'One third' as a decimal is 1 ÷ 3 = 0.333333..... ⇒ The decimal is recurring
'One fifth' as decimal is 1 ÷ 5 = 0.2 ⇒ The decimal terminates at the tenth value
'One seventh' as decimal 1 ÷ 7 = 0.142857142857.... ⇒ The decimal is recurring for every digit between 1 and 7
'One ninth' as decimal 1 ÷ 9 = 0.111111.... ⇒ The decimal is recurring
Answer: one fifth
Answer:
y-1=-2/3(x+9)
Step-by-step explanation:
y-1=-2/3(x-(-9))
y-1=-2/3(x+9)
Answer:
3x-6=9 is your equation x=5
Step-by-step explanation:
Step 1: Add 6 to both sides
Step 2: Divide both sides by 3
Complete question :
It is estimated 28% of all adults in United States invest in stocks and that 85% of U.S. adults have investments in fixed income instruments (savings accounts, bonds, etc.). It is also estimated that 26% of U.S. adults have investments in both stocks and fixed income instruments. (a) What is the probability that a randomly chosen stock investor also invests in fixed income instruments? Round your answer to decimal places. (b) What is the probability that a randomly chosen U.S. adult invests in stocks, given that s/he invests in fixed income instruments?
Answer:
0.929 ; 0.306
Step-by-step explanation:
Using the information:
P(stock) = P(s) = 28% = 0.28
P(fixed income) = P(f) = 0.85
P(stock and fixed income) = p(SnF) = 26%
a) What is the probability that a randomly chosen stock investor also invests in fixed income instruments? Round your answer to decimal places.
P(F|S) = p(FnS) / p(s)
= 0.26 / 0.28
= 0.9285
= 0.929
(b) What is the probability that a randomly chosen U.S. adult invests in stocks, given that s/he invests in fixed income instruments?
P(s|f) = p(SnF) / p(f)
P(S|F) = 0.26 / 0.85 = 0.3058823
P(S¦F) = 0.306 (to 3 decimal places)
Answer:
For First Solution: 
is the solution of equation y''-y=0.
For 2nd Solution:
is the solution of equation y''-y=0.
Step-by-step explanation:
For First Solution:
In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.
First order derivative:
2nd order Derivative:
Put Them in equation y''-y=0
e^t-e^t=0
0=0
Hence is the solution of equation y''-y=0.
For 2nd Solution:
In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.
First order derivative:
2nd order Derivative:
Put Them in equation y''-y=0
cosht-cosht=0
0=0
Hence is the solution of equation y''-y=0.
