Well,
We can see that there are 8 10's and 3 1's. So we can write 8 10's followed by 3 1's and join them with addition symbols.
83 = 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 1 + 1 + 1
Given:
The statement is "-3 raised to the power 0".
To find:
The value of the given expression.
Solution:
We know that 
raised to the power 
can be written as 
.
Any non zero number raised to the power 0 is always 1. It means,
, where 
.
-3 raised to the power 0 
Therefore, the value of the given statement is 1.
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!
I think it would be 0.23 repeating so basically 0.23232323
Answer:
u = 4.604 , s = 2.903
u' = 23.025 , s' = 6.49
Step-by-step explanation:
Solution:
- We will use the distribution to calculate mean and standard deviation of random variable X.
- Mean = u = E ( X ) = Sum ( X*p(x) )
u = 1*0.229 + 2*0.113 + 3*0.114 + 4*0.076 + 5*0.052 + 6*0.027 + 7*0.031 + 8*0.358.
u = 4.604
- Standard deviation s = sqrt ( Var ( X ) = sqrt ( E ( X^2) + [E(X)]^2
s = sqrt [ 1*0.229 + 4*0.113 + 9*0.114 + 16*0.076 + 25*0.052 + 36*0.027 + 49*0.031 + 64*0.358 - 4.604^2 ]
s = sqrt ( 8.429184 )
s = 2.903
- We will use properties of E ( X ) and Var ( X ) as follows:
- Mean = u' = E (Rate*X) = Rate*E(X) = $5*u =
u' = $5*4.605
u' = 23.025
- standard deviation = s' = sqrt (Var (Rate*X) ) = sqrt(Rate)*sqrt(Var(X)) = sqrt($5)*s =
s' = sqrt($5)*2.903
u' = 6.49
