Try 1(18) / 2
Just multiply 1 by 18 and divide 18 by 2.
Then we have 9!
Let's attack this problem using the z-score concept. The sample std. dev. here is (0.25 oz)/sqrt(40), or 0.040. Thus, the z score representing 3.9 oz. is
3.9 - 4.0
z = -------------- = -2.5
0.040
In one way or another we must find the area under the std. normal curve that lies to the left of z = -2.5. Use a table of z-scores or a calculator with built-in statistics functions. According to my TI-83 Plus calculator, that area is
0.006. One way of interpreting this that with so small a standard deviation, most volumes of coffee put into the jars are very close to the mean, 4 oz.
Group and factor
undistribute then undistribute again
remember
ab+ac=a(b+c)
this is important
6d^4+4d^3-6d^2-4d
undistribute 2d
2d(3d^3+2d^2-3d-2)
group insides
2d[(3d^3+2d^2)+(-3d-2)]
undistribute
2d[(d^2)(3d+2)+(-1)(3d+2)]
undistribute the (3d+2) part
(2d)(d^2-1)(3d+2)
factor that difference of 2 perfect squares
(2d)(d-1)(d+1)(3d+2)
77.
group
(45z^3+20z^2)+(9z+4)
factor
(5z^2)(9z+4)+(1)(9z+4)
undistribuet (9z+4)
(5z^2+1)(9z+4)
picture unclear because it is to find the sum of consecutive numbers 1 to 100, you multiply the number of sets (50) by the sum of each set (101): 101(50)=5050.{\displaystyle 101(50)=5050.} So, the sum of consecutive number 1 through 100 is 5,050 .
Answer:
100 x 100 x 100 number cubes
Step-by-step explanation:
It's basically a volume problem. The question is asking if you made a big cube filled with smaller cubes , specifically 1,000,000 of them, what are the dimensions of this big cube. Or in other words the volume is 1,000,000. Now how do we find the volume of a cube? length times width times height, and witht he cubes each of those are the same so we can call them all x.
Now we just set up the equation where we have the equation whatever one side is times itself three times, (or x cubed) it will equal 1,000,000 or x^3=1,000,000,000. Now you just take the cube root of 1,000,000 which gets us 100. so the length, width and height are 100 small cubes.
