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ANTONII [103]
3 years ago
11

How many downloads of the standard version were there?

Mathematics
1 answer:
Artemon [7]3 years ago
3 0

Answer:

standrad = 220 download

Step-by-step explanation:

x(number of standtrad download) y(number of high quality downloads)

y=3x

4.2y+2.7x=3366

4.2(3x)+2.7x=3366

12.6x+2.7x=3366

15.3x=3366

x=3366/15.3

x=220

y=3(220)=660

(brainliest plz :( )

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alisha [4.7K]
Yes because 4x3^x =12^x since the coefficient is added to the answer because it is in the 3.
3 0
3 years ago
What is the equation of this line in standard form? −8x+7y=25 −9x+8y=−23 −8x+9y=−23 −8x+9y=23
kvasek [131]
<span>−8x+7y=25 is already in standard form, altho some people would prefer to re-write it as

</span><span>−8x+7y-25 = 0.

You have shared 4 equations here.  Next time, please separate them with commas or semi colons, or type just 1 equation per line, for increased clarity.  Thanks.</span>
5 0
4 years ago
Some types of algae have the potential to cause damage to river ecosystems. Suppose the accompanying data on algae colony densit
Phantasy [73]

Answer:

y=-2.95836 x +234.56159

Step-by-step explanation:

We assume that th data is this one:

x: 50, 55, 50, 79, 44, 37, 70, 45, 49

y: 152, 48, 22, 35, 43, 171, 13, 185, 25

a) Compute the equation of the least-squares regression line. (Round your numerical values to five decimal places.)For this case we need to calculate the slope with the following formula:

m=\frac{S_{xy}}{S_{xx}}

Where:

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}

So we can find the sums like this:

\sum_{i=1}^n x_i =50+ 55+ 50+ 79+ 44+ 37+ 70+ 45+ 49=479

\sum_{i=1}^n y_i =152+ 48+ 22+ 35+ 43+ 171+ 13+ 185+ 25=694

\sum_{i=1}^n x^2_i =50^2 + 55^2 + 50^2 + 79^2 + 44^2 + 37^2 + 70^2 + 45^2 + 49^2=26897

\sum_{i=1}^n y^2_i =152^2 + 48^2 + 22^2 + 35^2 + 43^2 + 171^2 + 13^2 + 185^2 + 25^2=93226

\sum_{i=1}^n x_i y_i =50*152+ 55*48+ 50*22+ 79*35+ 44*43+ 37*171+ 70*13+ 45*185+ 49*25=32784

With these we can find the sums:

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=26897-\frac{479^2}{9}=1403.556

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=32784-\frac{479*694}{9}=-4152.22

And the slope would be:

m=-\frac{-4152.222}{1403.556}=-2.95836

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{479}{9}=53.222

\bar y= \frac{\sum y_i}{n}=\frac{694}{9}=77.111

And we can find the intercept using this:

b=\bar y -m \bar x=77.1111111-(-2.95836*53.22222222)=234.56159

So the line would be given by:

y=-2.95836 x +234.56159

7 0
3 years ago
You have the numbers 1-24 written on slips of paper. If you choose one slip at random, what is the probability that you will not
ra1l [238]

The probability of selecting a number not divisible by 3 is: 2/3

Step-by-step explanation:

There are two methods to solve the question.

  1. We can find the probability of numbers not divisible by 3
  2. We can find the probability of numbers divisible by 3 and then find the complement of it

We will use the second method:

Given:

There are 24 slips

S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}

n(S) = 24

Let A be the event that the slip number is divisible by 3

Then

A = {3,6,9,12,15,18,21,24}

n(A) = 8

The probability of number divisible by 3 is:

P(A) = \frac{n(A)}{n(S)}\\=\frac{8}{24}\\=\frac{1}{3}

The sum of the probability of an event's occurrence and non-occurrence is 1. So the probability of numbers divisible by 3 will be subtracted from 1 to find the probability of selecting a number not divisible by 3.

The probability of selecting a number not divisible by 3 will be:

=1-\frac{1}{3}\\=\frac{3-1}{3}\\=\frac{2}{3}

The probability of selecting a number not divisible by 3 is: 2/3

Keywords: Probability

Learn more about probability at:

  • brainly.com/question/9045597
  • brainly.com/question/9103248

#LearnwithBrainly

8 0
3 years ago
What is the value of x
fenix001 [56]

(2x+2) = (3x-52)

2x = 3x-54

-x = -54

x = 54

6 0
3 years ago
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