Answer:
it might take 11 hours
Step-by-step explanation:
i think
So you got
z²+y²=25² and
17²+x²=z²
x²+6²=y²
nice, so we have z² and y² all in terms of x
17²+x²+x²+6²=25²
2x²+289+36=625
2x²=300
x²=150
x=5√6
if we input back to find y and z
17²+x²=z², 287+150=z², √439 =z
x²+6²=y², 150+36=y², √186=y (not simplifying because no graphing calculator, so just eimplify yourself)
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