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ludmilkaskok [199]
3 years ago
15

Is this a good math essay?

Mathematics
1 answer:
astra-53 [7]3 years ago
7 0

Answer:

I'm just going to assume that your answer is correct since you didn't attach the question. If you are submitting this to your teacher then you are going to want to have correct spelling and grammar. The slang usage isn't professional, but if your teacher is fine with it then you should only worry about spelling. Hope this helped you!!

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HELP!
Studentka2010 [4]
So one way we can do this is- 

2(2n+7)+3n=

4n+14+3n=

7n+14

Another simpler way possible is-

2n+7+2n+7+3n=

RE-ORDER

2n+2n+3n+7+7=

COMBINE LIKE TERMS

7n+14








4 0
3 years ago
What is the mean of the set of data given below? 13 8 9 6 10 8​
olga_2 [115]

Answer:

Mean = The numbers added up and divided by the number of numbers.

Step-by-step explanation:

= (13 + 8 + 9 + 6 + 10 + 8) / 6

= 54 / 6

= 9

The mean is 9.

8 0
3 years ago
Can someone help me with this?
Wewaii [24]
<span>3x+4=y+5, then y+5=3x+4
It is called: reflexive property

-----------------------------
</span>
<span>x = y and y = 10, then x = 10
It is called: equality substitution property  
</span>

7 0
3 years ago
PLEASE HELP<br><br> FIND THE LENGTH OF RS.
Gwar [14]

Answer:

A

Step-by-step explanation:

Distance between 2 points

=  \sqrt{(x1 - x2)^{2}  + (y1 - y2)^{2} }

☆ (x₁, y₁) is the first coordinate while (x₂, y₂) I'd the second coordinate.

Length of RS

=  \sqrt{ {[ - 1 -( - 4)] }^{2}  + (9 - 1)^{2} }  \\  =  \sqrt{( - 1 + 4)^{2}  + 8^{2} }  \\  =  \sqrt{ {3}^{2}  + 64 }  \\  =  \sqrt{73}  \\ = 8.54 \: (3 \: s.f.)

Thus, RS is about 8.5 units.

8 0
3 years ago
Read 2 more answers
Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars
GuDViN [60]

Answer:

Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

Step-by-step explanation:

We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.

<em>Let X = incomes for the industry</em>

So, X ~ N(\mu=95,\sigma^{2}=5^{2})

Now, the z score probability distribution is given by;

         Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean income of firms in the industry = 95 million dollars

            \sigma = standard deviation = 5 million dollars

So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)

    P(X < 100) = P( \frac{X-\mu}{\sigma} < \frac{100-95}{5} ) = P(Z < 1) = 0.8413   {using z table]

                                                     

Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

5 0
3 years ago
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