Answer:
25
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 175cm
Standard deviation = 6 cm
Percentage of students below 163 cm
163 = 175 - 2*6
So 163 is two standard deviations below the mean.
By the Empirical rule, 95% of the heights are within 2 standard deviations of the mean. The other 100-95 = 5% are more than 2 standard deviations of the mean. Since the normal distribution is symmetric, 2.5% of them are more than 2 standard deviations below the mean(so below 163cm) and 2.5% are more than two standard deviations above the mean.
2.5% of the students have heights less than 163cm.
Out of 1000
0.025*1000 = 25
25 is the answer
Triangles = 180 so you’d use the equation ABC (60) + BAC (50) + ACB (x) = 180
Which equals out to ACB= 79
Answer:
h=48
Step-by-step explanation:
You should create an even proportion. So you could do seconds over heartbeats or the other way around. I decided to do the other way around. Since yu want to see h in m minutes and in a minute there are 60 seconds, I did 8/10 = h/60. Instead of cross multiplying, I saw that 10 times 6 is 60 so 8 x 6=h. 8 x 6 = 48 so h=48.
Answer:
A and C
Step-by-step explanation:
To determine which events are equal, we explicitly define the elements in each set builder.
For event A
A={1.3}
for event B
B={x|x is a number on a die}
The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as
B={1,2,3,4,5,6}
for event C
![C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1](https://tex.z-dn.net/?f=C%3D%5Bx%7Cx%5E%7B2%7D-4x%2B3%5D%5C%5Csolving%20%20x%5E%7B2%7D-4x%2B3%5C%5Cx%5E%7B2%7D-4x%2B3%3D0%5C%5Cx%5E%7B2%7D-3x-x%2B3%3D0%5C%5Cx%28x-3%29-1%28x-3%29%3D0%5C%5Cx%3D3%20or%20x%3D1)
Hence the set c is C={1,3}
and for the set D {x| x is the number of heads when six coins re tossed }
In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6
Hence the set D can be expressed as
D={0,1,2,3,4,5,6}
In conclusion, when all the set are compared only set A and set C are equal
