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2 years ago

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A bowl is the shape of a hemisphere and made out of thick clay. The diameter of the outside bowl is 44.6 cm. And the diameter of

the inside bowl is 41.6 cm.
a) If the bowl is filled up to the top, find how much soup could it hold?
b) Calculate the volume of clay that is needed to make a single bowl?
c) Six bowls are to be packed in a box side by side. The boxes dimensions are 22.5 cm by 90 cm by 135 cm. Sawdust packing is put in the box. What volume of the sawdust packing is needed?

1 answer:

aleksandr82 [10.1K]2 years ago

3 0

Answer:

Step-by-step explanation:

Ive already explained this somebody else asked if you find it you will have the answer

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5(n+2)-8=2n<br> what s the solution for x??

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5n+10-8=2n
5n+2=2n
-3n=2
n=-2/3

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2 years ago

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The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

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Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

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2 years ago

Suppose the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations. Th

madreJ [45]

Answer:

<em>H₀</em>: <em>μ</em>₁ = <em>μ</em>₂ vs, <em>Hₐ</em>: <em>μ</em>₁ > <em>μ</em>₂.

Step-by-step explanation:

A two-sample <em>z</em>-test can be performed to determine whether the claim made by the owner of pier 1 is correct or not.

It is provided that the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations.

The hypothesis to test whether the average weights of the fish in pier 1 is more than pier 2 is as follows:

<em>H₀</em>: The weights of fish in pier 1 is same as the weights of fish in pier 2, i.e. <em>μ</em>₁ = <em>μ</em>₂.

<em>Hₐ</em>: The weights of fish in pier 1 is greater than the weights of fish in pier 2, i.e. <em>μ</em>₁ > <em>μ</em>₂.

The significance level of the test is:

<em>α</em> = 0.05.

The test is defined as:

$z=\frac{(\bar x_{1}-\bar x_{2})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$

The decision rule for the test is:

If the <em>p</em>-value of the test is less than the significance level of 0.05 then the null hypothesis will be rejected and vice-versa.

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2 years ago

Which ordered pair is generated from the equation shown below? y = 3x - 1 A. (6, 18) B. (6, 17) C. (4, 12) D. (9, 8)

geniusboy [140]

Answer:

B 6,17

Step-by-step explanation: If you plug in the value of x from each of the choices only b gives the correct value of y. If you put 6 in the place of x in y= 3x-1 then y= 18-1 which is 17, and it does say 17 in 6,17.

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2 years ago

The volume of a barrel of oil is calculated as 257,166 cm³.

Firdavs [7]

What is the digit at the hundred thousands place? It is 2. Is the digit to the right of it less than 5? No, 5 is 5. So replace the digits to the right of 2 with 0, and change the 2 to 3.
Answer: 300,000

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2 years ago

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