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3 years ago

Instructions:Select the correct answer.

Mathematics

2 answers:

ICE Princess25 [194]3 years ago

8 0

Figure C.

The first and left quartiles are evenly skewed. (Meaning, there aren't any outliers present) And the median is directly in the middle of the data set. Thus, Figure C is the answer.

Brums [2.3K]3 years ago

6 0

I believe it's Figure C

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Complete the tables of values.

Vladimir [108]

Answer:

$a=1\\b=\frac{1}{16}\\\\c=\frac{1}{256}\\\\d=1\\e=\frac{4}{9}\\\\f=\frac{16}{81}$

Step-by-step explanation:

For the orange table:

Substitute the corresponding values of x into the function $f(x)=4^{-x}$ , then:

For a:

x=0

$f(0)=4^0=1$

For b:

x=2

$f(x)=4^{-2}=\frac{1}{16}$

For c:

x=4

$f(x)=4^{-4}=\frac{1}{256}$

For the blue table:

Substitute the corresponding values of x into the function $g(x)=(\frac{2}{3})^x$ , then:

For d:

x=0

$g(x)=(\frac{2}{3})^0=1$

For e:

x=2

$g(x)=(\frac{2}{3})^2=\frac{4}{9}$

For f:

x=4

$g(x)=(\frac{2}{3})^4=\frac{16}{81}$

3 0

3 years ago

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr

sergij07 [2.7K]

$f(t)=\begin{cases}0&\text{for }t$

Recall that

$u(t)=\begin{cases}0&\text{for }t$

Take it one piece at a time. For $t\ge0$ , we can scale $u(t)$ by -5:

$-5u(t)=\begin{cases}0&\text{for }t$

If we shift the argument by 1 and scale by -5, we have

$-5u(t-1)=\begin{cases}0&\text{for }t$

so if we subtract this from $-5u(t)$ , we'll end up with

$-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t$

For the next piece, we can add another scaled and shifted step like

$-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

so that

$-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

For the last piece, we add one more term:

$6u(t-7)=\begin{cases}0&\text{for }t$

and so putting everything together, we get $f(t)$ :

$f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)$
$f(t)\equiv-5u(t)-u(t-1)+12u(t-7)$

5 0

3 years ago

1. Determine if the set of ordered pairs is a relation or a function.

kykrilka [37]

Answer:

This is just a relation.

Step-by-step explanation:

It is easy to see if a set of ordered pairs is a relation or a relation that is a function.

If all the points are distinct, then you just have to look at the $x$ -coordinates.