Mode is most occurring
There is no mode for the peas as there is no score occurring more than once
Median is middle number
So cross of 4 and 5, and 7 and 8
Median is 6
Mean is average
4+5+6+7+8 divided by five
Mean is 6
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
Solve
hmmm
half life
well, 120 to 60 is half
60=120e^(-0.00043t)
solve for t
divide both sides by 120
1/2=e^(-0.00043t)
take ln of both sides
ln(1/2)=-0.00043t
divide both sides by -0.00043
ln(1/2)/-0.00043=t
t≈1611.97 years
about 1612 years
Answer:
box 1: monomial
box 2: binomial
Step-by-step explanation:
reasoning: Box 1’s volume is modeled by a monomial times a monomial, so it will be a monomial.
Box 2’s volume is modeled by a monomial times a binomial, so it will be a binomial.
Answer:
8 1/4
Step-by-step explanation:
20 2/4
-
12 3/4
__________
8 1/4