Ask question

Ask question

All categories

3 years ago

10

The box and whisker plot below shows the numbers of text messages received in one day by students in the seventh and eighth grad

es at lincoln middle school.
a.where do the two sets of data overlap?
b.find the IQR of each set.
c.Find the difference between the medians of the sets
d. what number multiplied by the IQR equals the difference between two sets?
please help!

2 answers:

mixer [17]3 years ago

7 0

Answer:

Answer below

Step-by-step explanation:

a. they overlap at 22 and 26

b. eighth grade IQR = 8 seventh grade IQR = 8

c. 12

d. 1.5

qaws [65]3 years ago

6 0

They overlap at 22-26

You might be interested in

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

3 years ago

Write the next Number in this pattern.10,11,20,21,30,31

stiv31 [10]

The next Numbers in the pattern is 40,41

6 0

3 years ago

Read 2 more answers

Find the value of m 5m=10: which is the answer 2,50,5

bagirrra123 [75]

$\huge\textbf{Hey there!}$

$\huge\boxed{\textbf{5m = 10}}\\\\\large\textbf{DIVIDE 5 to BOTH SIDES}\\\\\huge\boxed{\mathbf{\dfrac{5m}{5} = \dfrac{10}{5}}}\\\\\large\textbf{CANCEL out: }\bold{\dfrac{5}{5}}\large\textbf{ because it gives you 1}\\\large\textbf{KEEP: }\rm{\bf \dfrac{10}{5}}\large\textbf{ because it gives you the m-value}\\\large\textbf{NEW EQUATION: m = }\bf \rm{\dfrac{10}{5}}\\\\\large\textbf{SIMPLIFY IT!}\\\huge\boxed{\mathbf{m = 2}}\\\large\textbf{Therefore, your answer is: \huge\boxed{\mathsf{m = 2}}}\huge\checkmark$

$\huge\textbf{Good luck on your assignment \&}\\\huge\textbf{enjoy your day!}$

<h2>~ $\mathfrak{Amphitrite1040:)}$ </h2>

4 0

2 years ago

Keith is riding his bicycle. He rides at a speed of 13 kilometers per hour for 32.5 kilometers. For how many hours does he ride?

9966 [12]

Answer:

422.50

Step-by-step explanation:

just multiply 13 × 32.5=422.50

8 0

3 years ago

Which equation matches the table ?

natka813 [3]

B= a+4 because let’s say for the first column, b8/a4 it would be 8=4+4 which is true

7 0

3 years ago