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prisoha [69]
3 years ago
12

(2.2.5) In horse racing, one can make a trifecta bet by specifying which horse will come in first, which will come in second, an

d which will come in third, in the correct order. One can make a box trifecta bet by specifying which three horses will come in first, second and third, without specifying the order. a. In an eight-horse field, how many different ways can one make a trifecta bet
Mathematics
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

56 ways.

Step-by-step explanation:

This question is solved by using combinations and permutations chapters in the math book.

To put it simply, we need to find the number of 3 horse combinations we can make from 8 horses. This requires the combinations formula:

C(n,r)=\frac{n!}{r!(n-r)!}

here n is the total number of objects to choose from, and r is the number of objects we require in the combination or group.

Since there are 8 horses, n= 8

Since we need to choose only 3 of them, and order does not matter, r= 3

Solving the equation above using these inputs gives us 56 unique ways we can choose the three winners.

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The product of two consecutive odd integers is 72. Write a quadratic equation to find the integers,
zimovet [89]

<u>Answer:</u>

x^2 + x - 72 = 0

<u>Explanation:</u>

We are given product of two consecutive odd integers is equal to 72.

Let one integer be x , then since these are consecutive and odd so the other integer will be equal to x + 1.  

Their product is equal to 72. So,

x * (x + 1) = 72

=> x^2 + x = 72

=>  x^2 + x - 72 = 0 is the required quadratic equation.

5 0
3 years ago
When it is 11 a.m. in Johannesburg, it is 6 p.m. the same day in Tokyo, Japan. A flight leaves OR Tambo in Johannesburg at 6 a.m
kiruha [24]

Answer:

Dear User,

Answer to your query is provided below

The time will be 8am on Wednesday when flight arrive in Tokyo.

Step-by-step explanation:

Here, You can see that when 11 a.m. in Johannesburg, it is 6 p.m. the same day in Tokyo, Japan ; this means 7Hrs. gap.

Further, A flight leaves Johannesburg at 6 a.m. on Tuesday for Tokyo takes 18 hours in the air, with an additional 1 hour stop-over on land.

So, 6a.m. + 18hr.+1hr.+7hr. = 8am on Wednesday in Tokyo (Arrival)

4 0
3 years ago
Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

\implies F(x,y)=x^2y^3+\ln(x+y^2)=C

With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
3 years ago
Which of the following relations is a function?
Agata [3.3K]

Answer:

D

Step-by-step explanation:

Because in D, there no x-value that are repeating in the list.

For example A, the value 8 and -8 are being repeated twice, which makes this relation not a fuction.

5 0
3 years ago
Evaluate log416 -2 1/2 2
krok68 [10]

16 = 4²

So, by the definition of logarithms

\log_4{16}=2

5 0
4 years ago
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