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4 years ago

How would you do number 12 and 15?

Mathematics

1 answer:

Gnesinka [82]4 years ago

7 0

Answer:

$x_1=x_2=-\dfrac{2}{3}$

Step-by-step explanation:

For the quadratic equation $ax^2+bx+c=0$ the discriminant is defined as

$D=b^2-4ac$

and the quadratic formula for the roots gives us two roots:

$x_1=\dfrac{-b-\sqrt{D}}{2a}$

and

$x_2=\dfrac{-b+\sqrt{D}}{2a}$

For the equation $9x^2 +12x+4=0$ use quadratic formula to find roots:

$D=12^2-4\cdot 9\cdot 4=144-144=0$

So,

$x_1=x_2=\dfrac{-12\pm \sqrt{0}}{2\cdot 9}=-\dfrac{12}{18}=-\dfrac{2}{3}$

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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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3 years ago

Which congruence statement is true.!?

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$\\ \triangle ABC \cong \triangle DFE \\$

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3 years ago

Six distinct integers are picked from the set {1, 2, 3,…, 10}. How many selections are there, in which the second smallest integ

Ksivusya [100]

Answer:

1680 ways

Step-by-step explanation:

Total number of integers = 10

Number of integers to be selected = 6

Second smallest integer must be 3. This means the smallest integer can be either 1 or 2. So, there are 2 ways to select the smallest integer and only 1 way to select the second smallest integer.

2 ways 1 way

Each of the line represent the digit in the integer.

After selecting the two digits, we have 4 places which can be filled by 7 integers. Number of ways to select 4 digits from 7 will be 7P4 = 840

Therefore, the total number of ways to form 6 distinct integers according to the given criteria will be = 1 x 2 x 840 = 1680 ways

Therefore, there are 1680 ways to pick six distinct integers.

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4 years ago

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Ratling [72]

Answer:

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