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Sav [38]
2 years ago
5

5.

Mathematics
1 answer:
irina [24]2 years ago
7 0

Answer:

160 m²

384 m²

Step-by-step explanation:

The area of a trapezium is given by :

A = 1/2(a + b)h

h = height ; a and b = lengths

From the diagram :

A = 1/2(12 + 20)10

A = 1/2(32)10

A = 16 * 10

A = 160 m²

2.)

Area of parallelogram = base * height

Base = 24 ; h = 16

A = 24 * 16

A = 384 m²

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24g+40f this confusing
valina [46]

Answer:

16

Step-by-step explanation: because 24g reverts 40f so we have to subtract them

and get the answer

5 0
2 years ago
Use the figure to write and solve an equation for x.
Digiron [165]

Answer:

x = 33

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask :)

4 0
3 years ago
Can 3.65909090909 be expressed as a fraction whose denominator is a power of 10? Explain.
GuDViN [60]
\bf 3.659\textit{ can also be written as }\cfrac{3659}{1000}\textit{ therefore }3.6590909\overline{09}\\\\
\textit{can be written as }\cfrac{3659.0909\overline{09}}{1000}

notice above, all we did, was isolate the "recurring part" to the right of the decimal point, so the repeating 09, ended up on the right of it.

now, let's say, "x" is a variable whose value is the recurring part, therefore then

\bf \cfrac{3659.0909\overline{09}}{1000}\qquad \boxed{x=0.0909\overline{09}} \qquad \cfrac{3659+0.0909\overline{09}}{1000}\implies \cfrac{3659+x}{1000}

now, the idea behind the recurring part is that, we then, once we have it all to the right of the dot, we multiply it by some power of 10, so that it moves it "once" to the left of it, well, the recurring part is 09, is two digits, so let's multiply it by 100 then, 

\bf \begin{array}{llllllll}
100x&=&09.0909\overline{09}\\
&&9+0.0909\overline{09}\\
&&9+x
\end{array}\quad \implies 100x=9+x\implies 99x=9
\\\\\\
x=\cfrac{9}{99}\implies \boxed{x=\cfrac{1}{11}}\\\\
-------------------------------\\\\
\cfrac{3659.0909\overline{09}}{1000}\qquad \boxed{x=0.0909\overline{09}} \quad \cfrac{3659+0.0909\overline{09}}{1000}\implies \cfrac{3659+x}{1000}
\\\\\\
\cfrac{3659+\frac{1}{11}}{1000}

and you can check that in your calculator.
8 0
3 years ago
Someone please help meeeee
zepelin [54]

Answer:

The number is -1

Step-by-step explanation:

The equation for the first part is (2n+7)2. This is because "twice a number" is 2n and "7 is added" will make it 2n+7. The "sum is multiplied by two" Is multiplying everything by two. This will make it (2n+7)

The second part is  -8n+2. Since a number is multiplied by negative eight it will be -8n. Then it is added by two.

Both equation equal each other the overall equation will be (2n+7)2=-8n+2

You will distribute. making it 4n+14=-8n+2

Add 8n to both sides making it 12n+14=2

subtract 14 from both sides making it 12n=-12

Divide both sides by 12 . making it number=-1

4 0
3 years ago
For each given p, let ???? have a binomial distribution with parameters p and ????. Suppose that ???? is itself binomially distr
pshichka [43]

Answer:

See the proof below.

Step-by-step explanation:

Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."

Solution to the problem

For this case we can assume that we have N independent variables X_i with the following distribution:

X_i Bin (1,p) = Be(p) bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:

Z = \sum_{i=1}^N X_i

From the info given we know that N \sim Bin (M,q)

We need to proof that Z \sim Bin (M, pq) by the definition of binomial random variable then we need to show that:

E(Z) = Mpq

Var (Z) = Mpq(1-pq)

The deduction is based on the definition of independent random variables, we can do this:

E(Z) = E(N) E(X) = Mq (p)= Mpq

And for the variance of Z we can do this:

Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2

Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2

And if we take common factor Mpq we got:

Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]

And as we can see then we can conclude that   Z \sim Bin (M, pq)

8 0
3 years ago
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