We are given the data on the number of candies handed by neighborhood A and neighborhood B.
Let us first find the mean and variance of each neighborhood.
Mean:


Variance:


A. Null hypothesis:
The null hypothesis is that there is no difference in the mean number of candies handed out by neighborhoods A and B.

Research hypothesis:
The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.

Test statistic (t):
The test statistic of a two-sample t-test is given by

Where sp is the pooled standard deviation given by


So, the test statistic is -1.74
Critical t:
Degree of freedom = N1 + N2 - 2 = 6+6-2 = 10
Level of significance = 0.05
The right-tailed critical value for α = 0.05 and df = 10 is found to be 1.81
Critical t = 1.81
We will reject the null hypothesis because the calculated t-value is less than the critical value.
Interpretation:
This means that we do not have enough evidence to conclude that neighborhood A gives out more candies than neighborhood B.
Answer:
Rs1800
Step-by-step explanation:
20%*6000=1200
3000-1200=1800
Answer:
X=6
Step-by-step explanation:
It is under enlargement and reduction
Therefore we can equates sides because of the parallel lines

Cross multiply

dividing through by 12

simplify

Answer:
70.6% decrease
Step-by-step explanation:
4250-1250 = 3000
4250x = 3000
x=70.6%
70.6% decrease
:]
Answer:
Step-by-step explanation:
You need to find “Y.”
To find “Y”, you do
-3 - 4y = 12
+3 +3
which would be -4y = 15
divide both sides by -4, and you get
y = 15/-4
Then you plug 15/-4 into “Y”, so you get -3 - 4(15/-4) = 12
then multiply, and you get -3 + 15 = 12