3 to the 1st is 3,
3 to the 2nd is 9,
3 to the 3rd is 27,
3 to the 4th is 81,
3 to the 5th is 243,
3 to the 6th is 729.
Answer:
A) 34.13%
B) 15.87%
C) 95.44%
D) 97.72%
E) 49.87%
F) 0.13%
Step-by-step explanation:
To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:
![z=\frac{x-m}{s}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-m%7D%7Bs%7D)
Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:
![z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B90-100%7D%7B10%7D%3D-1%5C%5C%20z%3D%5Cfrac%7B100-100%7D%7B10%7D%3D0)
So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:
P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)
= 0.5 - 0.1587 = 0.3413
It means that the PERCENT of scores that are between 90 and 100 is 34.13%
At the same way, we can calculated the percentages of B, C, D, E and F as:
B) Over 110
![P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587](https://tex.z-dn.net/?f=P%28%20x%20%3E%20110%20%29%20%3D%20P%28%20z%3E%5Cfrac%7B110-100%7D%7B10%7D%29%3DP%28z%3E1%29%20%3D%200.1587)
C) Between 80 and 120
![P( 80](https://tex.z-dn.net/?f=P%28%2080%3Cx%3C%20110%20%29%20%3D%20P%28%20%5Cfrac%7B80-100%7D%7B10%7D%20%3Cz%3E%5Cfrac%7B120-100%7D%7B10%7D%29%3DP%28-2%3Cz%3C2%29%5C%5CP%28-2%3Cz%3C2%29%3DP%28z%3C2%29%20-%20P%28z%3C-2%29%20%3D%200.9772%20-%200.0228%20%3D%200.9544)
D) less than 80
![P( x < 80 ) = P( z](https://tex.z-dn.net/?f=P%28%20x%20%3C%2080%20%29%20%3D%20P%28%20z%3C%5Cfrac%7B80-100%7D%7B10%7D%29%3DP%28z%3C-2%29%20%3D%200.9772)
E) Between 70 and 100
![P( 70](https://tex.z-dn.net/?f=P%28%2070%3Cx%3C%20100%20%29%20%3D%20P%28%20%5Cfrac%7B70-100%7D%7B10%7D%20%3Cz%3E%5Cfrac%7B100-100%7D%7B10%7D%29%3DP%28-3%3Cz%3C0%29%5C%5CP%28-3%3Cz%3C0%29%3DP%28z%3C0%29%20-%20P%28z%3C-3%29%20%3D%200.5%20-%200.0013%20%3D%200.4987)
F) More than 130
![P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013](https://tex.z-dn.net/?f=P%28%20x%20%3E%20130%20%29%20%3D%20P%28%20z%3E%5Cfrac%7B130-100%7D%7B10%7D%29%3DP%28z%3E3%29%20%3D%200.0013)
p = plane's speed in calm air
w = wind's speed
p + w = speed across the ground of the plane flying with the wind
p - w = speed across the ground of the plane flying against the wind
d = rt is the standard distance equation, where d=distance, r=rate, t=time
d = 800 miles is given
800 = 4(p+w) :: given that it takes 4 hrs with the wind to travel 800 miles
800 = 5(p-w) :: given that it takes 5 hrs against the wind to travel 800 miles
so we have
4(p+w) = 800
5(p-w) = 800
4p + 4w = 800
5p - 5w = 800
multiply the first equation by 5 and the second by 4
20p + 20w = 4000
20p - 20w = 3200
add them
40p = 7200
p = 180
substituting for p = 180, we can find w
4p + 4w = 800
divide by 4
p + w = 200
180 + w = 200
So,
w = 20
checking our work.
4(180+20) = 800??
4(200) = 800
Yes
5(180-20) = 800??
5(160) = 800
Yes.
Answer:
The Plane flies at 180 mph in calm air.
The Wind is blowing at 20 mph.
Hope This Helps! Merry Xmas:)
Answer:
1/27000
Step-by-step explanation:
= 3^-3 × 10^-3
= 1/3³ × 1/10³
= 1/27 × 1/1000
= 1/27000
Hope this helped!
Answer:
C>-1/4
Step-by-step explanation:
Two have intersection it must be:
2x+c= x^2+3x, i.e,
x^2+x-c=0
Solution of its equation is:
![x_{1,2}=\frac{-1+-\sqrt{1^2+4c}}{4c}](https://tex.z-dn.net/?f=%20x_%7B1%2C2%7D%3D%5Cfrac%7B-1%2B-%5Csqrt%7B1%5E2%2B4c%7D%7D%7B4c%7D)
If we we want two solution we must have that:
1+4c>0
So c>-1/4.