Answer:
36%
Step-by-step explanation:
the <em>complement </em>of it raining on at least one of these days = it not raining on any day
this means that if it does not rain on at least one day, it does not rain on any day
thus,
1 - probability of it not raining at all = probability it rains at least one day
probability it doesn't rain on monday = 1- 20% = 80%
probability it doesn't rain on tuesday = 1 - 20% = 80%
probability it doesn't rain on monday AND it doesn't rain on tuesday = probability it doesn't rain at all
P(A and B) = P(A) * P(B) = 0.8 * 0.8 = 0.64
1 - 0.64 = 0.36 as our answer
Answer:
You add a bar till the 2 for the 9-10. You add bars till 6 for the 11- 12. You add bars till 8 for the 13-14. You add bars till the 4 for the 15-16.
Step-by-step explanation:
Answer:
the answer is 80
Step-by-step explanation:
104-24
80
Answer:
Step-by-step explanation:
Hello!
The chemistry instructor tested the hypothesis that the proportion of students that passed the introductory chemistry class is better with an embedded. If the known proportion for this population is 65%, the tested hypothesis is:
H₀: p=0.65
H₁: p>0.65
The calculated statistic is Z=2.52 and the associated p-value: 0.0059
Remember:
The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
In this case:
P(Z≥2.52)=0.0059
There is no significance level, the most common one is α: 0.05 so I'll use it as an example.
To make a decision using the p-value you have to compare it to the α.
If p- value>α then you support the null hypothesis (In this case, you can say that there is no change in the proportion of students that passed the introductory chemistry class with an embedded tutor.)
If p-value≤α your decision will be to reject the null hypothesis (In this case, there is significant evidence to say that there is an improvement in the success rate of the introductory chemistry class with an embedded tutor?
Since the p-value:0.0059 is less than the significance level 0.05, you will decide to reject the null hypothesis.
I hope you have a SUPER day!
