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3 years ago

D equals f plus fx, solve the equation for x

Mathematics

1 answer:

baherus [9]3 years ago

7 0

Answer:

question says D = f + fx,

then,

D = 2fx

so,

D/2f = x

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What is the coefficient of -7x-3x^2+5 A~7 B~3 C~-3 D~-7

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PLZ HELP 20 POINTS

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Pani-rosa [81]

Answer:

d) 28.472

Step-by-step explanation:

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3 years ago

Of all the soft drink consumers in a particular sales region, 30% prefer Brand A and 70% prefer Brand

victus00 [196]

Answer:

Option D -0.67

Step-by-step explanation:

Given : Preference of Brand A =30% ⇒ $P(A)=\frac{30}{100}=0.3$

Preference of Brand B =70% ⇒ $P(B)=\frac{70}{100}=0.7$

Prefer Brand A and are Female = 20% ⇒ $P(A/F)=\frac{20}{100}=0.2$

Prefer Brand B and are Female = 40% ⇒ $P(B/F)=\frac{40}{100}=0.4$

To find : Selected consumer is female, given that the person prefers Brand A $P(F/A)$

Solution : Using Bayes' theorem, which state that

$P(A/B)=\frac{P(B/A)P(A)}{P(B)}$

where, P(A) and P(B) are probabilities of observing A and B.

P(B/A)= is a conditional probability where event B occur and A is true

P(A/B)= also a conditional probability where event A occur and B is true.

Now, applying Bayes' theorem,

$P(F/A)=\frac{P(A/F)P(A)}{P(B)P(B/F)+P(A)P(A/F)}$

$P(F/A)=\frac{(0.2)(0.3)}{(0.7)(0.4)+(0.3)(0.2)}$

$P(F/A)=\frac{0.6}{0.28+0.6}$

$P(F/A)=\frac{0.6}{0.88}$

$P(F/A)=0.68$

Therefore, Option D is correct probability that a randomly selected consumer is female, given that the person prefers Brand A -0.67

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3 years ago

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Divide 12x 4 y 4 by 3x 2 y. A. 4x 6y 5 B. 4x 2y 3 C. 9x 6 D. 9x 2y 3

andrew-mc [135]

Answer:

$\frac{12x^4y^4}{3x^2y}=4x^2y^3$

Step-by-step explanation:

Consider an element $a\neq 0$

We know that $\frac{a^m}{a^n}=a^{m-n}$

Here, m and n are exponents and a is a base. Exponent refers to the no. of times a term is multiplied by itself. Another word for exponents is powers.

Other Properties of exponents:

$a^m\times a^n=a^{m+n}\\a^0=1\\\left ( a^m \right )^n=a^{mn}$

Here, we are required to divide $12x^4y^4$ by $3x^2y$ .

Consider $\frac{12x^4y^4}{3x^2y}$

We will solve $\frac{12}{3}\,,\,\frac{x^4}{x^2}\,,\,\frac{y^4}{y}$ separately.

Here,

$\frac{12}{3}=4\\\frac{x^4}{x^2}=x^{4-2}=x^2\\\frac{y^4}{y}=y^3$

So, $\frac{12x^4y^4}{3x^2y}=4x^2y^3$

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3 years ago