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r-ruslan [8.4K]
3 years ago
15

Evaluate 0.53n+5 when n=3 and when n=-3

Mathematics
2 answers:
melomori [17]3 years ago
4 0
<span>0.53(3)+5 = 1.59 + 5 = 6.59 when n = 3
</span>0.53(-3)+5 = -1.59 + 5 = 3.41 when n = -3
padilas [110]3 years ago
3 0
Simple...

you have: 0.53n+5

n=3

and 

n=-3

When n=3-->>

0.53(3)+5

1.59+5=6.59

When n=-3...

0.53(-3)+5

-1.59+5=3.41

Thus, your answer.
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Enter the difference between the rates of change for these two functions.
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Step-by-step explanation:

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Choose any two points. (0 , 2) and (-1 , -2)

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Find the area of cross section,the lateral surface area,the total surface area and volume of the following prisms.​
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Please answer this question
makvit [3.9K]

Step-by-step explanation:

Using the properties of logarithms, the left side of the equation becomes

\log{3x^3} - \log{x^2} = \log{3x}

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so we end up with

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2 years ago
Scores on a final exam taken by 1200 students have a bell shaped distribution with mean=72 and standard deviation=9
SVETLANKA909090 [29]

Answer:

a. 72

b. 816

c. 570

d. 30

Step-by-step explanation:

Given the graph is a bell - shaped curve. So, we understand that this is a normal distribution and that the bell - shaped curve is a symmetric curve.

Please refer the figure for a better understanding.

a. In a normal distribution, Mean = Median = Mode

Therefore, Median = Mean = 72

b. We have to know that 68% of the values are within the first standard deviation of the mean.

i.e., 68% values are between Mean $ \pm $ Standard Deviation (SD).

Scores between 63 and 81 :

Note that 72 - 9 = 63 and

72 + 9 = 81

This implies scores between 63 and 81 constitute 68% of the values, 34% each, since the curve is symmetric.

Now, Scores between 63 and 81 = $ \frac{68}{100} \times 1200 $

= 68 X 12 = 816.

That means 816 students have scored between 63 and 81.

c. We have to know that 95% of the values lie between second Standard Deviation of the mean.

i.e., 95% values are between Mean $ \pm $ 2(SD).

Note that 90 = 72 + 2(9) = 72 + 18

Also, 54 = 63 - 18.

Scores between 54 and 90 totally constitute 95% of the values. So, Scores between 72 and 90 should amount to $ \frac{95}{2} \% $ of the values.

Therefore, Scores between 72 and 90 = $ \frac{95}{2(100)} \times 1200 = \frac{95}{200} \times 1200  $

$ \implies 95 \times 12 $ = 570.

That is a total of 570 students scored between 72 and 90.

d. We have to know that 5 % of the values lie on the thirst standard Deviation of the mean.

In this case, 5 % of the values lie between below 54 and above 90.

Since, we are asked to find scores below 54. It should be 2.5% of the values.

So, Scores below 54 = $ \frac{2.5}{100} \times 1200 $

= 2.5 X 12 = 30.

That is, 30 students have scored below 54.

8 0
3 years ago
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