All categories

2 years ago

Subtract. (2p^2+q)-(p^2+q) Is the answer P^2 P^2+2q P^2-2q 3p^2

Mathematics

1 answer:

Fofino [41]2 years ago

4 0

Answer:

-p5q2 - p3q5 + 2p3 + 2pq2 + 2q3

____________________________

p3q2

Step-by-step explanation:

Simplify ——

Equation at the end of step 1 :

(p+q) 2

(((2•—————)-(q3))+——)-p2

(p3) q2

Step 2 :

p + q

Simplify —————

p3 (p + q) 2

(((2 • ———————) - q3) + ——) - p2

p3 q2

Step 3 : 2 • (p + q) 2

((——————————— - q3) + ——) - p2

p3 q2 q3 q3 • p3

q3 = —— = ———————

1 p3

You might be interested in

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st

Alex

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that $n = 15, p = \frac{1}{5} = 0.2$

What is the probability that he answers at least 10 questions correctly?

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001$

$P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011$

$P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0$

$P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0$

$P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0$

$P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0$

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111$

0.0111% probability that he answers at least 10 questions correctly

3 0

2 years ago

Ryan keeps his loose change in a milk jug. The jug contains 15 pennies, 10 nickels, 35

Rudik [331]

56% chance it won’t be a dime

5 0

3 years ago

Find the first three terms in the expansion , in ascending power of x , of (2+x)^6 and obtain the coefficient of x^2 in the expa

Nataly_w [17]

Answer:

The first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

coefficient of $x^{2}$ in the expansion of $(2+x - x^{2})^{6}$ = (240 - 192) = 48

Step-by-step explanation:

$(2+x)^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times x^{k} \times 2^{6 - k})$

= $6_{C_{0}} \times x^{0} \times 2^{6} + 6_{C_{1}} \times x^{1} \times 2^{5} + 6_{C_{2}} \times x^{2} \times 2^{4}$ + terms involving higher powers of x