Answer:
c. (12.12, 18.48)
Step-by-step explanation:
Hello!
The study variable is X: number of times a racehorse is raced during its career.
The average number is X[bar]= 15.3 and the standard deviation is S= 6.8 obtained from a sample of n=20 horses.
To estimate the population mean you need that the variable has a normal distribution, in this case, we have no information about its distribution so I'll assume that it has a normal distribution. With n=20 the most accurate statistic to use for the estimation is a Students-t for one sample, the formula for the interval is:
X[bar] ± 

[15.3 ± 2.093 *
]
[12.12; 18.48]
Using a significance level of 95% you'd expect that the true average of times racehorses are raced during their career is included in the interval [12.12; 18.48].
I hope it helps!
Let be Z the money ;
We have ( 40 / 100 ) x Z = $85 ;
( 4 / 10 ) x Z = $85 ;
( 2 / 5 ) x Z = $85 ;
Z = $( 85 x 5 ) ÷ 3 ;
Z = $425 ÷ 3 ;
Z = $141,66 ≈ $142 ;
Answer: 570
Step-by-step explanation: