All categories

3 years ago

Lena is recording the number of points she misses on each math quiz. Here are her results for the last nine quizzes.

Mathematics

2 answers:

sp2606 [1]3 years ago

4 0

Answer:

The range of the data is the difference between the largest and smallest value. In this case, the largest value is 8 and the smallest value is 1.

Hence, the range is $8-7=1$ .

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

The mode is the value that appears in a data set the most. When you arrange them in ascending order:

$1, 4, 6, 6, 6, 6, 6, 8, 8$

You can count that the value "6" appears five times, which is the most in this set.

Hence, the mode is 6.

ratelena [41]3 years ago

4 0

The mode is 6 because it’s the one you see the most
and the range is 7 because it’s the difference between the highest number which is 8 and the lowest number which is 1

You might be interested in

Im 10 years old and my sister is 4 years younger than me. 60 years later how old is my sister?

Morgarella [4.7K]

Your sister would be 66.
When you are 10, your sister would be 6 (10-4=6)
60 years later, you would be 70, and your sister would still be 4 years younger.

8 0

4 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago

A girl scout has (5x+18)

zhannawk [14.2K]

Step-by-step explanation:

(5x+18) - (3x+12)

5x+18=3x+12

5x-3x=2x

18-12=6

2x+6

3 0

4 years ago

Find the number of ways of arranging the numbers

Doss [256]

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

$\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}$

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

$0,1,2,0,1,2,0,1,2$

$2,0,1,2,0,1,2,0,1$

We have $3!=6$ possible patterns. Suppose for example that we choose the pattern

$0,1,2,0,1,2,0,1,2$

One possible way of following this pattern would be the arrangement

$3,1,2,6,4,5,9,7,8$

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

$3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216$