Answer:
![\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%7D)
Step-by-step explanation:
use the Pythagoras Theorem
Since you have the hypotenuse (6"), just go backwards. 62 = 36 = a2 + b2.
Volume of Cylinder = πr²h
Radius r = 14.5 feet
Volume, V = 4950 cubic feet
4950 = π*14.5²*h
π*14.5²*h = 4950
h = 4950 / (π*14.5²) Use your calculator with π function
h = 7.494
Therefore height ≈ 7.50 feet to the nearest tenth of a foot.
There are 24 possible answers, 6 are correct answers and 18 are incorrect answers. By use of the hypergeometric distribution the probabilities of each possible number of randomly selected correct answers can be found, as follows:
P(0) = 0.1379, P(1) = 0.3819, P(2) = 0.341, P(3) = 0.1213, P(4) = 0.017, P(5) = 0.0008, P(6) = 0.000007.
The expected or mean number of correct answers is found from:
The answer is: 1.4998
The answer is the first 1 !!
Here is how we see if it is true!
If Clayton drives 60 miles per hour, that means, <u>t</u>hat after one hour has passed, they should have traveled 60 miles!
If we look at the number of hours that have passed between 1:30 and 4:30, it is 3 hours.
If someone travels 60 miles every hour that has passed, and they traveled for 3 hours. To find how far they have traveled we just need to multiply 60 miles by 3 hours!
Let's solve that!
60 × 3 = 180 miles!
Clayton started driving at 1:30 pm and went 60 miles per hour.
Answer:
Step-by-step explanation:
A system of linear equations is one which may be written in the form
a11x1 + a12x2 + · · · + a1nxn = b1 (1)
a21x1 + a22x2 + · · · + a2nxn = b2 (2)
.
am1x1 + am2x2 + · · · + amnxn = bm (m)
Here, all of the coefficients aij and all of the right hand sides bi are assumed to be known constants. All of the
xi
’s are assumed to be unknowns, that we are to solve for. Note that every left hand side is a sum of terms of
the form constant × x
Solving Linear Systems of Equations
We now introduce, by way of several examples, the systematic procedure for solving systems of linear
equations.
Here is a system of three equations in three unknowns.
x1+ x2 + x3 = 4 (1)
x1+ 2x2 + 3x3 = 9 (2)
2x1+ 3x2 + x3 = 7 (3)
We can reduce the system down to two equations in two unknowns by using the first equation to solve for x1
in terms of x2 and x3
x1 = 4 − x2 − x3 (1’)
1
and substituting this solution into the remaining two equations
(2) (4 − x2 − x3) + 2x2+3x3 = 9 =⇒ x2+2x3 = 5
(3) 2(4 − x2 − x3) + 3x2+ x3 = 7 =⇒ x2− x3 = −1
