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2 years ago

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Of the films Empty Set Studios released last year, 60% were comedies and the rest were horror films. 75% of the comedies were pr

ofitable, but 75% of the horror moves were unprofitable. If the studio made a total of 40 films, and broke even on none of them, how many of their films were profitable?

1 answer:

Alborosie2 years ago

4 0

Answer: There are 22.5 films were profitable.

Step-by-step explanation:

Since we have given that

Number of total films = 40

Percentage of comedies = 60%

Number of comedies is given by

$0.6\times 40\\\\=24$

Percentage of horror films = 40%

Number of horror films is given by

$0.4\times 40\\\\=16$

Percentage of comedies were profitable = 75%

Number of profitable comedies is given by

$0.75\times 24=18$

Percentage of horror were unprofitable = 75%

Percentage of horror were profitable = 25%

Number of profitable horror films is given by

$0.25\times 18\\\\=4.5$

So, Total number of profitable films were

$18+4.5=22.5$

Hence, there are 22.5 films were profitable.

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For this case we have that the volume of the figure is composed of the volume of a prism and the volume of a pyramid:

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ANswer:

Option D

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2 years ago

Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

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We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

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