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Kruka [31]
3 years ago
11

How is the interquartile range calculated?

Mathematics
1 answer:
eimsori [14]3 years ago
8 0
C) Quartile 3 (q3) minus Quartile 1 (q1)
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If sin theta = 2/3 and tan theta < 0, what is the value of cos theta?
maria [59]

Here sin theta is positive and tan theta is negative , it means theta is in second quadrant where cos is negative too .

sin \Theta = 2/3 , cos \theta = - \sqrt{1-sin^2 \Theta} =- \sqrt{1-4/9}=-\sqrt{5}/3

So the correct option is the third option .

5 0
3 years ago
Read 2 more answers
To find the average of a set of numbers, add up all the items and divide by?
kati45 [8]

Answer:

you add up all of the items and then divide by the total number of items.

Step-by-step explanation:

For example,

Total/Number of Items = Average

Set: 1,6,4,5

(1+6+4+5)/4

16/4 = 4

4 is the average of the items above.

Good luck!!

4 0
3 years ago
Write the equation for what is 15% of 24 of what number. please include the equation too ​
matrenka [14]
Sorry i don’t know the answer maybe some one else knows
4 0
3 years ago
Solve the system y = -x + 7 and y = 0.5(x - 3)^2
ddd [48]

Answer:

x=-1;\ y=8\\x=5;\ y=2

Step-by-step explanation:

You can make both equations equal and solve for the variable x, as you can see below:

-x+7=0.5(x-3)^2

Keep on mind that:

(a-b)^2=a^2-2ab+b^2

Then:

-x+7=0.5(x-3)^2\\-x+7=0.5(x^2-2(x)(3)+3^2)\\-x+7=0.5x^{2}-3x+4.5\\0=0.5x^2-2x-2.5

Apply the quadratic formula:

x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}\\\\x=\frac{-(-2)+/-\sqrt{(-2)^2-4(0.5)(-2.5)}}{2(0.5)}\\x=-1\\x=5

Substitute each value into one of the originl equations to find y, then:

y=-(-1)+7=8\\y=-5+7=2

8 0
3 years ago
zack flips a coin and rolls number cube with sides labeled 1 to 6 . What is the probability that he gets heads and a number grea
maria [59]
In order to calculate the probability of this, you need to multiply the probability of each event. The chance of him rolling heads is 1/2, the chance of him getting a 4 is 1/6. Multiply them together to get 1/12 or 8.3333...%
Hope you understand!

6 0
3 years ago
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