Answer:
a) 28
b)
Class Limits Class Boundaries Midpoint f Rf Cf
10 - 37 9.5-37.5 23.5 7 0.096 7
38 - 65 37.5-65.5 51.5 25 0.342 32
66 - 93 65.5-93.5 79.5 26 0.356 58
94 - 121 93.5-121.5 107.5 9 0.123 67
122 - 149 121.5-149.5 135.5 5 0.068 72
150 - 177 149.5-177.5 163.5 0 0.0 72
178 - 205 177.5-20.5 191.5 1 0.014 73
Total 73 1.0
c)
The histogram is in attached file.
Step-by-step explanation:
a)
The width of class interval=range/desired number of classes
The width of class interval=(maximum-minimum)/7
The width of class interval=(200-10)/7
The width of class interval=190/7
The width of class interval=27.14=28 (rounded to next whole number)
Part(b) and part (c) is explained in attached word document.
-6/5 is the slope. you start at point (0,0) and then move down 6 units and to the right 5 units and then create one point. from that point, go down 6 units and to the right 5 units and repeat. when you reach the bottom of the graph stop. go back to (0,0) and then move up 6 units and to the left 5 units. connect the points with a ruler and add arrows to each side. label your line with "y=-6/5x" on the line
Answer:
d
Step-by-step explanation:
It is a 1/13 chance but even if you shuffle and replace it, its still a 1/13 chance.
1. 60,30,90 right triangle. y will be hypotenuse/2, x will be
hypotenuse*sqrt(3)/2. So x = 16*sqrt(3)/2 = 8*sqrt(3), approximately 13.85640646
y = 16/2 = 8
2. 45,45,90 right triangle (2 legs are equal length and you have a right angle).
X and Y will be the same length and that will be hypotenuse * sqrt(2)/2. So
x = y = 8*sqrt(2) * sqrt(2)/2 = 8*2/2 = 8
3. Just a right triangle with both legs of known length. Use the Pythagorean theorem
x = sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13
4. Another right triangle with 1 leg and the hypotenuse known. Pythagorean theorem again.
y = sqrt(1000^2 - 600^2) = sqrt(1000000 - 360000) = sqrt(640000) = 800 5. A 45,45,90 right triangle. One leg known. The other leg will have the same length as the known leg and the hypotenuse can be discovered with the Pythagorean theorem. x = 6. y = sqrt(6^2 + 6^2) = sqrt(36+36) = sqrt(72) = sqrt(2 * 36) = 6*sqrt(2), approximately 8.485281374
6. Another 45,45,90 triangle with the hypotenuse known. Both unknown legs will have the same length. And Pythagorean theorem will be helpful.
x = y.
12^2 = x^2 + y^2
12^2 = x^2 + x^2
12^2 = 2x^2
144 = 2x^2
72 = x^2
sqrt(72) = x
6*sqrt(2) = x
x is approximately 8.485281374
7. A 30,60,90 right triangle with the short leg known. The hypotenuse will be twice the length of the short leg and the remaining leg can be determined using the Pythagorean theorem.
y = 11*2 = 22.
x = sqrt(22^2 - 11^2) = sqrt(484 - 121) = sqrt(363) = sqrt(121 * 3) = 11*sqrt(3). Approximately 19.05255888
8. A 30,60,90 right triangle with long leg known. Can either have fact that in that triangle, the legs have the ratio of 1:sqrt(3):2, or you can use the Pythagorean theorem. In this case, I'll use the 1:2 ratio between the unknown leg and the hypotenuse along with the Pythagorean theorem.
x = 2y
y^2 = x^2 - (22.5*sqrt(3))^2
y^2 = (2y)^2 - (22.5*sqrt(3))^2
y^2 = 4y^2 - 1518.75
-3y^2 = - 1518.75
y^2 = 506.25 = 2025/4
y = sqrt(2025/4) = sqrt(2025)/sqrt(4) = 45/2
Therefore:
y = 22.5
x = 2*y = 2*22.5 = 45
9. Just a generic right triangle with 2 known legs. Use the Pythagorean theorem.
x = sqrt(16^2 + 30^2) = sqrt(256 + 900) = sqrt(1156) = 34
10. Another right triangle, another use of the Pythagorean theorem.
x = sqrt(50^2 - 14^2) = sqrt(2500 - 196) = sqrt(2304) = 48
Answer:
it is 6 because there is no other explanation to 1-f f=6!
Step-by-step explanation:
all i did was look 1 - f = 5 then f equaled 6!
