Answer:
Yes, we can assume that the percent of female athletes graduating from the University of Colorado is less than 67%.
Step-by-step explanation:
We need to find p-value first:
z statistic = (p⁻ - p0) / √[p0 x (1 - p0) / n]
p⁻ = X / n = 21 / 38 = 0.5526316
the alternate hypothesis states that p-value must be under the normal curve, i.e. the percent of female athletes graduating remains at 67%
H1: p < 0.67
z = (0.5526316 - 0.67) / √[0.67 x (1 - 0.67) / 38] = -0.1173684 / 0.076278575
z = -1.538681
using a p-value calculator for z = -1.538681, confidence level of 5%
p-value = .062024, not significant
Since p-value is not significant, we must reject the alternate hypothesis and retain the null hypothesis.
Answer:
y = -1/2x +11/2
Step-by-step explanation:
slope finding: -5+6/1-3 = -1/2
equation making: y+5 = -1/2(x-1)
rearranging: y = -1/2x +11/2
Answer:
I believe the answer is b
Step-by-step explanation: I think the answer is b because he gets $5 every week and he already has $25 the only thing we dont know is the weeks and you multiply 5 by the weeks so, 5(how much he gets each week) times x (The amount of weeks) plus 25(the amount he already has) equals y(the amount of everything put together)
Wait is it not at all 14.5 or 14 1/2 Because with the math I did I got 14 1/2
Answer with Step-by-step explanation:
We are given that
and 
are linearly independent.
By definition of linear independent there exits three scalar 
and 
such that
Where 
We have to prove that 
and 
are linearly independent.
Let 
and 
such that
...(1)
..(2)
..(3)
Because and are linearly independent.
From equation (1) and (3)
...(4)
Adding equation (2) and (4)
From equation (1) and (2)
Hence, and area linearly independent.
