Answer:
a) r ^ ¬q
b) p ^ꓥqꓥ^ r
c) r → p
d) p ^ ¬qꓥ^ r
e) (p ^ q) → r
f) r ↔ (q v p)
Step-by-step explanation:
ꓥ^ = AND Conjunction
vꓦ = OR disjunction
¬ = NOT Negation
↔ = Double Implication
→ = implication
Answer:
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the standard deviation for the sample
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean is less than 56, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
Use compound interest formula F=P(1+i)^n twice, one for each deposit and sum the two results.
For the P=$40,000 deposit,
i=10%/2=5% (semi-annual)
number of periods (6 months), n = 6*2 = 12
Future value (at end of year 6),
F = P(1+i)^n = 40,000(1+0.05)^12 = $71834.253
For the P=20000, deposited at the START of the fourth year, which is the same as the end of the third year.
i=5% (semi-annual
n=2*(6-3), n = 6
Future value (at end of year 6)
F=P(1+i)^n = 20000(1+0.05)^6 = 26801.913
Total amount after 6 years
= 71834.253 + 26801.913
=98636.17 (to the nearest cent.)
Answer:
B. Show that the ratios StartFraction U V Over X Y EndFraction and StartFraction W V Over Z Y EndFraction are equivalent, and ∠V ≅ ∠Y.