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2 years ago

Solve n/2+5=3 help me please thank you

2 answers:

6 0

Answer:

n/2+5=3

We simplify the equation to the form, which is simple to understand
n/2+5=3

Simplifying:
+ 0.5n+5=3

We move all terms containing n to the left and all other terms to the right.
+ 0.5n=+3-5

We simplify left and right side of the equation.
+ 0.5n=-2

We divide both sides of the equation by 0.5 to get n.
n=-4

Gemiola [76]2 years ago

5 0

Answer:

n=-1

Step-by-step explanation:

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Artist 52 [7]

X<4 hope this helped dude

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3 years ago

What is the sum of 3/4 and 18/19 closest to 0 1 2

Lisa [10]

2 because it is when you round 3/4 to 1 and 18/19 to 1. you then add 1+1 equals 2 . so my answer is correct.

8 0

2 years ago

The selling price of an item is $600. It is marked down by 30%, but this sale price is still marked up

Norma-Jean [14]

Answer:

7000

Step-by-step explanation:

divide 600 and 30 then multiply that answer by 350

7 0

2 years ago

The fair spinner shown in the diagram above is spun.

Katena32 [7]

Unable to provide a definitive answer due to lack of diagram.

Answer: 1/4

Step-by-step explanation:

Multiples of 6:

Multiples of 6 are numbers obtained when 6 is multiplied by an integer. Such as

(6*1) = 6, (6*2)=12, (6*3) = 18..... and so on.

Probability is calculated by finding the ratio of the required outcome and total possible outcomes.

Probability = required outcome / Total possible outcomes

P(multiple of 6) = number of multiples of 6 / total number of possible outcomes

If number of 6 multiples = 5 and

Total number of faces in the spinner = 20

Then :

P(multiple of 6) = 5/20 = 1/4

5 0

3 years ago

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

3 years ago