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Scorpion4ik [409]

3 years ago

8

On day one, a bacteria colony begins with 20 bacteria and the number of bacteria triples each day. Which explicit equation can b

e used to determine the number of bacteria after n days?
A. a n = 3(20)n
B. a n = 20(3)n
C. a n = 3(20)n – 1
D. a n = 20(3)n – 1

1 answer:

zmey [24]3 years ago

5 0

The answer is letter a

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Please help me out!

marishachu [46]

Answer: g = -4 and h =1 or g= -1 and h = -2

Step-by-step explanation:

If two exponents have the same base and are multiplied together, you will keep the base unchange and add the exponents.

So looking at the equation, the base is 2 for both and they are raised to the powers g and h, meaning that if g and h are added together they have to equal -3.

So find two numbers that their sum is -3.

For example, -1 and -2 add up to -3, -4 and 1 add up to -3.

4 0

3 years ago

14c^3d^2-21c^2d^3/14cd

levacccp [35]

14c³d²-21c²d³/14cd

= (196c^4d³-21c²d³)/ 14cd

= (196c³d²-21cd²)/14

= (28c³d²-3cd²)/2

I hope that's help !

3 0

3 years ago

The sum of twice a<br> number and 31

Andreyy89

Answer:

Let the unknown number be x ;

$2(x) +31\\=2x+31$

Step-by-step explanation:

3 0

3 years ago

(2m-3)(2m+3)(4m^2+9)

fredd [130]

Answer:

16 $m^{4}$ −81

Step-by-step explanation:

Hope this is correct and it helps :)

6 0

2 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

So

$P(X \geq 2) = 0.93803$

5 0

2 years ago