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Alex Ar [27]
3 years ago
5

Just wanted to say this..

Mathematics
1 answer:
slavikrds [6]3 years ago
7 0
Thank you so much, is it an app or on the internet?
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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 <em>x</em> in the interval 0 ≤ <em>x</em> ≤ 5, the shell at that point has

• radius = 5 - <em>x</em>, which is the distance from <em>x</em> to <em>x</em> = 5

• height = <em>x</em> ² + 2

• thickness = d<em>x</em>

and hence contributes a volume of 2<em>π</em> (5 - <em>x</em>) (<em>x</em> ² + 2) d<em>x</em>.

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?<br><br> 15 + (5 + 7) = (15 + 5) + 7
Ainat [17]

Associative Property !!


hope this helped!!!                                                

3 0
3 years ago
Read 2 more answers
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