The missing values represented by x and y are 8 and 20, that is
(x, y) = (8, 20)
The function y = 16 + 0.5x is a linear equation that can be solved graphically. This means the values of both variables x and y can be found on different points along the straight-line graph.
The ordered pairs simply mean for every value of x, there is a corresponding value of y.
The 2-column table has values for x and y which all satisfy the equation y = 16 + 0.5x. Taking the first row, for example, the pair is given as (-4, 14).
This means when x equals negative 4, y equals 14.
Where y = 16 + 0.5x
y = 16 + 0.5(-4)
y = 16 + (-2)
y = 16 - 2
y = 14
Therefore the first pair, just like the other four pairs all satisfy the equation.
Hence, looking at the options given, we can determine which satisfies the equation
(option 1) When x = 0
y = 16 + 0.5(0)
y = 16 + 0
y = 16
(0, 16)
(option 2) When x = 5
y = 16 + 0.5(5)
y = 16 + 2.5
y = 18.5
(5, 18.5)
(option 3) When x = 8
y = 16 + 0.5(8)
y = 16 + 4
y = 20
(8, 20)
From our calculations, the third option (8, 20) is the correct ordered pair that would fill in the missing values x and y.
To learn more about the straight line visit:
brainly.com/question/1852598
#SPJ1
Answer:
mean is equal to 5.12
Step-by-step explanation:
<em>We are given that 32% of college students work fulltime. We have to find the mean for the number of student s who are working full time in a sample of 16</em>
success rate, p = 32% = 0.32
Sample size is denoted as n = 16
The forumla of mean is given as
mean = sample size × success rate
mean = n × p
= 16 × 0.32
= 5.12
Answer:
Step-by-step explanation:
This is <em>a separable differential equation</em>. Rearranging terms in the equation gives
Integration on both sides gives
where 
is a constant of integration.
The steps for solving the integral on the right hand side are presented below.
Therefore,
Multiply both sides by 
By taking exponents, we obtain
Isolate 
.
Since 
when 
, we obtain an initial condition 
.
We can use it to find the numeric value of the constant .
Substituting 
for and 
in the equation gives
Therefore, the solution of the given differential equation is
