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3 years ago

I need to 2,4,6,8 But please send me the work too

Mathematics

1 answer:

Ugo [173]3 years ago

4 0

ok for number 2.) 2+8+1+5+1+3+1+7+5+4+3+1=41 and 41 divided by the number of number which is 12= 3.41666666667

the mean for number 2 is 3.41666666667.

for the median you put the number from least to greatest, like so:

1, 1, 1, 1, 2, 3, 3, 4, 5, 5, 7, 8.

The middle is 3 and 3 so 3+3 divided by 2 is 6 so the median is 6.

I cant do mode but range is 8-1 divided by 2

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Builders are covering a wall that is 4 meters tall and 7 meters wide with bricks. They will cover the entire wall except for a s

marin [14]

Answer:

24 m^2

Step-by-step explanation:

There are two areas to consider here: first, the area of the wall, and second, the area of the window. We answer this question by finding these two areas and subtracting the area of the window from the area of the wall.

Wall area: (4 m)(7 m) = 28 m^2

Window area: (2 m)(2 m) = 4 m^2

Net area of wall, to be covered in brick:

28 m^2 - 4 m^2 = 24 m^2

3 0

3 years ago

Y=4x. Y=40. What does x equal

egoroff_w [7]

Answer:

Your answer is 10!

Y = 4x

(40) = 4x

40 = 4(10)

4 0

2 years ago

Read 2 more answers

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).