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!
Answer:
^(3)+7x^(2)+4x-12
Step-by-step explanation:
I cant see the image...but I take it that the midpoint is (9,8) and the endpoint S is (10,10) and ur looking for the other endpoint R.
midpoint formula : (x1 + x2) / 2, (y1 + y2) / 2
(10,10)....x1 = 10 and y1 = 10
(x,y)....x2 = x and y2 = y
so we sub
(10 + x) / 2, (10 + y) / 2 = 9/8
(10 + x) / 2 = 9
10 + x = 9 * 2
10 + x = 18
x = 18 - 10
x = 8
(10 + y) / 2 = 8
10 + y = 8 * 2
10 + y = 16
y = 16 - 10
y = 6
so endpoint R has coordinates of (8,6) <===
The Following "Verified" Answer is wrong. On Financial Math the answer is...
C.
$1,285
Answer:y= -4
Step-by-step explanation:
1 Solve for xx in 7x-4y=-127x−4y=−12.
x=\frac{4(y-3)}{7}
x=
7
4(y−3)
2 Substitute x=\frac{4(y-3)}{7}x=
7
4(y−3)
into 9x-4y=-209x−4y=−20.
\frac{36(y-3)}{7}-4y=-20
7
36(y−3)
−4y=−20
3 Solve for yy in \frac{36(y-3)}{7}-4y=-20
7
36(y−3)
−4y=−20.
y=-4
y=−4
4 Substitute y=-4y=−4 into x=\frac{4(y-3)}{7}x=
7
4(y−3)
.
x=-4
x=−4
5 Therefore,
\begin{aligned}&x=-4\\&y=-4\end{aligned}
x=−4
y=−4
