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Natali [406]
3 years ago
6

Determine the value of n in the equation (n – 4) – 3 = 3 – (2n + 3).

Mathematics
2 answers:
VARVARA [1.3K]3 years ago
8 0
(N-4)-3=3-(2n+3)
•distribute the negative symbol into the parentheses on the right side of the equation and combine like terms. This would leave you with n-7= -2n. Bring the variables to the left and numbers to the right.
3n= 7
•divide 3 from both sides of the equation
N= 7/3 or N= 2.3
nlexa [21]3 years ago
3 0
Since the equation is (n-4)-3 = 3 - (2n+3):

By distributive property:
n-4-3 = 3 -2n -3

Variables on the right side, while constants on the left side: 
n+2n = 3
3n = 3

Divide both sides by 3:
n = 3/3
n = 1

Therefore, the value of n in the expression is 1.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

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7 and 1/5 -2 and 3/5 equal
Furkat [3]
7 1/5 - 2 3/5 = 
36/5 - 13/5 =
23/5 or 4 3/5 <==
6 0
2 years ago
Solve for the variable -3x 4 &gt; 5
Ierofanga [76]
-3x + 4 > 5
-3x > 5 - 4
-3x > 1
x < 1/-3
x < -1/3
7 0
3 years ago
5v + 7f =28.70 if f is 2.85
adell [148]
Hey!

First, let's write the problem.
5v+7\left(2.85\right)=28.7
Multiply 7 with 2.85.
5v+19.95=28.7
Subtract 19.95 from both sides.
5v+19.95-19.95=28.7-19.95
5v=8.75
Divide both sides by 5.
\frac{5v}{5}=\frac{8.75}{5}
v=1.75

Let me know if you have any questions regarding this problem!
Thanks!
-TetraFish
3 0
3 years ago
Read 2 more answers
One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independ
Illusion [34]

Question has missing details

One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is 0.05. What is the probability that at least one of 10 successive points indicates a problem when in fact the process is operating correctly?

Answer:

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

Step-by-step explanation:

Given

Let P = Probability that a point signals an error incorrectly = 0.05

Let Q = Probability that a point signals an error correctly

P + Q = 1 ---- Make Q the subject of formula

Q = 1 - P where P = 0.05

So, Q = 1 - P becomes

Q = 1 - 0.05

Q= 0.95

Solving for the probability that at least one of 10 successive points indicates a problem when in fact the process is operating.

If two events (P and Q) are independent

Then

P(P n Q) = P(P) * P(Q)

From De Morgan law;

P(P u Q) = 1 - P(P' n Q')

Where P(P u Q) represent the probability that at least one of 10 successive points

P(P' n Q') is calculated as follows;

P(P' n Q') = 0.95^10

P(P' n Q') = 0.59873693923837890625

So,

P(P u Q) = 1 - P(P' n Q') becomes

P(P u Q) = 1 - 0.59873693923837890625

P(P u Q) = 0.40126306076162109375

P(P u Q) = 0.4013 ----- Approximated

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

7 0
3 years ago
A teacher plans a simulation to estimate the probability that a student will pick a vowel out of a bag of 26 tiles, each with a
OleMash [197]

Answer:

No, because the probabilities of drawing a vowel versus a constant are not equally likely.

Step-by-step explanation:

Each number in a random number generator is equally likely.  If she assigns only the number 1 to vowels and only the number 2 to consonants, she will get results showing that they are equally likely.

In reality, 5 of the 26 letters are vowels (not including y).  The teacher should assign numbers 1 through 5 to vowels, and numbers 6 through 26 to consonants.  This will be a fair representation.

6 0
3 years ago
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