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3 years ago

What is the interquartile range of this data?

1 answer:

6 0

Hello!

To find the interquartile range you subtract the lower quartile by the higher quartile

33 - 25 = 8

The answer is C)8

Hope this helps!

You might be interested in

Show that cos2x=cosx

PSYCHO15rus [73]

cos (2x) = cos x

2 cos^2 x -1 = cos x using the double angle formula

2 cos ^2 x -cos x -1 =0

factor

(2 cos x+1) ( cos x -1) = 0

using the zero product property

2 cos x+1 =0 cos x -1 =0

2 cos x = -1 cos x =1

cos x = -1/2 cos x=1

taking the arccos of each side

arccos cos x = arccos (-1/2) arccos cos x = arccos 1

x = 120 degrees x=-120 degrees x=0

remember you get 2 values ( 2nd and 3rd quadrant)

these are the principal values

now we need to add 360

x = 120+ 360n x=-120+ 360n x = 0 + 360n where n is an integer

3 0

3 years ago

1) A business man gives N18 000,00 to his three children

pashok25 [27]

Answer: The least share is N450000

Step-by-step explanation:

Given that :

Amount shared = N18 00000

Sharing ratio = 5:4:3

Amount shared by each child :

Total ratio = (5 + 4 + 3 )= 12

First share :

(5 / 12) × 1800000 = 750000

Second share :

(4/12) × 1800000 = 600000

Third share:

(3/12) × 1800000 = 450000

The least amount of the three shares is :

N450000

4 0

3 years ago

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

likoan [24]

Answer:

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

Step-by-step explanation:

Consider, the given Quadratic equation, $x^2+4=6x$

This can be written as , $x^2-6x+4=0$

We have to solve using quadratic formula,

For a given quadratic equation $ax^2+bx+c=0$ we can find roots using,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ...........(1)

Where, $\sqrt{b^2-4ac}$ is the discriminant.

Here, a = 1 , b = -6 , c = 4

Substitute in (1) , we get,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1 \cdot (4)}}{2 \cdot 1}$

$\Rightarrow x=\frac{6\pm\sqrt{20}}{2}$

$\Rightarrow x=\frac{6\pm 2\sqrt{5}}{2}$

$\Rightarrow x={3\pm \sqrt{5}}$

$\Rightarrow x_1={3+\sqrt{5}}$ and $\Rightarrow x_2={3-\sqrt{5}}$

We know $\sqrt{5}=2.23607$ (approx)

Substitute, we get,

$\Rightarrow x_1={3+2.23607}$ (approx) and $\Rightarrow x_2={3-2.23607}$ (approx)

$\Rightarrow x_1={5.23607}$ (approx) and $\Rightarrow x_2=0.76393}$ (approx)

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

7 0

3 years ago

Read 2 more answers

According to the rational root theorem, the numbers below are some of the potential roots of f(x)=10x^3+29x^2-66x+27. Select all

Reil [10]

Answer:

-9/2, 3/5, 1

Step-by-step explanation:

Those are the answers on edginuity.

6 0

3 years ago

56% of 50 is what number

noname [10]

Divide 100 to 56 = 56/100= 0.56 then you wanna times it by 50 so then it would be.56x50=28

5 0

3 years ago

Read 2 more answers