Answer:1/4
Step-by-step explanation:Probability of first tails: 1/2, probability of second heads: 1/2, 1/2*1/2=1/4
You need the revenue of year 2 to calculate what percent you have to increase it to reach 3 million in year 3.
Let R2 be the revenue of year 2 and R3 the revenue of year 3.
The formula of percent increase is:
% increase = ( R3 - R2 ) / R2 *100
=> % increase = (R3 / R2 - 1) * 100
So, given R3 is 3 million
=> % increase = (3,000,000 / R2 - 1) * 100
So, you just have to replace R2 to get the solution.
Let's do one example:
R2 = 2 million
=> % increase = (3,000,000 / 2,000,000 - 1) * 100 = (0.5) * 100 = 50
So, if the revenue of year 2 was 2 million, he needs to increase it 50% to reach 3 million.
Answer:
-1 is the slope
Step-by-step explanation:
Answer:
The number of students we expect to have an interval that does not contain the true mean value is, ![255\times [\alpha\%]](https://tex.z-dn.net/?f=255%5Ctimes%20%5B%5Calpha%5C%25%5D)
.
Step-by-step explanation:
A [100(1 - α)%] confidence interval for true parameter implies that if 100 confidence intervals are created then [100(1 - α)] of these 100 confidence intervals will consist the true population parameter value.
Here α is the significance level. It is defined as the probability rejecting the claim that the true parameter value is not included in the 100(1 - α)% confidence interval.
It is provided that 255 students create the same confidence interval, correctly.
Then the number of students we expect to have an interval that does not contain the true mean value is,
For instance, if the students are creating a 95% confidence interval for mean then the number of students we expect to have an interval that does not contain the true mean will be:
The significance level is:
Number of students we expect to have an interval that does not contain the true mean will be: ![255\times [\alpha\%]=255\times 0.05=12.75\approx13](https://tex.z-dn.net/?f=255%5Ctimes%20%5B%5Calpha%5C%25%5D%3D255%5Ctimes%200.05%3D12.75%5Capprox13)
Thus, 13 of the 255 confidence intervals will not consist the true mean value.
What is great about this question is that you aren’t required to give numerical approximations of the rate of change over the interval. This means that you have a lot less work to do than you think.
Start by drawing a dot on each y-coordinate of every given x value in the interval table (-13, -10, -5, -2, and 0). Then, for each interval, draw a straight line between the two y-coordinates of the two x-values given in the interval. So, for interval A, you would draw a straight line from f(-13) to f(-10). This line would have a steep downward slope; this means that the average rate of change over this interval is negative, or less than zero. A satisfies the question.
Interval B has a positive slope between the two y-coordinates, so it doesn’t satisfy the question. Interval C has a line with a slope of 0, but does satisfy the question because it’s asking for average changes less than or equal to zero.
The answer for this problem would be intervals A and C, the answer that is selected.
