Answer:
Make sure you pass go!!!!!!!
Step-by-step explanation:
There is no solution since they are both adding $10 to their accounts every week.
Answer:
Margin of Error = 5.4088 ;
Confidence interval = (30.1 ; 40.9)
Interval estimate are almost the same
Step-by-step explanation:
Given that :
Population standard deviation, σ = 9.3
Sample size, n = 8
Xbar = 35.5
Confidence level = 90%
The confidence interval:
Xbar ± Margin of error
Margin of Error = Zcritical * σ/sqrt(n)
Zcritical at 90% = 1.645
Margin of Error = 1.645 * 9.3/sqrt(8) = 5.4088
Confidence interval :
Xbar ± Margin of error
35.5 ± 5.4088
Lower boundary = (35.5 - 5.4088) = 30.0912 = 30.1
Upper boundary = (35.5 + 5.4088) = 40.9088 = 40.9
(30.1 ; 40.9)
T distribution =. (30.5 ; 40.5)
Normal distribution = (30.1, 40.9)
9514 1404 393
Answer:
$1790.99
Step-by-step explanation:
<u>Given</u>:
$1625 is invested at an annual rate of 1.95% compounded quarterly for 5 years
<u>Find</u>:
the ending balance
<u>Solution</u>:
The compound interest formula applies.
FV = P(1 +r/n)^(nt) . . . Principal P at rate r for t years, compounded n per year
FV = $1625(1 +0.0195/4)^(4·5) = $1625(1.004875^20) ≈ $1790.99
The account ending balance would be $1790.99.
Answer:
Step-by-step explanation:
Let X= each ticket for adults and Y= each ticket for kids
x+y=548
6.50x+3.50y=2881
subtract x from each side leaving you with y= -x+548
Plug in for Y
6.50x+3.50(-x+548)=2881
6.50x - 3.50x + 1918= 2881
3x+1918=2881 ....subtract
3x=963 .....divide by 3
X= 321 tickets for adults
321+y=548 ...subtract 321
Y=227 childrens tickets