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2 years ago

The data for numbers of times per week 20 students at Stackamole High eat vegetables are shown below:

Mathematics

1 answer:

Step2247 [10]2 years ago

7 0

The dot plot uses dots to for data representation.

The dot plot is right skewed
There is an outlier, 9.
The best measure of center is median

(a) The description

From the dot plot (see attachment), we have the following observations,

There are more data at the left-hand side, than the right-hand side of the dot plot.
There is only one data element at the right-hand side

This means that the dot plot is right skewed

(b) Outlier

From the dot plot, we can see that 9 is relatively far from the other data elements.

This means that 9 is outlier

(c) Measure of center

In (b), we established that the dot plot has an outlier.

When there is an outlier, the best measure of center to use is the median.

Hence, the median will be used as the measure.

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Describe one kind of diagram you might draw to help you solve a problem

Mila [183]

You can use a bar graph or histogram to compare amounts of items You can also draw arrays to help with multiplication problems.
In algebra, you can graph equations to see points of sets.

Hope this helps!

5 0

3 years ago

Read 2 more answers

Find the LCM and the GCF of 100 and 120.

Advocard [28]

Answer:

Step-by-step explanation:

LCM is least common multiple. It means the least multiple that both 100 and 120 have in common. 600 is the LCM. 20 is the GCF.

4 0

2 years ago

5. Show that the following points are collinear. a) (1, 2), (4, 5), (8,9)

Irina-Kira [14]

Label the points A,B,C

A = (1,2)
B = (4,5)
C = (8,9)

Let's find the distance from A to B, aka find the length of segment AB.

We use the distance formula.

$A = (x_1,y_1) = (1,2) \text{ and } B = (x_2, y_2) = (4,5)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (2-5)^2}\\\\d = \sqrt{(-3)^2 + (-3)^2}\\\\d = \sqrt{9 + 9}\\\\d = \sqrt{18}\\\\d = \sqrt{9*2}\\\\d = \sqrt{9}*\sqrt{2}\\\\d = 3\sqrt{2}\\\\$

Segment AB is exactly $3\sqrt{2}$ units long.

Now let's find the distance from B to C

$B = (x_1,y_1) = (4,5) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(4-8)^2 + (5-9)^2}\\\\d = \sqrt{(-4)^2 + (-4)^2}\\\\d = \sqrt{16 + 16}\\\\d = \sqrt{32}\\\\d = \sqrt{16*2}\\\\d = \sqrt{16}*\sqrt{2}\\\\d = 4\sqrt{2}\\\\$

Segment BC is exactly $4\sqrt{2}$ units long.

Adding these segments gives

$AB+BC = 3\sqrt{2}+4\sqrt{2} = 7\sqrt{2}$

----------------------

Now if A,B,C are collinear then AB+BC should get the length of AC.

AB+BC = AC

Let's calculate the distance from A to C

$A = (x_1,y_1) = (1,2) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (2-9)^2}\\\\d = \sqrt{(-7)^2 + (-7)^2}\\\\d = \sqrt{49 + 49}\\\\d = \sqrt{98}\\\\d = \sqrt{49*2}\\\\d = \sqrt{49}*\sqrt{2}\\\\d = 7\sqrt{2}\\\\$

AC is exactly $7\sqrt{2}$ units long.

Therefore, we've shown that AB+BC = AC is a true equation.

This proves that A,B,C are collinear.

For more information, check out the segment addition postulate.

7 0

2 years ago

How do I differentiate this?

disa [49]

Answer:

$\frac{8}{\left(-x+2\right)^2}$ & $x = 1, x = 3$

Step-by-step explanation:

I'm sure you are familiar with the product rule,

If y = u*v => dy/dx = u * dv/dx + v * dy/dx <----- product rule

In this case:

$u=3x+2,\:v=\left(2-x\right)^{-1},\\=>\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

Now remember the sum rule:

$\frac{d}{dx}\left(3x+2\right) = \frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right),\\\frac{d}{dx}\left(3x\right) = 3,\\\frac{d}{dx}\left(2\right) = 0\\\frac{d}{dx}\left(3x+2\right) = 3$

For this second bit we apply the chain rule:

$\frac{d}{dx}\left(\left(2-x\right)^{-1}\right) = -\frac{1}{\left(2-x\right)^2}\frac{d}{dx}\left(2-x\right),\\\frac{d}{dx}\left(2-x\right) = -1,\\\\=> -\frac{1}{\left(2-x\right)^2}\left(-1\right)\\=> \frac{1}{\left(2-x\right)^2}$

If we substitute these values back into the expression... $\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

...we get the following:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)$

The rest is just pure simplification:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}$

Now let's equate this to equal 8 for the second bit and solve for x:

$\frac{8}{\left(-x+2\right)^2}=8,\\\frac{8}{\left(-x+2\right)^2}\left(-x+2\right)^2=8\left(-x+2\right)^2,\\8=8\left(-x+2\right)^2,\\\left(-x+2\right)^2=1,\\x = 1, x = 3$

4 0

2 years ago

Based on past experience, a bank believes that 4% of the people who receive loans will not make payments on time. The bank has r

Nina [5.8K]

Answer:

Mean = 0.04

Step-by-step explanation: given that P = 4%

n = 300

the mean of the sampling distribution of the proportion of clients in this group who may not make timely payments will be

4/ 100 = 0.04

4 0

3 years ago