To find 20% of 950 you would set it up as a proportion. When doing percentages the prevent is always out of 100 so the first step would be 20/100. You are trying to find a number out of 950 so the second part would be ?/950. Now you want to cross multiply and divide. 20*950=19,000 then you divide it by 100 (your other number) 19,000/100=190. So 20% of 950 is 190.
Answer:
should be - 8
Step-by-step explanation:
-2*-2=4 4*-2=-8
Answer:
0.9256
Step-by-step explanation:
Given that a convenience store owner claims that 55% of the people buying from her store, on a certain day of the week, buy coffee during their visit
Let X be the number of customers who buy from her store, on a certain day of the week, buy coffee during their visit
X is Binomial (35, 0.55)
since each customer is independent of the other and there are two outcomes.
By approximation to normal we find that both np and nq are >5.
So X can be approximated to normal with mean = np = 19.25
and std dev = 
Required probability = prob that fewer than 24 customers in the sample buy coffee during their visit on that certain day of the week
=
(after effecting continuity correction)
= 0.9256
Answer:
Check the explanation
Step-by-step explanation:
One way ANOVA
The null and alternative hypothesis for this one way ANOVA is given as below:
Null hypothesis: H0: There is no significant difference in the averages of the scores for the quizzes, exams and final only.
Alternative hypothesis: There is a significance difference in the averages of the scores for the quizzes, exams and final only.
The ANOVA table with calculations can be seen in the attached images below:
In the attached image below, we get the p-value for this one way ANOVA test as 0.0221. We do not reject the null hypothesis if the p-value is greater than the given level of significance and we reject the null hypothesis if the p-value is less than the given level of significance or alpha value.
In the attached image below, we are given that the p-value = 0.0221 and level of significance or alpha value = 0.05, that is p-value is less than the given level of significance. So, we reject the null hypothesis that there is no significant difference in the averages of the scores for the quizzes, exams and final only. This means we conclude that there is a significance difference in the averages of the scores for the quizzes, exams and final only.