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2 years ago

(7 point)

Mathematics

2 answers:

Vladimir [108]2 years ago

5 0

The number of social media post of Noami has a greater spread when compared to that of Amir.

<h3 /><h3>What is the range and intequartile range of the social media posts?</h3>

Range is the difference between the highest and lowest values of a set of observations

Range = highest value - lowest value

The range of Amir's social media posts = 19 - 15 = 4
The range of Noami's social media posts = 10 - 0 = 10

The inter quartile range is the difference between the third quartile and the first quartile of a data set.

IQR = Q3 - Q1

The IQR of Amir's social media posts: 7

The IQR of Naomi's social media posts = 10

To learn more about range, please check: brainly.com/question/12372689

Yuki888 [10]2 years ago

3 0

Answer:

Naomi"s data set has more spread since her range is 10 and her IQR is 3.

Step-by-step explanation:

Range IQR

12,14,15,15,16,19 16-14=2

0,2,2,4,5,10 5-2=3

19-12=7 10-0=10

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4 years ago

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lozanna [386]

In each of the problems, one of the equations is solved for y. Just plug in what y is equal to into the other equation.

1.) Plug -3 in for y.
-3 = 7x - 10
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2.) Plug -8 in for y.
-8 = -2x - 12
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x = -2

9.) Plug -8x in for y.
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5 0

3 years ago

Find the value of the expression when b= 9 c= 5 and d=10 ............ ½d + c² - b

Minchanka [31]

Answer:

1/2d+c^2-b = 21

Step-by-st ep explanation:

= 1/2(10) + 5^2 -9

= 5 + 25 - 9

= 30 - 9

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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

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soldi70 [24.7K]

Answer:

Cos A = x/z

Step-by-step explanation:

You need to memorize the definitions of all the trig functions and then these types of problems become easy! OK?

Cos = adjacent side/hypotenuse

Cos A = x/z

7 0

3 years ago