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3 years ago

Evaluate 3a2 + 4b2 for a = 2 and b = 3. O 168 48 O 36

Mathematics

2 answers:

Juliette [100K]3 years ago

8 0

Answer:

Step-by-step explanation:

a=2

b=3

3a^2+4b^2

3(2)^2+4(3)^2

3(2)(2)+4(3)(3)

3(4)+4(9)

12+36

Hope this helps ;) ❤❤❤

podryga [215]3 years ago

7 0

48 done hope this helps

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alisha [4.7K]

Answer: C

Step-by-step explanation:

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3 years ago

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If f(x) = 4x - 12, what is f(2)? O A. -4 O'B. 4 O c. -20 D. 8 Helpppp plzzz

photoshop1234 [79]

Answer:

-4

Step-by-step explanation:

f(x) = 4x - 12

Let x=2

f(2) = 4*2 - 12

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3 years ago

How do you Divide 88 by1/4+20=?

Artyom0805 [142]

88 divided by 1/4 can be placed like this: 88/1 divided by 1/4, which should lead to an answer of 22/1 or 22. Then you add 20, so 22 plus 20 would equal 42. 42 is the correct answer.

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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.

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

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4 years ago

Triangle ABC is congruent to A'BC' by the HL theorem What single rigid transformation maps ABC onto A'BC'?

larisa [96]

Answer:

b. reflection

Step-by-step explanation:

just took the test

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3 years ago

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