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2 years ago

Can you help me get this answer

Mathematics

1 answer:

scoundrel [369]2 years ago

3 0

Answer:

y = f(x-1) - 3

is the correct function

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Identify the set of X-values for which f(X) is greater than or equal to 0.

Allushta [10]

The given function is greater than or equal to 0 for the interval given as follows:

{x | x ≤ -3 or -1 ≤ x ≤ 0 or x ≥ 2}.

<h3>When a function is positive and when it is negative?</h3>

A function is positive when it is above the x-axis.

A function is negative when it is below the x-axis.

For this problem, with a closed interval, we have that the function is above for:

x ≤ -3, -1 ≤ x ≤ 0, x ≥ 2.

Hence the interval notation is:

{x | x ≤ -3 or -1 ≤ x ≤ 0 or x ≥ 2}.

More can be learned about functions at brainly.com/question/24808124

#SPJ1

6 0

1 year ago

Brian purchased 4 items that were $1.00 each. The sales tax rate is 5.5%. How much was Brian's total bill including sales tax?

ad-work [718]

The answer will be $4.22

4 0

3 years ago

I really need help. My teacher said that I could use any number.

katen-ka-za [31]

The keyword is Rise over run
So the 2nd y value minus the first y value and divide it by the 2nd x value minus
(y2-y1)/(x2-x1)
(10-7)/5-3)
the first x value and you should get 3/2
*dont worry the if the number is under and not above it’s not a saying to do to the power of but it’s just saying the order in which to put the x and y values.

4 0

3 years ago

Find 118% of 19 could really use some help late bloomer

Zanzabum

Hello,

to find the percent of a number what we have to do is to multiply the number by de percent that we want to know and divide by 100, so:

$19* \frac{118}{100} =22.42$

Answer: 118% of 19 is 22.42

7 0

3 years ago

Read 2 more answers

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$