Answer:
5
Step-by-step explanation:
Given :
Observed values = 5, 10, 15
Expected freqencies = 10
χ² = Σ(observed - Expected)² / Expected
χ² = (5-10)²/10 + (10-10)²/10 + (15-10)²/10
χ² = 25/10 + 0/10 + 25/10
χ² = 2.5 + 0 + 2.5
χ² = 5
The Gcf of 18m x 18m and 7m is 1
Answer:
1/4
Step-by-step explanation:
11/44 = 1/4
Divide the numerator and the denominator by 11
11 ÷ 11 / 44 ÷ 11
1/4
Hope this helped!
Have a supercalifragilisticexpialidocious day!
_ divided by 7 = 3 * 1. Just do 7*3=21. so 21 divided by 7 = 3 * 1
Answer:
8333.33
Step-by-step explanation:
Given that the rate at which radioactive waste is entering the atmosphere at time t is decreasing and is given by Pe^-kt, where P is the initial rate. Use the improper integral
∫^[infinity]_0 Pe^-kt dt
with P = 50 to find the total amount of the waste that will enter the atmosphere for each value of k.
k=0.06
Hence the total amount of waste that can enter the atmosphere when k =0.06 is given by

Substitute limits
=8333.33