Answer:
Step-by-step explanation:
![3p- 2c^3- 3c^3p +2 \\ \\ = (3p - 3 {c}^{3} p) +( 2 - 2c^3) \\ \\ = 3p(1 - c^3) + 2(1 - c^3) \\ \\ = (1 - c^3)(3p + 2) \\ \\ = (1 - c)(1 + c + {c}^{2} )(3p + 2) \\ \\ = \bold{ \red{(3p + 2) (1 - c)(1 + c + {c}^{2} )}}](https://tex.z-dn.net/?f=3p-%202c%5E3-%203c%5E3p%20%2B2%20%20%5C%5C%20%20%5C%5C%20%20%3D%20%283p%20-%203%20%7Bc%7D%5E%7B3%7D%20p%29%20%2B%28%202%20-%202c%5E3%29%20%5C%5C%20%20%5C%5C%20%20%3D%203p%281%20-%20c%5E3%29%20%2B%202%281%20-%20c%5E3%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%281%20-%20c%5E3%29%283p%20%2B%202%29%20%5C%5C%20%20%5C%5C%20%20%3D%20%281%20-%20c%29%281%20%2B%20c%20%2B%20%20%7Bc%7D%5E%7B2%7D%20%29%283p%20%2B%202%29%20%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%5Cbold%7B%20%5Cred%7B%283p%20%2B%202%29%20%20%281%20-%20c%29%281%20%2B%20c%20%2B%20%20%7Bc%7D%5E%7B2%7D%20%29%7D%7D)
Answer: 75 students
Step-by-step explanation:
15% of 500= 75
Answer:
24
Step-by-step explanation:
Let's see the factor pairs of 60 first.
1,60
2,30
3,20
4,15
5,12
6,10
10,6
12,5
15,4
20,3
60,1
1,60 and 60,1 is not possible for xy1 and 1xy. It is because any one digit*one digit=not 60. The remaining pairs are...
2,30
3,20
4,15
5,12
6,10
10,6
12,5
15,4
20,3
30,2
Next, start with 2,30 and 30,2
The one digit factor pairs of 30 is 5 and 6.
So...
256 is possible
265 is possible
562 is possible
652 is possible
3,20
4,15
5,12
6,10
10,6
12,5
15,4
20,3
+4
Next, 3,20 and 20,3
20=4*5
453,543,345,354
4,15
5,12
6,10
10,6
12,5
15,4
+8
4,15 and 15,4
4 number that are possible(5*3=15)
5,12
6,10
10,6
12,5
+12
5,12 and 12,5
3,4 and 2,6
526,562,534,543,435,345,265,625
6,10
10,6
+20
10=5*2
+4
20+4=24
(Sorry it's long)
Answer:
![(1024.69,\ 1127.31)](https://tex.z-dn.net/?f=%281024.69%2C%5C%201127.31%29)
Step-by-step explanation:
We know that the sample size was:
![n = 300](https://tex.z-dn.net/?f=n%20%3D%20300)
The average was:
![{\displaystyle {\overline {x}}}=1,076](https://tex.z-dn.net/?f=%7B%5Cdisplaystyle%20%7B%5Coverline%20%7Bx%7D%7D%7D%3D1%2C076)
The standard deviation was:
![\s = 345](https://tex.z-dn.net/?f=%5Cs%20%3D%20345)
The confidence level is
![1-\alpha = 0.99](https://tex.z-dn.net/?f=1-%5Calpha%20%3D%200.99)
![\alpha=1-0.99\\\alpha=0.01](https://tex.z-dn.net/?f=%5Calpha%3D1-0.99%5C%5C%5Calpha%3D0.01)
The confidence interval for the mean is:
![{\displaystyle{\overline {x}}} \± Z_{\frac{\alpha}{2}}*\frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=%7B%5Cdisplaystyle%7B%5Coverline%20%7Bx%7D%7D%7D%20%5C%C2%B1%20Z_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%2A%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
Looking at the normal table we have to
![Z_{\frac{\alpha}{2}}=Z_{\frac{0.01}{2}}=Z_{0.005}=2.576](https://tex.z-dn.net/?f=Z_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%3DZ_%7B%5Cfrac%7B0.01%7D%7B2%7D%7D%3DZ_%7B0.005%7D%3D2.576)
Therefore the confidence interval for the mean is:
![1,076\± 2.576*\frac{345}{\sqrt{300}}](https://tex.z-dn.net/?f=1%2C076%5C%C2%B1%202.576%2A%5Cfrac%7B345%7D%7B%5Csqrt%7B300%7D%7D)
![1,076\± 51.31](https://tex.z-dn.net/?f=1%2C076%5C%C2%B1%2051.31)
![(1024.69,\ 1127.31)](https://tex.z-dn.net/?f=%281024.69%2C%5C%201127.31%29)
This means that <em>the mean planned spending of all Americans who take long summer vacations in 2009 is between $ 1024.69 and $ 1127.31</em>
Answer:
u waste ur points but m good
Step-by-step explanation: