Which equation can be used to find the surface area of the cylinder?

An advertising agency notices that approximately 1 in 50 potential buyers of a product sees a given magazine ad, and 1 in 8 sees

zheka24 [161]

Answer:

Probability = 0.12025

Step-by-step explanation:

P (Am) = 1/50 = 0.02 {Magazine ad}

P (At) = 1/8 = 0.125 {Television ad}

P (Am ∩ At ) = 1/100 = 0.01 {Both ads}

P (Am U At) = P (Am) + P (At) - P (Am ∩ At )

= 0.02 + 0.125 - 0.01

P (Am U At) = 0.135 {Person sees either ad}

P (Am' ∩ At') = 1 - P (Am U At)

P (Am' ∩ At') = 1 - 0.135 = 0.865 {Person sees none ad}

Prob (Purchase) = Prob (Purchase with ad) + Prob (purchase without ad)

P (P/ A) = 1/4 = 0.25 , P (P / A') = 1/10 = 0.1

P (P) = (0.25) (0.135) + (0.1) (0.865)

= 0.03375 + 0.0865

0.12025

7 0

2 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

Please help me with this!!

Troyanec [42]

Answer:

As the height increases, the temperature decreases.

It is decreasing 5°C per kilometer.

At the height 8 km the temperature is 0°.

5 0

3 years ago

If you want to know the strength of an earthquake to a _____ ?

dsp73

It should be C. richer scale

4 0

3 years ago

What is the common denominator as a fraction of 1/5

andrezito [222]

The answer would be 30 if u use the gcf

4 0

3 years ago

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