The period is equal to 2pi/n, where n is the coefficient of t. In this case pi/2. Therefore the period in this example is 4. Frequency is equal to 1/period. Hence the frequency for this problem is 1/4
The procedure is to make the difference of the terms that occupy the same position (column and row):
| - 6 - 4 | | - 5 5 | | - 6 + 5 - 4 - 5 | | -1 - 9 |
| 6 0 | - | - 4 -1 | = | 6 + 4 0 + 1 | = | 10 1 |
| 6 4 | | 6 - 4 | | 6 - 6 4 + 4 | | 0 8 |
Answer: option B.
Answer:
Yes
Step-by-step explanation:
f(x)=3 x-2
y= 3 x-2
x= 3y - 2
3 y -2 = x
3 y -2 + 2 = x + 2
<u>3 y </u> = <u>x </u> + <u>2</u>
3 3 3
<u>3 y </u> = <u>x </u> + <u>2 </u>
3 3 3
y = <u>x </u> + <u>2 </u> replace y with f ^ -1(x)
<em> </em> 3 3
f ^-1 (x) = <u>x</u> + <u>2 </u>
3 3
X = 4/7 = 0.571 I believe so
Answer:
The answer is "0.765 and 0.2353".
Step-by-step explanation:
Please find the complete question in the attached file.
In point a:
P(a substantive term only)
P(major health insurance only) 
P(both)
P(renewal) =P(insurance and renewal term only)+P (substantial and renewable health insurance only)+P (both and renew)
In point b:
In reality, the probability of having both life and major medical insurance provided the policyholder would renew next year
