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ludmilkaskok [199]
1 year ago
10

Horizontal plane A and vertical plane B intersect at a line. Line n is vertical on plane B and forms a right angle with horizont

al line m on plane A. Line l is diagonal on plane A. Are the lines in the diagram perpendicular, parallel, skew, or none of these? l and m: l and n: m and n:
Mathematics
2 answers:
Greeley [361]1 year ago
5 0

Answer:

L and M - None of these

L and N - Skew

M and N - Perpendicular

Step-by-step explanation:

Mars2501 [29]1 year ago
3 0

Question:

Are the lines in the diagram perpendicular, parallel, skew, or none of these?

Answer:

l and m:

<em>none of these</em>

l and n:

<em>skew</em>

m and n:

<em>perpendicular</em>

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Little Mexican restaurant sells only two kinds of beef burritos: Mucho beef and Mucho Mucho beef. Last week in the restaurant so
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Step-by-step explanation:

16+22=38

So then youll check out the price thats worth each. 231$ So what is 38 divided by 231$? Thats your answer

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Compare the values of the 2s and 5s in 55,220
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The five digit number, 55,220 contains to 5's and two 2's.  The 5's are in the ten thousand and one thousand columns. This means that one five represents 50,000 and the second represents 5, 000.  The two's are in the hundreds and tens columns meaning one 2 represents 200 and the other 2 represents 20. In a direct comparison of these numbers the 5's equal 55,000 and the 2's equal 220.
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2 years ago
If x and y are two positive real numbers such that x 2 +4y 2 =17 and xy =2, then find the value of x- 2y. a. 3 b. 4 c. 8 d. 9
Dafna11 [192]

Answer: The value of x- 2y is a. \pm 3.

Step-by-step explanation:

Given:  x and y are two positive real numbers such that x^2+4y^2=17   and xy= 2 .

Consider (x-2y)^2=x^2-2(x)(2y)+(2y)^2\ \ \ [(a+b)^2=(a^2-2ab+b^2)]

=x^2-4xy+4y^2

=x^2+4y^2-4(xy)

Put  x^2+4y^2=17   and xy= 2 , we get

(x-2y)^2=17-4(2)=17-8=9

\Rightarrow\ (x-2y)^2=9

Taking square root on both sides , we get'

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7 0
1 year ago
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,
Gre4nikov [31]

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.

8 0
2 years ago
What is the scale factor of polygon qrstu to poligon defgh
koban [17]

Answer:

A

Step-by-step explanation:

Since the two polygons are similar, corresponding sides will have a constant ratio. So, the ratio between ST and FG will be the same as the scale factor between the two shapes.

\dfrac{FG}{ST}=\dfrac{15}{36}=\dfrac{5}{12}

Therefore, the correct answer is choice A. Hope this helps!

5 0
2 years ago
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