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2 years ago

Simplify 4^3 • 4^5 a 4^8 b 16•15 c 4^15 d 16^8

Mathematics

1 answer:

user100 [1]2 years ago

3 0

Answer:

4^8

Step-by-step explanation:

keep base (which is 4) and add exponents when multiplying

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Determine whether the function below is an even function, an odd function, both, or neither.

Ad libitum [116K]

A function is even if, for each x in the domain of f, f (- x) = f (x). The even functions have reflective symmetry through the y-axis. A function is odd if, for each x in the domain of f, f (- x) = - f (x). The odd functions have rotational symmetry of 180º with respect to the origin.
f (x) = (x + 5) ^ 2
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f (x) = x ^ 2 - 10x + 25
the function does not comply with the definitions.
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answer
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2 years ago

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36x^3a+45xa^2=9xa( ) complete the statement.

Dafna1 [17]

9xa is the common factor.

You can complete the statement by dividing the two terms of the expression of the left side by 9xa, and you will get:

36x^3a + 45xa^2 = 9xa ( 4x^2 + 5a)

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3 years ago

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2 years ago

Solve the system by substitution.<br> -2x-10y=-2<br> -4y=x

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2 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

2 years ago