If BD bisects angle ABC, m angle DBC = 79° , and m angle ABC = (9x

Lazar drives to work every day and passes two independently operated traffic lights. The probability that both lights are green

Natalija [7]

Conditional probability is a measure of probability of an event given that another event has occurred.
P ( A\ B ) = P ( A ∩ B ) / P ( B ) - the conditional probability of A given B
P ( A ∩ B ) = 0.41; P ( B ) = 0.59
P ( A \ B ) = 0.41 / 0.59 = 0.6949152 ≈ 0.69
Answer: B ) 0.69

7 0

2 years ago

Read 2 more answers

The two-way frequency table below shows the preferred communication method of employees at a company, based on years of employme

Anna35 [415]

Answer:

The percentage of employees with 8 or more years at the company that reported that email is their preferred method of communication is 30.48%. I suppose there was a small typing mistake in option B.

Step-by-step explanation:

The proportion is the number of desired outcomes divided by the number of total outcomes.

The percentage is the proportion multiplied by 100.

In this question:

210 employees with 8 of more year.

Of those, 64 have the email as their preferred method of communication.

64/210 = 0.3048

0.3048*100 = 30.48%

The percentage of employees with 8 or more years at the company that reported that email is their preferred method of communication is 30.48%. I suppose there was a small typing mistake in option B.

3 0

2 years ago

Read 2 more answers

A teacher promised a movie day to the class that did better, on average, on their test. The box plot shows the results of the te

masha68 [24]

I say it is C. Basically I just eliminate the potential answered down. A could not be the one since not all student in 2nd period got 100% and average students got below 95%. B cannot be it since both box plot have the same median. I do not think D is it because of how the answer is told. "The 4th period class should get the reward. Their lowest score is an outlier, and should be thrown out," it sound childish and makeing a joke to put "and should be thrown out." I may be wrong but that is my opinion. The relatively the best and reasonable answer is C.
<span />

4 0

2 years ago

Read 2 more answers

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

A school pays $1,825 for 150 shirts. This includes the $25 flat-rate shipping cost.

yKpoI14uk [10]

Answer:

a)

We know that the total cost for the 150 shirts (plus a flat-rate of $25) is $1,825.

First, let's find how much costs each shirt.

First, the total cost of the shirts alone (neglecting the flat-rate) is:

$1,825 - $25 = $1,800

Then each shirt costs equal to the quotient of the cost and the number of shirts:

$1,800/150 = $12

Each shirt costs $12.

Then if you order T shirts, the total cost will be the flat-rate of $25 plus T times $12. The equation is:

C(T) = $25 + T*$12

b) C represents the cost

T represents the number of shirts ordered.

c) The initial value is the constant value, in this case, is $25

The rate of change is the coefficient that multiplies the independent variable (T), in the equation the rate of change is $12.

7 0

2 years ago

If BD bisects angle ABC, m angle DBC = 79° , and m angle ABC = (9x - 4)° , find the value of x. ​

If BD bisects angle ABC, m angle DBC = 79° , and m angle ABC = (9x - 4)° , find the value of x.