Write an equation perpendicular to y= 1 -x+ 9 that passes through the point (-2,-2). 5

20 people applied for a job. Everyone either has a school certificate or diploma or even both. If 14 have school certificates an

STatiana [176]

Given:

Either has a school certificate or diploma or even both = 20 people

Having school certificates = 14

Having diplomas = 11

To find:

The number of people who have a school certificate only.

Solution:

Let A be the set of people who have school certificates and B be the set of people who have diplomas.

According to the given information, we have

$n(A)=14$

$n(B)=11$

$n(A\cup B)=20$

We know that,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=14+11-n(A\cap B)$

$20=25-n(A\cap B)$

Subtract both sides by 25.

$20-25=-n(A\cap B)$

$-5=-n(A\cap B)$

$5=n(A\cap B)$

We need to find the number of people who have a school certificate only, i.e. $n(A\cap B')$ .

$n(A\cap B')=n(A)-n(A\cap B)$

$n(A\cap B')=14-5$

$n(A\cap B')=9$

Therefore, 9 people have a school certificate only.

3 0

3 years ago

You have 2 5/8 pizzas to share with three friends, how much pizza will each friend get?

Mazyrski [523]

2 5/8 divided by 3

(2*8)+5/8 divided by 3

16+5/8 divided by 3

(3*7/2^3)(1/3)

3*7/2^3*3

7/2^3

7/8

each friend would get 7/8

5 0

3 years ago

Read 2 more answers

What is the effect on the graph of the function f(x) = x^2 when f(x) is changed to f(x − 6)

olya-2409 [2.1K]

Answer:

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing $f(x) = x^2$ and $\:f\left(x\right)=\left(x-6\right)^2$ is represented as being shifted 6 units to the right as compare to the function $f(x) = x^2$ .

Step-by-step explanation:

When we Add or subtract a positive constant, let say c, to input x, it would be a horizontal shift.

For example:

Type of change Effect on y = f(x)

$y = f(x - c)$ horizontal shift: c units to right

So

Considering the function

$f(x) = x^2$

The graph is shown below. The first figure is representing $f(x) = x^2$ .

Now, considering the function

$\:f\left(x\right)=\left(x-6\right)^2$

According to the rule, as we have discussed above, as a positive constant 6 is added to the input, so there is a horizontal shift, 6 units to the right.

The graph of $\:f\left(x-6\right)=\left(x-6\right)^2$ is shown below in second figure. It is clear that the graph of $\:f\left(x-6\right)=\left(x-6\right)^2$ is shifted 6 units to the right as compare to the function $f(x) = x^2$ .

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing $f(x) = x^2$ and $\:f\left(x\right)=\left(x-6\right)^2$ is represented as being shifted 6 units to the right as compare to the function $f(x) = x^2$ .