All categories

3 years ago

x + 3y = 7 x - 3y = 1 Solve the system of equations. A) x = 4, y = 1 B) x = 1, y = 4 C) x = - 2 3 , y = 3 D) x = 3, y = - 2 3

2 answers:

6 0

X+3y=7
x-3y=1
add them together

x+3y=7
x-3y=1 +
2x+0y=8

2x=8
divide 2
x=4
subsitute
x+3y=7
4+3y=7
subtract 4
3y=3
divid 3
y=1

x=4
y=1

answer is A

Bumek [7]3 years ago

6 0

X + 3y = 7
x - 3y = 1

x + 3y = 7
- 3y -3y

x = 7 - 3y

x - 3y = 1
7 - 3y - 3y = 1
7 -6y = 1
-7 -7

-6y = -6
y = 1

Now solve for x

X + 3y = 7
x + 3 = 7
x = 4

Now we have our values:-
x = 4 and y = 1

Answer: A) x = 4, y = 1

You might be interested in

The sum of five times a number and -6 is -2

k0ka [10]

5n + (-6) = -2
add 6 to -2
5n = -2 + 6
5n = 4
divide both sides by 5
n = 4/5

4 0

3 years ago

Read 2 more answers

The first term of a geometric series is 3, and the sum of the first term and the second term is 15. What is the sum of the first

Ivan

Sum of n terms in a geometric progression is given by Sn = a1 (1 - r^n) / (1 -r)

3( 1 - 4^6) / (1-4)

= 4,095

5 0

3 years ago

A cylindrical pressure vessel has a height of 9 feet and a diameter of 6 feet. How much gas can the pressure vessel hold when fu

notka56 [123]

since the diameter of the base of the cylinder is 6 feet, then its radius is half that, or 3 feet.

$\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h~~
\begin{cases}
r=radius\\
h=height\\[-0.5em]
\hrulefill\\
r=3\\
h=9
\end{cases}\implies V=\pi (3)^2(9)\implies V=81\pi$

5 0

3 years ago

Read 2 more answers

please help!! if you’re good at these i have a few more of them can you continue to help me i’m begging !!

olga_2 [115]

Answer:

Step-by-step explanation:

y = 1/2x + 3

7 0

3 years ago

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used 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.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

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

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

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

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

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0

3 years ago