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3 years ago

A baseball team played 130 games in a season. If it won sixteen more games

Mathematics

1 answer:

egoroff_w [7]3 years ago

4 0

Answer:

The team has 92 wins and 38 losses.

Step-by-step explanation:

In a season 130 games are played by a baseball team.

Let us assume that the team wins x games and losses y games.

Therefore, we can write, x + y=130....... (1)

It is given that if the team won sixteen more games twice as many games as it lost.

So, we can conclude that x =2y+16 ......(2)

Hence, solving equations (1) and (2) we get, (2y+16) + y =130

⇒ 3y = 114

⇒ y =38

And from equation (2), x = 2×38 + 16 = 92

Therefore, the team has 92 wins and 38 losses. (Answer)

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You are creating a model of a bookcase for a dollhouse.

Goshia [24]

Answer:

If the model is 3 inches tall, the model bookcase should be <em>2 inches</em> wide.

Step-by-step explanation:

When using similar figures, set up a proportion of the model measurements to the actual measurements:

$\frac{3}{48}=\frac{x}{32}$

Then cross multiply and solve for x:

$48x=96\\x=\frac{96}{48}\\x=2$

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3 years ago

A man has a 10m X 10m square garden. In the center is a 2m X 2m square patch which he cannot use. He divides his usable space in

ra1l [238]

Answer:

4m X 6m

Step-by-step explanation:

This is because if there is a 2m square in the middle, there is 8 m of usable space left along both sides of the garden.

Because the 2m square was in the middle, there is 8/2 = 4m along each width of the 4 small rectangles.

Becuase 4m is the width, there is 10 - 4 = 6m along each rectangle's length.

8 0

3 years ago

A population has the following characteristics. (a) A total of 75% of the population survives the first year. Of that 75%, 25% s

Artemon [7]

Answer:

$= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right] \left \begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 1 \ \leq age \leq 2 }\\{2 \ \leq age \leq 3}\end{array}\right$

i.e after the first year ;

there 1344 members in the first age class

84 members for the second age class; and

28 members for the third age class

Step-by-step explanation:

We can deduce that the age distribution vector x represents the number of population members for each age class; Given that in each class of age there are 112 members present.

The current age distribution vector is as follows:

$x = \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right] \left[\begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 0 \ \leq age \leq 2 }\\{0 \ \leq age \leq 3}\end{array}\right]$

Also , the age transition matrix is as follows:

$L = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]$

After 1 year ; the age distribution vector will be :

$x_2 =Lx_1 = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right] \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right]$

$= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right] \left \begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 1 \ \leq age \leq 2 }\\{2 \ \leq age \leq 3}\end{array}\right$

6 0

3 years ago

Solve for (f+g)(5) if f(x)=6x+3 and g(x)=x-4 show you work please

Diano4ka-milaya [45]

There are two ways to do this.

The first way is to algebraically find (f+g)(x) first and plug in x = 5 later. Doing that method leads us to
(f+g)(x) = f(x) + g(x)
(f+g)(x) = 6x+3 + x-4
(f+g)(x) = 7x-1
(f+g)(5) = 7(5)-1
(f+g)(5) = 34

OR

you can compute f(5) and g(5) first, then add up those sub-results to get
f(x) = 6x+3
f(5) = 6(5)+3
f(5) = 33
g(x) = x-4
g(5) = 5-4
g(5) = 1
Adding up these results gives: (f+g)(5) = f(5) + g(5) = 33+1 = 34

Either way, the final answer is 34

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4 years ago

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hoa [83]

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3 years ago