Just wanted to say this..

Use the Shell Method to find the volume of the solid obtained by rotating region under the graph of f(x)=x^2+2 for 0≤x≤5 about x

Irina18 [472]

For each x in the interval 0 ≤ x ≤ 5, the shell at that point has

• radius = 5 - x, which is the distance from x to x = 5

• height = x ² + 2

• thickness = dx

and hence contributes a volume of 2π (5 - x) (x ² + 2) dx.

Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:

$\displaystyle 2\pi \int_0^5 (5-x)(x^2+2)\,\mathrm dx=2\pi\int_0^5 (10-2x+5x^2-x^3)\,\mathrm dx=\boxed{\frac{925\pi}6}$

4 0

3 years ago

Justin has $500 in a savings account at the beginning of the summer. He wants to have at least $300 in the account by the end of

SSSSS [86.1K]

Answer:

10 weeks

Step-by-step explanation:

1 because 20 times 10 is 200 and if he starts with 500 then he can withdraw 20 dollars a week for 10 weeks to have 300 left by the end of the summer, hope this helps :)

3 0

3 years ago

How do I factor 2x^3−18x and x ^3+6x^2+8x

Nana76 [90]

Answer:

2x(x − 3)(x + 3)

x(x + 4)(x + 2)

Step-by-step explanation:

2x³ − 18x

x(2x² − 18)

2x(x² − 9)

2x(x − 3)(x + 3)

x³ + 6x² + 8x

x(x² + 6x + 8)

x(x + 4)(x + 2)

4 0

3 years ago

Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat

stiks02 [169]

Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, $\overline x _1$ = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, $\overline x _2$ = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

$\left (\bar{x}_{2}- \bar{x}_{1} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

$t_{(0.025, \, 18)}$ = 1.734

$C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}$

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

$t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}$

$t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147$

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

4 0

3 years ago

Which property is illustrated? 15 + (5 + 7) = (15 + 5) + 7

Ainat [17]

Associative Property !!

hope this helped!!!

3 0

3 years ago

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