6*6-(3*3)-(2*2) is the answer
Set h to 640 and solve for t:
640 = -490t^2 + 1120t
Subtract 640 from both sides:
-490t^2 + 1120t - 640 = 0
The formula to solve a quadratic equation is:
x = -b -/+ sqrtroot (b^2-4ac)/(2a) where a = -490, b = 1120 and c = -640
Solve:
x = -1120 -/+ sqrtroot (1120^2-4(-490)(-640) )/ 2(-490)
x = 8/7 = 1.1428 = 1.14
Time was 1.14 seconds
Answer:
The total number of sites are 15 .
Step-by-step explanation:
As shown in question.
The number of sites in 2 shield darter = 3
The number of sites in 3 shield darter = 1
The number of sites in 4 shield darter = 4
The number of sites in 6 shield darter = 2
The number of sites in 7 shield darter = 3
The number of sites in 8 shield darter = 1
The number of sites in 9 shield darter = 1
Thus
Total number of sites = Number of sites in 2 shield darter + Number of sites in 3 shield darter + Number of sites in 4 shield darter + Number of sites in 6 shield darter + Number of sites in 7 shield darter + Number of sites in 8 shield darter + Number of sites in 9 shield darter
Putting the values in above
= 3 + 1 + 4 + 2 + 3 + 1 +1
= 15
Therefore the total number of sites are 15 .
Answer:
c. 0.1151
Step-by-step explanation:
d) the probability that, on average, fish are larger than 86 centimeters in length.The area under part of a normal probability curve is directly proportional to probability and the value is calculated as
z = (x₁−x) /σ
where z = propability of normal curve
x₁ = variate mean = 86cm
x = mean of 80cm
σ = standard deviation = 5cm
applying the formula,
z= (86-80)/5
z = 6/5 =1.2
Using a table of partial areas beneath the standardized normal curve (see Table of normal curve, a z-value of 1.2 corresponds to an area of 0.3849 between the mean value. but, because the standard curve has 0.5, then will minus 0.3849 from 0.5= 0.5 - 0.3849 = 0.1151
Thus the probability of a fish are larger than 86 centimeters in length is 0.1151
