#24 please I need help this is due 8 hours from now!
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
This is permutation, since order matters. The formula for us is
₁₈P₅ = which simplifies to
₁₈P₅ =
The factorial of 13 cancels out on the top and bottom leaving you with
₁₈P₅ = 18 × 17 × 16 × 15 × 14
which comes to 1,028,160 ways
Another way to look at it is: the first 5 people of 18 finish and the others you don't care about. Once the first place person is first, there are only 4 of the 18 left to finish in second place. Then there are only 3 left to finish in third place, etc. So if we use that reasoning, we don't even need to use the formula, we can just say
18 * 17 * 16 * 15 * 14 and those are the first 5 people of 18 to finish.
Answer: The answer is 381.85 feet.
Step-by-step explanation: Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.
This situation is framed very nicely in the attached figure, where
BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?
From the right-angled triangle WGB, we have
and from the right-angled triangle WAB, we have'
Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.
Thus, the height of the building across the street is 381.85 feet.
ANSWER:
[a] A linear regression line has an equation of the form Y = a + bX, where X is the explanatory variable and Y is the dependent variable. The slope of the line is b, and a is the intercept (the value of y when x = 0).
[b] In the equation of a straight line (when the equation is written as “y = mx + b”), the slope is the number “m” that is multiplied on the x, and “b” is the y-intercept (that is, the point where the line crosses the vertical y-axis). This useful form of the line equation is sensibly named the “Slope-intercepts form”.
NOTE: See picture attached.
