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3 years ago

Which is an equation of the line that passes through (-4,-2) and (4,2)?

1 answer:

N76 [4]3 years ago

4 0

The equation of a line is y = mx + b. Calculate the slope (m) from the 2 points.

m = (-2-2)/(-4-4) = -4/-8 = 1/2

We could stop here as only one answer has a slope of 1/2, namely A). But if you wanted to show that the y-intercept (b) was 0, you could use either of the points and substitute it into the equation of the line and solve for b.

(-4,-2):
-2 = 1/2(-4) + b
-2 = -2 + b
0 = b

(4,2):
2 = 1/2(4) + b
2 = 2 + b
0 = b

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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:

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

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:

The 2nd value can only be either 2 or 3

The 12th value can take any of the range 33 to 57

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

3 years ago

I want to know about perimeter and area

grandymaker [24]

Add each sides' centimeters together to get the perimeter. multiply length x width to get the area. I hope this helped

6 0

3 years ago

Please help for gods sake:(

worty [1.4K]

Answer:

answer is b)

Step-by-step explanation:

please mark me brainliest

6 0

3 years ago

Read 2 more answers

A closed box has a square base with side length l feet and height h feet. Given that the volume of the box is 29 cubic feet, exp

qwelly [4]

Answer:

The surface area of the box in terms of l is $2l^2 +116/l$

Step-by-step explanation:

The length of the box = $l$

The height of the box = $h$

The volume of the box can be obtained by multiplying the surface area of the base by the height of the box.

The surface area of the base of the box is a square, so it will be obtained by multiplying $l \times l = l^2$

hence volume = $l^2 \times h =29$

Notice that we are told to give our answer in terms of l, so we will make h the Subject of the formula in the above equation.

$h =29/l^2$

In the next phase, we are to find the surface area of the box.

The box has a square base, hence it will be made up of

2 squares with 4 rectangles. This is assuming the top is closed.

This means the surface area will be

$2(l^2) +4 (l\times h)$

recall $h = 29/l^2$

Hence, surface area =

$2(l^2) +4(l \times 29/l^2)\\2l^2 +116/l$

The surface area of the box in terms of l is $2l^2 +116/l$

4 0

3 years ago

You have five $1 bills, four $5 bills, six $10 bills, and three $20 bills you select a bill at random, without replacing the bil

Lostsunrise [7]

Answer:

Option B.

Step-by-step explanation:

It is given that,

Number of $1 bills = 5

Number of $5 bills = 4

Number of $10 bills = 6

Number of $20 bills = 3

So,

Total number of bills = 5 + 4 + 6 + 3 = 18

We need to find the probability of a 1 then a ten (without replacement).

Probability of selecting $1 bill in first draw $=\dfrac{5}{18}$

After one draw, the number of remaining bills is 18 - 1 = 17.

Probability of selecting $10 bill in second draw $=\dfrac{6}{17}$

The probability of a 1 then a ten is

$P=\dfrac{5}{18}\times \dfrac{6}{17}=\dfrac{5}{51}$

Therefore, the correct option is B.

6 0

3 years ago