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Umnica [9.8K]
3 years ago
9

Given that ∠XQR = 180° and ∠LQM = 180°, which equation could be used to solve problems involving the relationships between ∠XQM

and ∠RQM?

Mathematics
2 answers:
Levart [38]3 years ago
5 0
They are congruent: angle XQM = angle RQM or measure of this equat to measure of that.
They are straight angles: There measure is 180.
Angle XQM = Angle RQM = 180

Orlov [11]3 years ago
3 0

Answer: (136 − 2a) + (3a + 39) = 180

Step-by-step explanation: Supplementary angles add to 180°

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Factor the expression below using the greatest common factor 6p3- 2p2 - 8p
miss Akunina [59]

Answer:

The Factor of expression using the greatest common factor is 2p(3p^2- p-4)

Step-by-step explanation:

Consider the provide expression.

6p^3- 2p^2 - 8p

We need to Factor the expression using the greatest common factor.

First look at the coefficients of the variable.

The coefficients are the factor of 2.

Variable p is the greatest common factor in the provided expression.

2p\times 3p^2- 2p\times p- 2p\times 4

2p(3p^2- p-4)

Hence, the Factor of expression using the greatest common factor is 2p(3p^2- p-4).

7 0
3 years ago
Y−3=−1/2(x+4) I need help
frozen [14]

Answer: The slope is - 1/2.

Step-by-step explanation:

4 0
3 years ago
The mean incubation time of fertilized eggs is 19 days. Suppose the incubation times are approximately normally distributed with
kotykmax [81]

Answer:

Step-by-step explanation:

Since the incubation times are approximately normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = incubation times of fertilized eggs in days

µ = mean incubation time

σ = standard deviation

From the information given,

µ = 19 days

σ = 1 day

a) For the 20th percentile for incubation times, it means that 20% of the incubation times are below or even equal to 19 days(on the left side). We would determine the z score corresponding to 20%(20/100 = 0.2)

Looking at the normal distribution table, the z score corresponding to the probability value is - 0.84

Therefore,

- 0.84 = (x - 19)/1

x = - 0.84 + 19 = 18.16

b) for the incubation times that make up the middle 97​% of fertilized eggs, the probability is 97% that the incubation times lie below and above 19 days. Thus, we would determine 2 z values. From the normal distribution table, the two z values corresponding to 0.97 are

1.89 and - 1.89

For z = 1.89,

1.89 = (x - 19)/1

x = 1.89 + 19 = 20.89 days

For z = - 1.89,

- 1.89 = (x - 19)/1

x = - 1.89 + 19 = 17.11 days

the incubation times that make up the middle 97​% of fertilized eggs are

17.11 days and 20.89 days

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3 years ago
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charle [14.2K]
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-2(bx-5)=16
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3 years ago
Most _____ jobs are worked 10 to 20 hours per week.
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1. Contractual
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