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3 years ago

What is the area of the figure?

Mathematics

1 answer:

muminat3 years ago

7 0

I don't know for sure but I think the answer is 84 because it is 7×12

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If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No

Effectus [21]

If A and B are independent, then $P(A\cap B)=P(A)P(B)$ .

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$

$P(A\cup B)=0.5+0.2-0.5\cdot0.2$

$\boxed{P(A\cup B)=0.6}$

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

$P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)$

$P(A^c\cap B^c)=1-0.6$

$\boxed{P(A^c\cap B^c)=0.4}$

c. DeMorgan's law can be used here too:

$P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)$

$P(A^c\cup B^c)=1-0.5\cdot0.2$

$\boxed{P(A^c\cup B^c)=0.9}$

4 0

2 years ago

Can some one please help me <br> will give 94 points for it

NNADVOKAT [17]

You can solve this by using "similar triangles".

In triangle ABC, we are looking for side AC which is x. Side AC is similar to side DF in triangle EDF.

You can solve for side x by picking two sides in triangle ABC and their corresponding sides in triangle EDF. This is what I mean:
$\frac{AC}{BC} = \frac{DF}{EF}$
Substitute for the values of AC, BC, DF and EF:
$\frac{x}{4} = \frac{11}{8} \\ \\ 8x = 4 \times 11 \\ \\ x = \frac{44}{8}$
$x = 5.5 \: units$
To solve for y, do the same thing. Pick two sides on triangle ABC and their corresponding sides in triangle DEF.
$\frac{AB}{BC} = \frac{DE}{EF}$
Substitute for the values and solve:
$\frac{3}{4} = \frac{y}{8}$
$4y = 24 \\ \\ y = 6 \: units$

We have the value x to be 5.5 units and y to be 6 units.

7 0

3 years ago

Read 2 more answers

I just need help on number 14. Thank you very much.

lisabon 2012 [21]

I believe you only need to know one angle.

For example, if you know angle 1, you can calculate angle 3. Angle 2 = angle 3 and angle 4 = angle 1.

Also, Angle 5= angle 1 and so on...

4 0

3 years ago

Help pleaseeeeeeeeeeeeeeeeeeeeeee

bixtya [17]

Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

<u>Step-by-step explanation:</u>

(1) (12, 18, 27, ...)

The common ratio is:

$r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$

3 0

3 years ago

What’s the correct answer for this?

Yanka [14]

The answer would be 13.9 kg of gravel used per square meter

6 0

2 years ago

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