Answer:
Part A = Jillian is not correct, bottom right answer
Part B = The first blank is fraction fs
Step-by-step explanation:
Step-by-step explanation:
i cant do this
Answer:
2 Units
Step-by-step explanation:
THere may be easier ways to do it, but I would just convert them to rectangular and find the distance between those.
(3, -60) becomes (3cos(-60), 3sin(-60)) = (3/2, -3*sqrt(3)/2)
(-5, 105) becomes (-5cos(105), -5sin(105)) = (5/2, -5sqrt(3)/2)
Now just use the distance formula, or in other words think of the two points as being connected by a horizontal and vertical line to form a right triangle. The idea is to find the hypotenuse of that right triangle.
sqrt((3/2 - 5/2)^2 + (-3sqrt(3)/2 + 5sqrt(3)/2)^2)
sqrt((-2/2)^2 + (2*sqrt(3)/2)^2)
sqrt(1 + 3)
sqrt(4)
2
ANSWER:
Reserve rate = 1.03%
Step by step explanation:
Initial amount = $6000
Total sum = $2,00,000
A G.P(geometric progresssion) series is formed with initial value = 6000
Since the depositing cycle continues infinitely, therefore the sum of G.P formula will be taken for infinite values
S= a\1-r
where s= 2,00,000 (sum)
a=6000 (initial amount)
r= ratio (reserve rate)
2,00,000= 6000\1-r
r = 1.03 %
Answer:
Correct option: B. Less between group variability
Step-by-step explanation:
The Analysis of Variance (ANOVA) test is performed to determine whether there is a significant difference between the different group mean.
The hypothesis is defined as:
<em>H₀</em>: There is no difference between the group means, i.e. <em>μ</em>₁ = <em>μ</em>₂ = ... = <em>μ</em>ₙ
<em>Hₐ</em>: At least one of the mean is different from the others, i.e. <em>μ</em>
≠ 0.
The test statistic is defined as:

If the null hypothesis is true then the test statistic will be small and if it is false then the test statistic will be large.
In this case it is provided that the null hypothesis is true.
This implies that:

Implying that the sum of squares for between group variability is less than within group variability.
Thus, if the null hypothesis is true there will be less between group variability.