Answer:
The equation is: <em>y</em> = 15·<em>x</em>
Step-by-step explanation:
It is provided that at the Speedy Bike Works, 15 bicycles are produced each hour.
Consider the table below.
Number of Hours: 1 3 6 10
Number of Bicycles Produced: 15 45 90 150
Compute the ratio of number of bicycles produced and number of hours for every data above as follows:
The ratio of the number of bicycles produced and number of hours is same for every data value.
Thus, the relationship between the number of bicycles produced and number of hours is proportional.
The equation for the relationship is:
<em>y</em> = 15·<em>x</em>
<em>y</em> = number of bicycles produced
<em>x</em> = number of hours
Answer:
Mean of the data set in the dot plot would be: 3.6
Step-by-step explanation:
As we know that Mean from the dot plot can be obtained by:
- adding the numbers and then
- divide the resulting sum by the number of addends.
Please check the attached figure where the dot plot is also plotted.
From the dot plot, it is clear that
There are 3 dots at 1.
There are 4 dots at 2.
There are 3 dots at 3.
There are 4 dots at 4.
There are 5 dots at 5.
There are 3 dots at 6.
All we have to do is to add the dots and divide the sum by the number of addend dots.
In other words:
There are 3 dots at 1 ⇒ 1+1+1
There are 4 dots at 2 ⇒ 2+2+2+2
There are 3 dots at 3 ⇒ 3+3+3
There are 4 dots at 4 ⇒ 4+4+4+4
There are 5 dots at 5 ⇒ 5+5+5+5+5
There are 3 dots at 6 ⇒ 6+6+6
As there are total 22 dots.
And the sum of all the dots with respect to their plot number = 79
i.e. 1+1+1+2+2+2+2+3+3+3+4+4+4+4+5+5+5+5+5+6+6+6 = 79
Thus
Mean of the data set in the dot plot = 79/22
= 3.6
Therefore, Mean of the data set in the dot plot would be: 3.6
so, let's keep in mind that
so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.
we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.
![\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%2By%2B3y%3D78%5C%5C%20x%2B4y%3D78%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%200.25x%2B0.4y%2B0.6%283y%29%3D35.1%5C%5C%200.25x%2B0.4y%3D1.8y%3D35.1%5C%5C%200.25x%2B2.2y%3D35.1%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20x%2B4y%3D78%5C%5C%5C%5C%200.25x%2B2.2y%3D35.1%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20x%2B4y%3D78%5Cimplies%20%5Cboxed%7Bx%7D%3D78-4y%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20substitution%20on%20the%202nd%20equation%7D%7D%7B0.25%5Cleft%28%20%5Cboxed%7B78-4y%7D%20%5Cright%29%2B2.2y%3D35.1%7D%5Cimplies%2019.5-y%2B2.2y%3D35.1)
![\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill](https://tex.z-dn.net/?f=%5Cbf%201.2y%3D15.6%5Cimplies%20y%3D%5Ccfrac%7B15.6%7D%7B1.2%7D%5Cimplies%20%5Cblacktriangleright%20y%3D13%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20x%3D78-4y%5Cimplies%20x%3D78-4%2813%29%5Cimplies%20%5Cblacktriangleright%20x%3D26%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20z%3D3y%5Cimplies%20z%3D3%2813%29%5Cimplies%20%5Cblacktriangleright%20z%3D39%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B25%5C%25%7D%7B26%7D%5Cqquad%20%5Cstackrel%7B40%5C%25%7D%7B13%7D%5Cqquad%20%5Cstackrel%7B60%5C%25%7D%7B39%7D~%5Chfill)
Answer:
the probability that two 18 year old boys chosen at random will have heights greater than 185cm is 0.403
Step-by-step explanation:
P( x > 193) = 0.15
= 1- p(x less than or equal 193)
= 1 -p( z < (x- u) /sigma)
= 1- p( z< (193 - 187)/ sigma)
= 1- p( z< 6/ sigma)
P(z< 6/sigma) = 1 - 0.15
P(z < 6/sigma)= 0.85
6/sigma =1.036
Sigma= 6/1.036
Sigma= 5.79
P( x> 185) = 1- p( x< 185)
= 1- p (z < (185- 187)/5.79)
= 1- p( z< -0.345)
= 1- 0.365
= 0.635
P (x> 185) = 0.635 × 0.635
=0.403
