The answer is going to be 7,434,000. hope that helped
Answer:
90% confidence interval is (43.25, 61.41)
Step-by-step explanation:
Given that,
n = 17
Mean,
= 52.33
Standard deviation,
= 21.44
∝ = 0.10
Now,
Confidence interval =
± ![t_{\frac{\alpha }{2}, n-1} [\frac{\sigma}{\sqrt{n} } ]](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%2C%20n-1%7D%20%5B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%5D)
= 52.33 ± 1.7646 [ 21.44 / √17 ]
= 52.33 ± 9.0791
So,
52.33 + 9.0791 = 61.4091
52.33 - 9.0791 = 43.2509
So,
90% confidence interval is (43.25, 61.41)
Also,
Lower end-point = 43.25
Upper end-point = 61.41
I believe the answer is 26a+5b+4ds
0.032x+0.041 (1500-x)=54.3
Solve for x
0.032x+61.5-0.041x=54.3
0.032x-0.041x=54.3-61.5
-0.009x=-7.2
X=7.2/0.009
X=800 invested at 3.2%
1500-800=700 invested at 4.1%
That’s what u got I don’t know if there’s another answer