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1 year ago

What must be the value of x so that lines a and b are parallel lines cut by transversal f?

Mathematics

2 answers:

ExtremeBDS [4]1 year ago

5 0

we know that

if the lines a and b are parallel lines cut by transversal f

then

$(6x-36)=96$ --------> by alternate exterior angles

solve for x

$(6x-36)=96\\ 6x=96+36\\6x=132\\x=(132/6)\\x=22\ degrees$

therefore

<u>the answer is</u>

the value of x is $22\ degrees$

chubhunter [2.5K]1 year ago

4 0

The answer is 22 i did it on ed
and i got it correct

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A study investigated the job satisfaction of teachers allowed to choose supplementary curriculum for their classes versus teache

ludmilkaskok [199]

The answer is b. the data shows that the authors connot make a determination either way with this data.

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1 year ago

In a statewide soccer competition, 36 games were played. Each team participating in the competition played once with every other

Tomtit [17]

Answer:

9 teams

Step-by-step explanation:

If the total games played was 36 and no team played each other twice, we need to ensure there isn't any double counting.

36 = (n-1) + (n-2) + (n-3) ... + (n-(n-1))

using this knowledge, we can then count up:

1+2+3+4+5+6+7+8 = 36

If our highest number is 8, then we know there must be 9 teams, because no team can play themselves.

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2 years ago

Read 2 more answers

A quality control technician works in a factory that produces computer monitors. Each day, she randomly selects monitors and tes

jeka94

Answer:

There is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 300

p = 5% = 0.05

Alpha, α = 0.05

Number of dead pixels , x = 24

First, we design the null and the alternate hypothesis

$H_{0}: p = 0.05\\H_A: p > 0.05$

This is a one-tailed(right) test.

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{24}{300} = 0.08$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.08-0.05}{\sqrt{\frac{0.05(1-0.05)}{300}}} = 2.384$

Now, we calculate the p-value from excel.

P-value = 0.00856

Since the p-value is smaller than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

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1 year ago

A. identify the parent function of the given graph.

Paha777 [63]

The parent function is f(x) = x^3

The domain are all x values (-infinity, infinity)

The range are all y values (-infinity, infinity)

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2 years ago

Read 2 more answers

From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def

jasenka [17]

Answer:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X 0 1 2 3

P(X) 0.92 0.03 0.03 0.02

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

In order to find the variance we need to find first the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

8 0

2 years ago