2 [-3x+2y=8]
6x-2y=-20

-6x + 4y = 16
6x - 2y = -20
2y = -4
y = -2

use the first eq (u can use any) then substitute the value of y

-3x + 2(-2) = 8
-3x -4 = 8
-3x = 12
x = -4

6 0

3 years ago

The sum of the digits of a two-digit number is 5. If the number is multiplied by 3, the result is 42. Write and solve a system o

Elenna [48]

A two digit number has a tens digit and a ones digit.
Let's say x = tens digit and y = ones digit

"The sum of the digits is 5"
x + y = 5

The next phrase is "the number multiplied by 3 is 42" but we need to represent the number using the digits. So they need to be multiplied first by their place value and added together. [Example: 34 = 3(10) + 4(1)]

The number is: 10x + y
3(10x + y) = 42

The system of equations: (two equations for two unknowns)
x + y = 5
30x + 3y = 42

Then you can use substitution or elimination to combine and solve.
I'll use elimination, multiply the entire top equation by -3 and add the equations together. y will cancel out

-3x - 3y = -15
30x + 3y = 42
------------------
27x + 0 = 27
x = 1

then plug x = 1 into either equation to find y

1 + y = 5
y = 4

remember the x and y represent digits so the number xy is 14

6 0

3 years ago

Solve for x <br>40 = x/ -4/7<br>

Lena [83]

Answer:

$40 = \frac{x}{ - \frac{4}{7} } \\ 40 \times ( - \frac{4}{7} ) = x \\ x = - \frac{160}{7}$

3 0

2 years ago

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

myrzilka [38]

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that 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.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is $p = \frac{1}{5} = 0.2$

10 questions.

This means that $n = 10$

A) What is the probability that he will answer all questions correctly?

This is $P(X = 10)$

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

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024$

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so $P(X = 0)$

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

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

Since $P(X = 0) = 0.1074$ , from item b.

$P(X \geq 1) = 1 - 0.1074 = 0.8926$

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264$

$P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055$

$P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008$

$P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001$

$P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0$

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328$

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

8 0

3 years ago