographers. They take pictures of seniors for high school yearbooks. They charge a standard fee for taking

swat32

Yes that is correct. And?

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3 years ago