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anastassius [24]
3 years ago
6

Find the mean, median, and interquartile for the data set below. Round your answer to the nearest tenth. 27,16,8,5,19,14,22,31,1

3,5,23,16,8
Mathematics
1 answer:
posledela3 years ago
6 0
For the mean you add everything up and divide by how many numbers there are so....
15.9 is you mean.
The median is the middle number so...
16 is your median.
Your IQR is 14.5

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Jade buys a blouse and a skirt for 34 of their original price. She pays a total of $31.50 for the two items. If the original pri
solniwko [45]
It’s what the other person said
7 0
2 years ago
Help me please, thank you.
Ivan
The correct answer is -2
4 0
3 years ago
Solve c =(a² + 3b)/4 for b.<br><br>Bruv take all my points I literally blow at math :'(​
Sati [7]

Answer:

b=\frac{4c-a^{2} }{3}

Step-by-step explanation:

First: multiply both sides by 4. 4 times c is 4c and the other sides cancels out as you are doing the inverse operation of division

Then, you have 4c=a^{2} +3b

Subtract both sides by a squared

DO NOT TAKE THE SQUARE ROOT! This is because it is a squared plus 3b so you have to do the inverse of addition

From that you get 4c-a^{2} =3b

Finally, divide both sides by 3.

You get  b=\frac{4c-a^{2} }{3}

4 0
3 years ago
Kobe has collected 750 football cards and 660 baseball cards. He wants to divide them into piles so that each pile has only one
Harrizon [31]

Answer: 30 cards

Step-by-step explanation:

To find the greatest possible number of cards in each pile such that there is the same number of cards in each​ pile we need to find the greatest common factor of 750 and 660.

Using Euclid division method , we have

750=1(660)+90\\\\660=7(90)+30\\\\90=3(30)+0

Hence, the greatest common factor of 750 and 660 = 30

Thus, there will be 30 cards in each pile.

4 0
3 years ago
Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
3 years ago
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