Answer:
See solutions below
Step-by-step explanation:
a) Given the expressions
x³y and x⁴y³
Find their factors
x³y = (x * x * x * y)
x⁴y³ = (x * x * x * y) * x * y * y
GCF = x*x*x*y
GCF = x³y
The expressions in parenthesis is the GCF. Hence the GCF of x³y and x⁴y³ is x³y
b) For 48m⁶n⁵p⁴ and 20m⁴n⁵p⁶
48m⁶n⁵p⁴ =12 * (4 * m⁴ * n⁵ * p⁴) * m²
20m⁴n⁵p⁶ = 5 * (4 * m⁴ * n⁵ * p⁴) * p²
GCF is the expression common to both terms
GCF = 4 * m⁴ * n⁵ * p⁴
GCF = 4m⁴n⁵p⁴
Answer:
c. Assuming that the mean weights of wolves in the populations are different, the probability of obtaining a test statistic that is greater than 2.771 or less than −2.771 is 0.01.
Step-by-step explanation:
Test if the sample means were different:
This means that we have a two-tailed test.
t=2.771 and a p-value of 0.01.
0.01 is the p-value of t = 2.771 multiplied by 2, as we have a two-tailed test(counting both more than 2.771 and less than -2.771). This represents the probabiility of obtaining a test statistic that is greater than 2.771 or less than −2.771 is 0.01, and the correct answer is given by option c.
I believe it would be 242 cm^2
Answer:
I did for you but I don't know you will understand my HANDWRITING or not .
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Answer:
2/3, 1,2,4
Step-by-step explanation:
