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3 years ago

There are 454 fuels are there in 12rows

2 answers:

kramer3 years ago

8 0

That means there is about 38 fuels in each row because 454 divided by 12 = 37.83333333333333333 (exact number)

Please mark as brainliest, if you found this helpful.

statuscvo [17]3 years ago

6 0

Answer= 38 (454 divides by 12 is 37,3333333333) round it up to 38.

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Which expression is equal to 2X^2+7<br> (Simplify)

almond37 [142]

This expression cannot be simplified any further.

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3 years ago

Heather buys a dog bowl priced at $8. If the sales tax is 10%, how much tax will Heather<br> pay?

Fantom [35]

Answer:

I think the answer is 2 dollars

Step-by-step explanation:

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3 years ago

Which of the following options is a root of the function? f(x) = (3x + 5) (x2 - 6x + 9)2

Alika [10]

Answer:

The roots of the function are x = -5/3 and 3. Select these.

Step-by-step explanation:

The roots of the function are the solutions to the function. They are also called the x-intercepts or zeros. To find them, factor the function completely and solve using the zero product property.

f(x) = (3x+5)(x2-6x+9)2 becomes f(x) = (3x+5)(x-3)^2

Set each factor equal to 0 and solve for x.

(3x+5)=0

3x = -5

x = -5/3

(x-3)^2 = 0

x - 3=0

x = 3

3 0

3 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

3 years ago

A square base a side length of x+3. What is its area

zepelin [54]

Well it would be (x+3)(x+3)
using the FOIL method you end up with:
x^2 + 6x + 9

:D

6 0

3 years ago