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3 years ago

Is x^2-1 a polynomial

Mathematics

1 answer:

77julia77 [94]3 years ago

4 0

Answer:

Yes it's a binomial.

Step-by-step explanation:

A polynomial is an expression with more than one term that often has variables and exponents on the variable. Since this has two terms separated by a subtraction sign with one term being a variable to the exponent of 2 then this is a polynomial. It is called a binomial. More than 2 terms are referred exclusively as a polynomial with ___ terms.

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Pat's income is 20 % more than Adam. How much percent is Adam's income less than Pat's?

pantera1 [17]

Answer: 20%

Step-by-step explanation: The same number applies here. Think about it this way, let's say you have a friend named Sofia and she has 3 dollars more than you, how many dollars do you have less than Sofia? 3 dollars. That's the easiest way I can explain it.

4 0

3 years ago

4. two lines are perpendicular when

Gre4nikov [31]

c c a a are the right answers just took it

8 0

3 years ago

Read 2 more answers

15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0

3 years ago

5 1/6 + (7 1/3)l <60

agasfer [191]

Answer:

$l$

Step-by-step explanation:

Given the inequality

$5\dfrac{1}{6}+7\dfrac{1}{3}l$

First, convert mixed numbers to improper fractions:

$5\dfrac{1}{6}=\dfrac{5\cdot 6+1}{6}=\dfrac{31}{6}\\ \\7\dfrac{1}{3}=\dfrac{7\cdot 3+1}{3}=\dfrac{22}{3}$

Now the inequality looks like

$\dfrac{31}{6}+\dfrac{22}{3}l$

Multiply it by 6:

$\dfrac{31}{6}\cdot 6+\dfrac{22}{3}l\cdot 6$

7 0

3 years ago

The Taylors have recorded their weekly grocery expenses for the past 12 weeks and determined that the mean weekly expense was $

SVETLANKA909090 [29]

Answer:

$84.80

Step-by-step explanation:

find the difference between accurate price and written price: 74 - 47 = 27

find written total (before Mrs. Taylor found mistake) by undoing the mean calculation: 82.55 x 12 = $990.60

the total expenses is $27 more than what they thought: 990.60 + 27 = 1017.60

divide accurate total by 12 to find mean: 1017.60 / 12 = $84.80

hope this helps :)

4 0

2 years ago