Answer:
E) we will use t- distribution because is un-known,n<30
the confidence interval is (0.0338,0.0392)
Step-by-step explanation:
<u>Step:-1</u>
Given sample size is n = 23<30 mortgage institutions
The mean interest rate 'x' = 0.0365
The standard deviation 'S' = 0.0046
the degree of freedom = n-1 = 23-1=22
99% of confidence intervals 
(from tabulated value).
using calculator
Confidence interval is
the mean value is lies between in this confidence interval
(0.0338,0.0392).
<u>Answer:-</u>
<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>
Alright, so 3f-g=4 and f+2g=5.
3f-g=4
f+2g=5
Multiplying the first equation by 2 and adding it to the second, we get 7f=13 and by dividing both sides by 7 we get f=13/7. Since f+2g=5, then we can plug 13/7 in for f to get 13/7+2g=5. Next, we subtract 13/7 from both sides to get 2g=3+1/7=22/7 (since 3*7=21 and 21+1=22). DIviding both sides by 2, we get 22/14=g. Plugging that into f/39g, we get (13/7)/(22*39/14)
= (13/7)/(858/14)
= (13/7)*(14/858)
=182/6006
= 91/3003 (by dividing both numbers by 2)
= 13/429 (by dividing both numbers by 7)
= 1/33 (by dividing both numbers by 13)
Hello there, to answer your question...
<h2>1.95 is 1.5% of 130.</h2><h2 />
I hope the information I provided has helped.
Answer:
OMG you are so pretty
Step-by-step explanation:
I need points bro, so imma just type this rq cuz I’m mid test thank u tho
