All categories

3 years ago

Solve the equation 3x-4y=16 for x

Mathematics

2 answers:

dolphi86 [110]3 years ago

5 0

Hi there!

One little tip of advice I would like to give you is that when solving equations with 2 variables, and you're looking for only one, the second will always end up in your answer. So, since we're solving for x, the y will end up in our equation. :D

Let's begin!

$3x - 4y = 16$

Step 1) Add 4y to both sides.

$3x - 4y + 4y = 16 + 4y 3x = 4y + 16$

Step 2) Divide both sides by 3.

$\frac{3x}{3} = \frac{4y + 16}{3}$

$x = \frac{4}{3} y + \frac{16}{3}$

Final answer -

$x = \frac{4}{3} y + \frac{16}{3}$

If you wanted to see this graphed, however, see the attached file. :)

Hope this helps!

Message me if you need anything else! I'd be happy to help! :D

Alex Ar [27]3 years ago

4 0

3x-4y=16
add 4y to both sides
3x=4y+16
divide both sides by 3
x=(4y+16)/3

You might be interested in

Which number produces a rational number when added to 0.25

Sonja [21]

Answer:

the number that produces a rational number when added to 0.25 is 2/9

3 0

3 years ago

Simplify 79.5-2.58 x (3.63-1.63) to the third power

trapecia [35]

Answer:

Step-by-satep explanation:

3 0

3 years ago

alex spent 3/7 of his money he gave 1 /4 of the remainder to his sistet he haf $120 left how much money did he in the beginning

Sonbull [250]

Let his total money be x.
He spent (3/7)*x of his money.
He gave 1/4(x-3/7x) to his sister.
which is : (1/4)*(4/7)*x = (1/7)*x
Now,
How much money he have in the beginning =
(3/7)*x+(1/7)*x+120=x
(4/7)*x+120=x
120=x-(4/7)*x
120=(3/7)*x
x=120*7/3 [cross multiplication]
x=280.

So, Alex had $280 in the beginning.

6 0

3 years ago

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

2 years ago

What's the range of 37000 and 45000

Galina-37 [17]

Range: Subtract <span>37000 from 45000 and you'll have it.</span>

8 0

3 years ago