Answer:
y=41
Step-by-step explanation:
Let us start by writing an equation. <em>4 plus y divided by 5 equals to 9</em> :
Great! Now let us solve for the variable, <em>y</em>, by isolating the terms:
<em>I hope this helps! Let me know if you have any questions :)</em>
Answer:
Take a look of the image below, we can think on this problem as a problem of two triangle rectangles.
We can see that both triangles share the adjacent cathetus, then the height of the flagpole is just the difference between the opposite cathetus.
Remember the relation:
Tan(a) = (opposite cathetus)/(adjacent cathetus)
So, if we define H as the height of the cliff
X as the distance between the observer and the cliff
and h as the height of the flagopole
we can write:
tan(40°) = H/X
tan(45°) = (H + h)/X
Notice that we have two equations and 3 variables (we should have the same number of equations than variables) then here is missing information, and we can't get an exact solution for the height of the flagpole.
But we can write it in terms of the height of the cliff H, or in terms of the distance between the observer and the cliff.
We want to find the value of h.
If we take the quotient between both equations, we get:
Tan(45°)/Tan(40°) = (H + h)/H
1.192 = (H + h)/H
1.192*H = H + h
1.192*H - H = h
0.192*H = h
So the height of the flagpole is 0.192 times the height of the cliff.
Answer: a. 1.981 < μ < 2.18
b. Yes.
Step-by-step explanation:
A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.
First, we calculate mean of the sample:
2.08
Now, we estimate standard deviation:
s = 0.1564
For t-score, we need to determine degree of freedom and :
df = 12 - 1
df = 11
= 1 - 0.95
α = 0.05
0.025
Then, t-score is
= 2.201
The interval will be
±
2.08 ±
2.08 ± 0.099
The 95% two-sided CI on the mean is 1.981 < μ < 2.18.
B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.