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3 years ago

2. Which algebraic expression represents the following phrase?

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

3 0

The algebraic expression which represents the square of the difference of s and 6 is (s - 6)²

Step-by-step explanation:

Let us represent some words by mathematics expressions

Square a number ⇒ x²
Square the sum of two numbers ⇒ (x + y)²
Sum of the squares of two numbers ⇒ x² + y²
Square of difference of two numbers (x - y)²
The difference of square two numbers x² - y²

∵ The expression is the square of the difference of s and 6

- That means find the difference between s and 6 at first,

then square the difference

∵ The difference of s and 6 = s - 6

- Square this difference means but the difference in a bracket

and then square the bracket

∴ The square of the difference = (s - 6)²

The algebraic expression which represents the square of the difference of s and 6 is (s - 6)²

Learn more:

You can learn more about the algebraic expressions in brainly.com/question/10771256

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Answer:
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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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Answer:

The slope is 6/5 and the y intercept is -13/5

Step-by-step explanation:

6x-5y=13

Solve for y

Subtract 6x from each side

6x-6x-5y=-6x+13

-5y = -6x+13

Divide by -5

-5y/-5 = -6x/-5 +13/-5

y = +6/5x -13/5

This is in slope intercept form y = mx+b where m is the slope and b is the y intercept

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Answer:

Step-by-step explanation:

Givens

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Let the third piece = 5x

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Solution

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Answer

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