Answer: W(t) = 4t + 2
Step-by-step explanation:
y = ax + b
Because you have already typed two pages, the y-intercept(b) will be 2
The slope(a) is the number of pages typed per hour, which is 4.
So if you put it together:
W(t) = 4t + 2
Answer:
The answer is 6x+4
Step-by-step explanation:
2(3x+4) multiply both constant and variable by the value outside of the parenthesis
and you'll get 6x+4
if you need help then before you look for the answer online be sure to ask your teacher for help so you can understand and want have to look for the answer online
4x + 8y = 20
- 4x + 2y = -30
find y first
in order to find y eliminate x
add the two equations together
4x+-4x+8y+2y= 20+-30
4x-4x+8y+2y= 20-30
10y= -10
divide both sides by 10 to get y by itself
10y/10= -10/10
y= -1
find x by using the substitution method
- 4x + 2y = -30
-4x+2(-1)=-30
-4x-2= -30
move -2 to the other side
sign changes from -2 to +2
-4x-2+2= -30+2
-4x= -28
divide both sides by -4
-4x/-4= -28/-4
x= 7
answer:
(7,-1)
Answer:
Step-by-step explanation:
Let x be the random variable representing the number of miles that each person walked each day for 6 months. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,
z = (x - µ)/σ
Where
x = sample mean
µ = population mean
σ = standard deviation
For Rueben,
µ = 5
σ = 1.1
the probability that Rueben walked more than 6.1 miles is expressed as
P(x > 6.1) = 1 - P( x ≤ 6.1)
For x = 6.1,
z = (4 - 6.1)/1.1 = - 1.91
Looking at the normal distribution table, the probability corresponding to the z score is 0.02807
P(x > 6.1) = 1 - 0.02807 = 0.97193
P(x > 6.1) = 0.97 × 100 = 97%
For Victor,
µ = 4.4
σ = 1.4
the probability that Victor walked less than 5.8 miless is expressed as
P(x < 5.8)
For x = 5.8,
z = (5.8 - 4.4)/1.4 = 1
Looking at the normal distribution table, the probability corresponding to the z score is 0.8413
P(x < 5.8) = 0.84 = 84%
Answer:
It is a relation and a function.
Step-by-step explanation:
The expression: is a relation because it connects x values of the Real number set with y values of the Real number set via this associated equation.
It is also a function,because this relation satisfies that for each x-value in the starting set called Domain (which in this case is as we said the Real number set), there is ONLY one value in the END set (called the Range) associated it.