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2 years ago

The triangular prism shown has triangular _____face(s) and lateral face(s).

Mathematics

1 answer:

dmitriy555 [2]2 years ago

5 0

Answer:

The triangular prism shown has 2 triangular face(s) and 3 lateral face(s).

The area of one triangular face is 7.5 mm²

The surface area of the triangular prism is 135 mm²

Step-by-step explanation:

A triangular face of the prism is the face that has a three sided figure while the lateral faces are faces with four-sided figure. As seen from the triangular prism, it has 2 triangular faces since it is made up of 2 right angled triangles and has 3 lateral faces since it is made up of 3 rectangles which are 4 sided each.

Area of one triangular face = $\frac{1}{2} * base * height$

Given base = 2.5 mm and height = 6 mm

Area of one triangular face = $\frac{1}{2} *2.5 * 6\\$

$= 3*2.5\\= 7.5mm^{2}$

Surface area of triangular prism = bh + pH

b= base of the triangle = 2.5 mm

h- height of the triangle = 6 mm

p= perimeter of the triangle = base + height + slant height = 2.5+6+6.5

p = 15 mm

H= height of the prism = 8mm

Surface area of triangular prism = 2.5(6)+15(8)

Surface area of triangular prism = 15 + 120

Surface area of triangular prism = 135mm²

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Hey there! :D

So, we know that Jane received $9 out of the $30.

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A recipe calls for 5 ounces of vinegar and 10 ounces of oil. Daren is making dressing with 9 ounces of vinegar using the same ra

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Step-by-step explanation:

Set this up as ratio of vinegar to oil in fraction form:

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Answer:

7 is (14x)°

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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