Answer:
0.47 and 11.53
Step-by-step explanation:
h = 60t − 5t²
27 = 60t − 5t²
5t² − 60t + 27 = 0
Quadratic formula:
x = [ -b ± √(b² − 4ac) ] / 2a
t = [ -(-60) ± √((-60)² − 4(5)(27)) ] / 2(5)
t = (60 ± √3060) / 10
t = 0.47 or 11.53
Can't help if I don't know what the question is
1. false because the GCF of 101 and 102 is 1 whereas the GCF of 33 and 99 is 33.
2. idk
3. false because the GCF of 1 and 2 is 1
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 152.5
For the alternative hypothesis,
µ ≠ 152.5
This is a two tailed test.
Since no population standard deviation is given, the distribution is a student's t.
Since n = 231
Degrees of freedom, df = n - 1 = 231 - 1 = 230
t = (x - µ)/(s/√n)
Where
x = sample mean = 148.9
µ = population mean = 152.5
s = samples standard deviation = 27.4
t = (148.9 - 152.5)/(27.4/√231) = - 2
We would determine the p value using the t test calculator. It becomes
p = 0.047
Since alpha, 0.05 > thanthere sufficient evidence to conclude that the self-efficacy of adults who have experienced childhood trauma differs from that in the general population of individuals the p value, 0.047, then we would reject the null hypothesis. Therefore, At a 5% level of significance, there is sufficient evidence to conclude that the self-efficacy of adults who have experienced childhood trauma differs from that in the general population of individuals
Answer:
a= -3, b= 4 and c= -5
ab + c = (-3)4+(-5) = -12 -5 = -17
