A veterinarian collects data on the number of times race horses are raced during their careers. The veterinarian finds that the average number of races a horse enters is ¯ x = 15.3 , with a standard deviation of s = 6.8 in a sample of n = 20 horses. A 95% confidence for μ , the average number of times a horse races, is given by.

a. (12.32, 18.28).
b. (10.95, 19.65).
c. (12.12, 18.48).
d. (11.38, 19.22).

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!

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752