60 mL = 60/1000 L
= 0.06 L
1 container of resin = 0.06L
19 containers of resin = 0.06 x 19
= 1.14 L
1.14 L of hardener should be added to 19 containers of resin.
Step-by-step explanation:
Given that the graph shows the normal distribution of the length of similar components produced by a company with a mean of 5 centimeters and a standard deviation of 0.02 centimeters.
A component is chosen at random, the probability that the length of this component is between 4.98 centimeters and 5.02
=P(|z|<1) (since 1 std dev on either side of the mean)
=2(0.3418)
=0.6826
=68.26%
The probability that the length of this component is between 5.02 centimeters and 5.04 centimeters is
=P(1<z<2) (since between 1 and 2 std dev from the mean)
=0.475-0.3418
=0.3332
=33.32%
Answer:
I think it might be D but feel free to correct me if I'm wrong
y = 
x + 3 → B
subtract 5x from both sides
- 3y = - 5x - 9 ( multiply through by ( - 1 ) )
3y = 5x + 9 ( divide all terms by 3 )
y = x + 3 ( in slope- intercept form )
F(x)=sqrt(x)
D(x)=sqrt((f(x)-0)^2+(x-4)^2) [distance of (4,0) from any point on graph]
=sqrt((√x-0)^2+(x-4)^2)
=sqrt(x+(x-4)^2)
To minimize distance, we equate D'(x)=0, or equivalently, we equate the derivative of D(x)^2 to zero.
The latter is easier to derive, and gives the same results.
We will do both.
D'(x)=(1/2)(2(x-4)+1)/sqrt(x+(x-4)^2) [using chain rule]
D'(x)=0 => (2(x-4)+1)=0 => x-4=-1/2 => x=7/2
d(D²(x))/dx
=d(x+(x-4)^2)/dx
=2(x-4)+1
=> 2(x-4)+1=0 => x=7/2
So the point on the graph is (sqrt(7/2),7/2)=(1.8708,3.5) approx.
