I NEED HELP!!!<br> FIRST ONE TO DO IT GETS BRAINLIEST

Using the data in the table, which statement is true?

horsena [70]

Answer: Choice C. mean > median

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Explanation:

Multiply each measure with their corresponding frequency.

8*1 = 8
10*3 = 30
14*2 = 28

Add up those products: 8+30+28 = 66

Then divide by the total frequency n = 1+3+2 = 6 to get 66/6 = 11 as the mean.

mean = 11

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Since we have n = 6 values in this list, this means the median is between slot n/2 = 6/2 = 3 and slot 4.

Note how that places us in the middle row because 1+3 = 4 encapsulates both of those slots mentioned. So the median is 10.

Or you could list out the values in roster notation {8, 10, 10, 10, 14, 14} to see that {10,10} occupy the middle most slots. So the median is (10+10)/2 = 20/2 = 10.

median = 10

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The mode is simply the most frequent value. The table shows that mode = 10 since it occurs 3 times, compared to 8 showing up 1 time and 14 showing up twice.

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We have the following summary

mean = 11
median = 10
mode = 10

With that in mind, let's go through the answer choices.

We can see that mean < mode is false, since it should be mean > median, so cross choice A off the list.
mean = median is also false, so choice B is crossed off as well.
mean > median is true since 11 > 10 is true. Choice C is the answer. Note how this being true directly contradicts choice B, which is another reason to see why choice B is false.
median > mode is false because 10 > 10 is false. It should be median = mode. Choice D is crossed off the list.

4 0

3 years ago

X ^ (2) y '' - 7xy '+ 16y = 0, y1 = x ^ 4

nignag [31]

Standard reduction of order procedure: suppose there is a second solution of the form $y_2(x)=v(x)y_1(x)$ , which has derivatives

$y_2=vx^4$
${y_2}'=v'x^4+4vx^3$
${y_2}''=v''x^4+8v'x^3+12vx^2$

Substitute these terms into the ODE:

$x^2(v''x^4+8v'x^3+12vx^2)-7x(v'x^4+4vx^3)+16vx^4=0$
$v''x^6+8v'x^5+12vx^4-7v'x^5-28vx^4+16vx^4=0$
$v''x^6+v'x^5=0$

and replacing $v'=w$ , we have an ODE linear in $w$ :

$w'x^6+wx^5=0$

Divide both sides by $x^5$ , giving

$w'x+w=0$

and noting that the left hand side is a derivative of a product, namely

$\dfrac{\mathrm d}{\mathrm dx}[wx]=0$

we can then integrate both sides to obtain

$wx=C_1$
$w=\dfrac{C_1}x$

Solve for $v$ :

$v'=\dfrac{C_1}x$
$v=C_1\ln|x|+C_2$

Now

$y=C_1x^4\ln|x|+C_2x^4$

where the second term is already accounted for by $y_1$ , which means $y_2=x^4\ln x$ , and the above is the general solution for the ODE.

4 0

3 years ago

Help i need somebody to help

solong [7]

Answer:

she selling five dollars

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Someone help me get outta summer school slope? And y intercept?

KiRa [710]

Okay bud I’m not sure how about it or what if we are all the good ones you have that to me and my brother wants to go buy a new one and then go back to church class today or what do you want it to be in a big deal if we can get some stuff done then I’ll do the work I have for that you don’t want to me I don’t know how you want it I know but it’s like I don’t know why I’m not

7 0

2 years ago

Find the quotient of 2/3 and 4/5

Lorico [155]

Answer:

$\frac{5}{6}$