Answer:
y=1/2x-1
Step-by-step explanation:
Answer:
discrete, the class boundaries are;
16 - 25
26 - 35
36 - 45
46 - 55
56 - 65
66 = 75
continuous, the class boundaries are;
16 to less than 26
26 to less than 36
36 to less than 46
46 to less than 56
56 to less than 66
66 to less than 76
Step-by-step explanation:
Given that;
Range of the dataset = maximum value - minimum value = 74 - 16 = 58
total number of data set n = 50
Rule : ≥ n
substitute value of n
≥ 50
k × log(2) = log( 50 )
k × 0.3010 = 1.69897
k = 1.69897 / 0.3010
k = 5.64
Hence, we have a total of 6 number of classes needed to be constructed.
so,
class width = 74 - 16 / 6 = 58 / 6 = 9.67
If the Data is discrete, the class boundaries are;
16 - 25
26 - 35
36 - 45
46 - 55
56 - 65
66 = 75
If the Data is continuous, the class boundaries are;
16 to less than 26
26 to less than 36
36 to less than 46
46 to less than 56
56 to less than 66
66 to less than 76
Answer:
x = 11/2
Step-by-step explanation:
→Take the LCM
<h3>Hope it helps you ツ</h3>
1. first parabola is up, so a are going to be positive
y = a(x-r)(x-s),
roots -7,-3, so we can write y=a(x+7)(x+3) [1]
we can take point (-5,-2), substitute into equation [1], and find value of a
-2=a(-5+7)(-5+3)
-2=a*2*(-2)
a=1/2
so first parabola has an equation
y=1/2(x+7)(x+3)
2. second parabola
y=a(x-r)(x-s) roots are 2 and 4 so we ca write
y=a(x-2)(x-4) take point (3,4) and substitute into this equation
4 =a(3-2)(3-4)
4=-a
a=-4, and it will look down
y=-4(x-2)(x-4)
I wanna say it is C, a translation of five units to the right.