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3 years ago

Which of the following is a binomial? (2 points) c2 + c + 6 c2 − 16 −8c c3 + 4c2 − 12c + 7

2 answers:

7 0

In maths, for an expression to be called a binomial, this expression must:
1- have only two terms
2- exponents of the terms must only be positive integers (not fractions or negative integers)
3- if the two terms have the same variable, then the exponent of the variable in each term must be different from that of the other term

Note:
an expression with only one term is called monomial while an expression with only three terms is called trinomial

Comparing the definition of binomial to the given choices, we will find that the correct choice is:
c^2 - 16

Svet_ta [14]3 years ago

7 0

The binomial is c2-16

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Find and simplify the difference quotient of the form, <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh

Anastasy [175]

With $f(x)=-10x+9$ , we have

$f(x+h) = -10(x+h) + 9 = -10x + 9 - 10h$

Then the difference quotient is

$\dfrac{f(x+h)-f(x)}h = \dfrac{(-10x+9-10h)-(-10x+9)}h = \dfrac{-10h}h = -10$

since <em>h</em> ≠ 0.

7 0

3 years ago

Prove sin^2A + cos^2A = 1

Nady [450]

Step-by-step explanation:

${sin}^{2} A + {cos}^{2} A = 1 \\ lhs = {sin}^{2} A + {cos}^{2} A \\ = {sin}^{2} A + 1 - {sin}^{2} A \\ = 1 \\ = rhs \\ \therefore \: {sin}^{2} A + {cos}^{2} A = 1 \\$

8 0

3 years ago

Prove that u(n) is a group under the operation of multiplication modulo n.

katrin2010 [14]

Answer:

The answer is the proof so it is long.

The question doesn't define u(n), but it's not hard to guess.

Group G with operation ∘

For all a and b and c in G:

1) identity: e ∈ G, e∘a = a∘e = a,

2) inverse: a' ∈ G, a∘a' = a'∘a = e,

3) closed: a∘b ∈ G,

4) associative: (a∘b)∘c = a∘(b∘c),

5) (optional) commutative: a∘b = b∘a.

Define group u(n) for n prime is the set of integers 0 < i < n with operation multiplication modulo n.

If n isn't prime, we exclude from the group all integers which share factors with n.

Identity: e = 1. Clearly 1∘a = a∘1 = a. (a is already < n).

Closed: u(n) is closed for n prime. We must show that for all a, b ∈ u(n), the integer product ab is not divisible by n, so that ab ≢ 0 (mod n). Since n is prime, ab ≠ n. Since a < n, b < n, no factors of ab can equal prime n. (If n isn't prime, we already excluded from u(n) all integers sharing factors with n).

Inverse: for all a ∈ u(n), there is a' ∈ u(n) with a∘a' = 1. To find a', we apply Euclid's algorithm and write 1 as a linear combination of n and a. The coefficient of a is a' < n.

Associative and Commutative:

(a∘b)∘c = a∘(b∘c) because (ab)c = a(bc)

a∘b = b∘a because ab = ba.

5 0

3 years ago

Read 2 more answers

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago

Un rollo de cable mide más de 150 metros y menos de 200 metros. ¿Cuál es su longitud exacta, sabiendo que se puede dividir en tr

lapo4ka [179]

Answer:

<h2>The cable measures 180 meters.</h2>

Step-by-step explanation:

To solve this problem, we need to find the Least Common Factor between 15 and 9.

15 9 | 3

5 3 | 3

5 1 | 5

So, the Least Common Factor would be 3x3x5 = 45. But, we need a multiple of this number which must be between 150 and 200. So,

45 x 4 = 180.

Therefore, the cable measures 180 meters.

5 0

3 years ago