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emmainna [20.7K]
3 years ago
12

Classify each number below as an integer or not.

Mathematics
1 answer:
fomenos3 years ago
6 0

Answer:

1. yes

2. no

3. yes

4. no

5. yes

Step-by-step explanation:

The method i use is if the number is not a fraction than it is an integer. besides 6 for some reason.

Here is the catch if the fraction can be simplified into a whole number than it is an integer.

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A researcher is investigating variables that might be associated with the academic performance of high school students. She exam
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Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

5 0
4 years ago
Let f(x)=x^2f ( x ) = x 2. Find the Riemann sum for ff on the interval [0,2][ 0 , 2 ], using 4 subintervals of equal width and t
sladkih [1.3K]

Answer:

A_L=1.75

Step-by-step explanation:

We are given:

f(x)=x^2

interval = [a,b] = [0,2]

Since n = 4 ⇒ \Delta x = \frac{b-a}{n} = \frac{2-0}{4}=\frac{1}{2}

Riemann sum is area under the function given. And it is asked to find Riemann sum for the left endpoint.

A_L= \sum\limits^{n}_{i=1}\Delta xf(x_i) = \frac{1}{2}(0^2+(\frac{1}{2})^2+1^2+(\frac{3}{2})^2)=\frac{7}{4}=1.75

Note:

If it will be asked to find right endpoint too,

A_R=\sum\limits^{n}_{i=1}\Delta xf(x_i) =\frac{1}{2}((\frac{1}{2})^2+1^2+(\frac{3}{2})^2+2^2)=\frac{15}{4}=3.75

The average of left and right endpoint Riemann sums will give approximate result of the area under f(x)=x^2 and it can be compared with the result of integral of the same function in the interval given.

So, (A_R+A_L)/2 = (1.75+3.75)/2=2.25

\int^2_0x^2dx=x^3/3|^2_0=8/3=2.67

Result are close but not same, since one is approximate and one is exact; however, by increasing sample rates (subintervals), closer result to the exact value can be found.

3 0
3 years ago
(i) Represent these two sets of data by a back-to-back stem-and-leaf diagram.
alexgriva [62]
<h3>Answer: </h3>

{\begin{tabular}{lll}\begin{array}{r|c|l}\text{Leaf (Ali)} & \text{Stem} & \text{Leaf (Kumar)}\\\cline{1-3} 7 & 4 & 1\ 2\ 3\ 6\ 6\ 9\ 9 \\  9\ 8 & 5 & 2\ 2\ 3\\  5\ 5 & 6 & \\  7\ 2\ 0 & 7 & 8\ 8\ 9\\  9\ 9\ 8\ 4\ 3\ 3\ 3\ 1\ 1 & 8 & 2\ 2\ 4\ 5\\  9\ 8\ 1 & 9 & 0\ 2\ 5\\  \end{array} \\\\ \fbox{\text{Key: 7} \big| \text{4} \big| \text{1 means 4.7 for Ali and 4.1 for Kumar}} \end{tabular}}

=========================================================

Explanation:

The data set for Ali is

8.3, 5.9, 8.3, 8.9, 7.7, 7.2, 8.1, 9.1, 9.8, 5.8,

8.3, 4.7, 7.0, 6.5, 6.5, 8.4, 8.8, 8.1, 8.9, 9.9

which when on a single line looks like this

8.3, 5.9, 8.3, 8.9, 7.7, 7.2, 8.1, 9.1, 9.8, 5.8, 8.3, 4.7, 7.0, 6.5, 6.5, 8.4, 8.8, 8.1, 8.9, 9.9

Let's sort the values from smallest to largest

4.7, 5.8, 5.9, 6.5, 6.5, 7.0, 7.2, 7.7, 8.1, 8.1, 8.3, 8.3, 8.3, 8.4, 8.8, 8.9, 8.9, 9.1, 9.8, 9.9

Now lets break the data up into separate rows such that each time we get to a new units value, we move to another row

4.7

5.8, 5.9

6.5, 6.5

7.0, 7.2, 7.7

8.1, 8.1, 8.3, 8.3, 8.3, 8.4, 8.8, 8.9, 8.9

9.1, 9.8, 9.9

We have these stems: 4, 5, 6, 7, 8, 9 which represent the units digit of the values. The leaf values are the tenths decimal place.

For example, a number like 4.7 has a stem of 4 and leaf of 7 (as indicated by the key below)

This is what the stem-and-leaf plot looks like for Ali's data only

\ \ \ \ \ \ \ \ \text{Ali's data set}\\\\{\begin{tabular}{ll}\begin{array}{r|l}\text{Stem} & \text{Leaf}\\ \cline{1-2}4 & 7 \\ 5 & 8\ 9 \\ 6 & 5\ 5 \\ 7 & 0\ 2\ 7 \\ 8 & 1\ 1\ 3\ 3\ 3\ 4\ 8\ 9\ 9 \\ 9 & 1\ 8\ 9\\ \end{array} \\\\ \fbox{\text{Key: 4} \big| \text{7 means 4.7}} \\ \end{tabular}}

The stem-and-leaf plot condenses things by tossing out repeated elements. Instead of writing 8.1, 8.1, 8.3 for instance, we can just write a stem of 8 and then list the individual leaves 1, 1 and 3. We save ourselves from having to write two more copies of '8'

Through similar steps, this is what the stem-and-leaf plot looks like for Kumar's data set only

\ \ \ \ \ \ \ \ \text{Kumar's data set}\\\\{\begin{tabular}{ll}\begin{array}{r|l}\text{Stem} & \text{Leaf}\\ \cline{1-2}4 & 1\ 2\ 3\ 6\ 6\ 9\ 9 \\ 5 & \ 2\ 2\ 3\  \  \  \   \\ 6 & \\ 7 & 8\ 8\ 9 \\ 8 & 2\ 2\ 4\ 5\\ 9 & 0\ 2\ 5\\ \end{array} \\\\ \fbox{\text{Key: 4} \big| \text{1 means 4.1}} \\ \end{tabular}}

Kumar doesn't have any leaves for the stem 6, so we will have that section blank. It's important to have this stem so it aligns with Ali's stem plot.

Notice that both stem plots involve the same exact set of stems (4 through 9 inclusive).

What we can do is combine those two plots into one single diagram like this

{\begin{tabular}{lll}\begin{array}{r|c|l}\text{Leaf (Ali)} & \text{Stem} & \text{Leaf (Kumar)}\\\cline{1-3} 7 & 4 & 1\ 2\ 3\ 6\ 6\ 9\ 9 \\  8\ 9 & 5 & 2\ 2\ 3\\  5\ 5 & 6 & \\  0\ 2\ 7 & 7 & 8\ 8\ 9\\  1\ 1\ 3\ 3\ 3\ 4\ 8\ 9\ 9 & 8 & 2\ 2\ 4\ 5\\  1\ 8\ 9 & 9 & 0\ 2\ 5\\  \end{array} \\  \end{tabular}}

Then the last thing to do is reverse each set of leaves for Ali (handle each row separately). The reason for this is so that each row of leaf values increases as you further move away from the stem. This is simply a style choice. This is somewhat similar to a number line, except negative values aren't involved here.

This is what the final answer would look like

{\begin{tabular}{lll}\begin{array}{r|c|l}\text{Leaf (Ali)} & \text{Stem} & \text{Leaf (Kumar)}\\\cline{1-3} 7 & 4 & 1\ 2\ 3\ 6\ 6\ 9\ 9 \\  9\ 8 & 5 & 2\ 2\ 3\\  5\ 5 & 6 & \\  7\ 2\ 0 & 7 & 8\ 8\ 9\\  9\ 9\ 8\ 4\ 3\ 3\ 3\ 1\ 1 & 8 & 2\ 2\ 4\ 5\\  9\ 8\ 1 & 9 & 0\ 2\ 5\\  \end{array} \\\\ \fbox{\text{Key: 7} \big| \text{4} \big| \text{1 means 4.7 for Ali and 4.1 for Kumar}} \end{tabular}}

The fact that Ali is on the left side vs Kumar on the right, doesn't really matter. We could swap the two positions and end up with the same basic table. I placed Ali on the left because her data set is on the top row of the original table given.

The thing you need to watch out for is that joining the stem and leaf for Ali means you'll have to read from right to left (as opposed to left to right). Always start with the stem. That's one potential drawback to a back-to-back stem-and-leaf plot. The advantage is that it helps us compare the two data sets fairly quickly.

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2 years ago
1) What percentage of 18 is 9
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Answer:

2%

Step-by-step explanation:

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