KE = 1/2(m * v^2)
KE = 1/2(1350 * 12^2)
KE = 1/2(1350 * 144)
KE = 1/2(194,400)
KE = 97,200 Joules
Answer:
x=3
Step-by-step explanation:
Given,
1 = 1/(x-2)
Multiply both sides by (x-2),
1*(x-2) = 1/(x-2) * (x-2)
x-2 = 1
Adding 2 on both sides,
x-2+2 = 1+2
x = 3
Answer:
Total amount Tara earned for babysitting for h hours = 8h
Step-by-step explanation:
Amount earned per hour for babysitting = $8.00
Number of hours of babysitting = h hours
Total amount Tara earned for babysitting for h hours =
Amount earned per hour for babysitting × Number of hours of babysitting
= 8 × h
= 8h
Total amount Tara earned for babysitting for h hours = 8h
Unit form: 3 tens
Standard form: 30
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!
