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12345 [234]
2 years ago
15

Please help!!! I will vote BRAINLIEST What is the value of k?

Mathematics
1 answer:
Natali [406]2 years ago
4 0
<h3>Answer:  k = 10</h3>

======================================================

Work Shown:

Use the remote interior angle theorem.

The interior angles (4k+5) and (6k+10) add to the exterior angle 115. Note the interior angles are not adjacent to the exterior angle. This is where the "remote" aspect comes in.

So,

(4k+5) + (6k+10) = 115

10k + 15 = 115

10k = 115 - 15

10k = 100

k = 100/10

k = 10

We can stop here.

If you want to keep going, then,

4k+5 = 4*10+5 = 45

6k+10 = 6*10+10 = 70

Note how 45+70 = 115 to help confirm our answer.

You might be interested in
The vertex of this parabola is at (-5, -2). When the x-value is -4, the
Advocard [28]

Answer:

The answer to your question is letter A

Step-by-step explanation:

Data

Vertex = (-5, -2)

Point = (-4, 2)

This is a vertical parabola the opens upwards.

The equation is

                             (x - h)² = 4p(y - k)

Substitution

                             (-4 + 5)² = 4p (2 + 2)

Simplification

                                      1² = 4p(4)

                                      1 = 16p

                                      p = 1/16

Equation of the parabola

                          (x + 5)² = 4/16(y + 2)

                          x² + 10x + 25 = 1/4y + 1/2

Conclusion

The coefficient of the squared expression is 1

7 0
3 years ago
The expression 5(2x - 1) - 3x(2x - 1) when factorized fully is equal to​
Nadusha1986 [10]
Here is my work! the answer is (-3x+5)(2x-1)

5 0
2 years ago
The goat population on a farm is decreasing at a rate of 7.25% each year. If there were 75 goats originally, how many goats will
Natasha2012 [34]

Answer:

53

Step-by-step explanation:

x/75=7.25/100

Cross Multiply

100x = 543.75

Simplify

x=5.4375

Multiply By The Number Of Years

5.4375 x 4 = 21.75

Subtract From The Original Number

75-21.75 = 53.25

<h2>Round it down (You cant have a quarter o a sheep. It would be dead)</h2><h2>53</h2>
3 0
3 years ago
Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 45
vovikov84 [41]

Answer:

a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544

b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390

ii. There are no statistical significant evidence that the prep course helped

d. i. The 95% confidence interval for the change in average test scores is  3.47467 < μ₁ - μ₂ < 14.52533

ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

Step-by-step explanation:

The mean of the standardized test = 1,000

The number of students test to which the test is administered = 453 students

The mean score of the sample of students, \bar{x} = 1013

The standard deviation of the sample, s = 108

a. The 95% confidence interval is given as follows;

CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}

At 95% confidence level, z = 1.96, therefore, we have;

CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}

Therefore, we have;

1,022.94559 < μ < 1,003.0544

b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. The parameters of the students taking the test are;

The number of students, n = 503

The number of hours preparation the students are given, t = 3 hours

The average test score of the student, \bar{x} = 1019

The number of test scores of the student, s = 95

At 95% confidence level, z = 1.96, therefore, we have;

The confidence interval, C.I., for the difference in mean is given as follows;

C.I. = \left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore, we have;

C.I. = \left (1013- 1019  \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}

Which gives;

-18.955390 < μ₁ - μ₂ < 6.955390

ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped

d. The given parameters are;

The number of students taking the test = The original 453 students

The average change in the test scores, \bar{x}_{1}- \bar{x}_{2} = 9 points

The standard deviation of the change, Δs = 60 points

Therefore, we have;

C.I. = \bar{x}_{1}- \bar{x}_{2} + 1.96 × Δs/√n

∴ C.I. = 9 ± 1.96 × 60/√(453)

i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533

ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

5 0
2 years ago
Eight less than the product of 4 and a number is equal to 9.​
Len [333]

Answer:

4x - 8 = 9

Step-by-step explanation:

Hope this helps! :)

Please choose me! I was first! :) PLS BRAINLIEST I NEED ONE MORE! :(

5 0
3 years ago
Read 2 more answers
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