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2 years ago

4n g(n) = 3n² +41 g h(n) = 2n - 5 = Find (goh)(4)

Mathematics

1 answer:

xxTIMURxx [149]2 years ago

3 0

g * h = 6n squared + 77

6*4^2 + 77 = 125

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Solve and reduce to lowest terms:<br> 6 / 11 x 1 / 3

zysi [14]

Answer:

2/11

Step-by-step explanation:

6/11 x 1/3 =

6/33 =

2/11

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

5 0

3 years ago

Read 2 more answers

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

3 years ago

The number of fans who attended a softball game increased from 1,200 for the first game to 1,500 for the second game. How many f

erma4kov [3.2K]

1875 fans will attend

6 0

3 years ago

Plz help ill give you brainlist

Vladimir [108]

It's 6.9. Just do the circumference divided by pi

6 0

3 years ago

Read 2 more answers

Se quiere construir un muro de 4 m de alto, 12 m de largo y 10 cm de espesor. ¿Cuántos ladrillos de 8 cm de alto, 20 cm de largo

Llana [10]

Answer:

3000

Step-by-step explanation:

Let's start by finding the volume of the wall. The volumen of the wall can be considered as the volume of a rectangular prism. The volume of a rectangular prism is given by:

$V_w=w*l*h\\\\Where:\\\\w=Width=10cm=0.1m\\l=Length=12m\\h=Height=4m$

So the volume of the wall is:

$V_w=0.1*12*4=4.8m^3$

Now, we can find the volume of the brick using the same method since a brick can be considered as a rectangular prism as well:

$V_b=w*l*h\\\\For\hspace{3}the\hspace{3}brick\\\\w=10cm=0.1m\\l=20cm=0.2m\\h=8cm=0.08m$

Hence:

$V_b=(0.1)*(0.2)*(0.08)=0.0016m^3$

In order to know how many bricks are required to build the wall, we just need to fill the wall volume with the number of bricks of this volume. So:

$V_w=nV_b\\\\Where\\\\n=Number\hspace{3}of\hspace{3}bricks$

Solving for n:

$n=\frac{V_w}{V_b} =\frac{4.8}{0.0016} =3000$

Therefore, we need 3000 bricks to build that wall.

Translation:

Comencemos por encontrar el volumen del muro. El volumen del muro puede considerarse como el volumen de un prisma rectangular. El volumen de un prisma rectangular viene dado por:

$V_w=w*l*h\\\\Donde:\\\\w=Espesor=10cm=0.1m\\l=Largo=12m\\h=Alto=4m$

Entonces el volumen del muro es:

$V_w=0.1*12*4=4.8m^3$

Ahora, podemos encontrar el volumen del ladrillo utilizando el mismo método, ya que un ladrillo también puede considerarse como un prisma rectangular:

$V_b=w*l*h\\\\Para\hspace{3}el\hspace{3}ladrillo\\\\w=10cm=0.1m\\l=20cm=0.2m\\h=8cm=0.08m$

Por lo tanto:

$V_b=(0.1)*(0.2)*(0.08)=0.0016m^3$

Para saber cuántos ladrillos se requieren para construir el muro, solo necesitamos llenar el volumen del muro con la cantidad de ladrillos de este volumen. Entonces:

$V_w=nV_b\\\\Donde\\\\n=Numero\hspace{3}de\hspace{3}ladrillos$

Resolviendo para n:

$n=\frac{V_w}{V_b} =\frac{4.8}{0.0016} =3000$

Por lo tanto, necesitamos 3000 ladrillos para construir ese muro.

4 0

3 years ago