Answer:
0.33
Step-by-step explanation:
Let A denote an event. The odds of the horse finishing second is 33 to 67.
From the information, the total outcomes of event A is:
![Total \ Outcomes, A=Favorable +Unfavorable\\\\=33+67\\\\=100](https://tex.z-dn.net/?f=Total%20%5C%20Outcomes%2C%20A%3DFavorable%20%2BUnfavorable%5C%5C%5C%5C%3D33%2B67%5C%5C%5C%5C%3D100)
Hence, there are 100 total outcomes. The probability of the horse finishing second is calculated as:
![p=\frac{n(A_f)}{n(A_u)}\\\\\\=p=\frac{33}{100}\\\\=0.33](https://tex.z-dn.net/?f=p%3D%5Cfrac%7Bn%28A_f%29%7D%7Bn%28A_u%29%7D%5C%5C%5C%5C%5C%5C%3Dp%3D%5Cfrac%7B33%7D%7B100%7D%5C%5C%5C%5C%3D0.33)
Hence, the probability of finishing second is 0.33
Answer:
A) v = T/2 - 5
Step-by-step explanation:
T = 10+2v
2v = T-10
v = (T-10)/2 = T/2-5
If x +23 = 9, then, x = 9-23.
So x = -14
Given equation of hyperbola is
![\frac{(x-1)^{2}}{25}-\frac{(y+3)^{2}}{9}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-1%29%5E%7B2%7D%7D%7B25%7D-%5Cfrac%7B%28y%2B3%29%5E%7B2%7D%7D%7B9%7D%3D1)
Given hyperbola is the horizontal hyperbola
The standard form of the horizontal hyperbola is
![\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-h%29%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D-%5Cfrac%7B%28y-k%29%5E%7B2%7D%7D%7Bb%5E%7B2%7D%7D%3D1)
When we compare these two equations, we get
h = 1, a = 5 , k = -3 and b = 3
Length of the transverse axis = 2a
Plug in the value of 'a' as 5
So,Length of the transverse axis = 2(5) = 10
14.94 cm ( to 2 dec. places )
We require the sum totals of the 2 groups before calculating the mean of all 14
using mean = ![\frac{sum}{count}](https://tex.z-dn.net/?f=%5Cfrac%7Bsum%7D%7Bcount%7D)
Children's sum
12.6 =
( cross-multiply )
sum = 6 × 12.6 = 75.6
Adult's sum
16.7 =
( cross- multiply )
sum = 8 × 16.7 = 133.6
Total sum of all 14 = 75.6 + 133.6 = 209.2
mean =
= 14.94 cm