Ask question

Ask question

All categories

grandymaker [24]

2 years ago

8

Consider the following pair of equations: y = −2x + 8 y = x − 1 Explain how you will solve the pair of equations by substitution

. Show all the steps and write the solution in (x, y) form.

1 answer:

Rzqust [24]2 years ago

6 0

Y = -2x +8
y = x-1

y=y

-2x+8 = x-1

-3x = -1 -8

-3x = -9

3x = 9

x = 3

You might be interested in

Triangle PQR is formed by the three squares A, B, and C:

Levart [38]

The first choice if ()2 means to the power of two if not then it is none of them.

3 0

2 years ago

Read 2 more answers

Jake has five cats and then he finds ten more cats how many does jake have after he loses 11 cats

AleksandrR [38]

jake has 5 cats

then he finds 10 cats

jake now has 5+10 = 15 cats

now he loses 11 cats

15 -11 =4

jake now has 4 cats

8 0

3 years ago

Read 2 more answers

What do you do if the divisor (the number you're dividing

Annette [7]

Answer:

you can use decimals

Step-by-step explanation:

7 0

2 years ago

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)

<u>Lower half:</u>

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

<u>Upper half</u>

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

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

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

<u></u>

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:

<u>Lower half:</u>

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

From this, the updated values of the box is:

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

From the question, the median is:

$Median = 22$ and $N = 14$

To calculate the median, we make use of:

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

$Median = \frac{14 + 1}{2}th$

$Median = \frac{15}{2}th$

$Median = 7.5th$

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

$Mode = 18$

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

$7th = 18$

The 8th item is calculated as thus:

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

$22= \frac{1}{2}(18 + 8th)$

Multiply through by 2

$44 = 18 + 8th$

$8th = 44 - 18$

$8th = 26$

The updated values of the box is:

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

From the question.

$Mean = 26$

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

So, we have:

$26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}$

Collect like terms

$26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}$

$26= \frac{ 2nd + 12th+304}{14}$

Multiply through by 14

$14 * 26= 2nd + 12th+304$

$364= 2nd + 12th+304$

This gives:

$2nd + 12th = 364 - 304$

$2nd + 12th = 60$

From the updated box,

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

We know that:

<em>The 2nd value can only be either 2 or 3</em>

<em>The 12th value can take any of the range 33 to 57</em>

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

$2nd = 3$

$12th = 57$

i.e.

$2nd + 12th = 60$

$3 + 57 = 60$

So, the complete box is:

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

6 0

2 years ago

Can anyone please help me why i got 3/4 marks on this question what am i missing?

Lesechka [4]

I’m pretty sure it’s because you got the price wrong

4 0

2 years ago

Read 2 more answers