The LCM of 6 and 4 is 12. The LCM of 6 and 8 is 24. The LCM of 6 and 10 is 30.
To find the LCM, we find the prime factorization of each number.
6 = 2*3
4 = 2*2
They have a 2 in common, so this is one factor of the LCM. We then take the uncommon factors, 3 and 2, giving us
2*3*2 = 12.
6 = 2*3
8 = 2*2*2
LCM = 2 (in common) * 3 * 2 * 2 (uncommon) = 24.
6 = 3*2
10 = 5*2
LCM = 2 (in common) * 3 * 5 (uncommon) = 30.
Answer:
42 miles ≈ 68 km
35 km ≈ 22 miles
150 miles ≈ 252 km
Step-by-step explanation:
1 mile ≈ 1.6 km - just look at the picture
Answer:
Youre answer would be y
≥
7
13
y≥7/13
Step-by-step explanation:
Move all terms not containing
y
y
to the right side of the inequality.
13
y
≥
7
13y≥7
Divide each term by
13
13
and simplify.
Divide each term in
13
y
≥
7
13y≥7
by
13
13
.
13
y
13
≥
7
13
13y13≥713
Reduce the expression by cancelling the common factors.
y
≥
7
13
y≥713
Hello,
P(x)=150*1.07^x
==>ln(P(x)/150)=x*ln(1.07)
==>x=(ln(P(x)-ln(150))/ln(1.07)
If you know how much is P(x) then you have the answer.
Answer:
Following are the solution to the given question:
Step-by-step explanation:
The population std. dev of the dog weight=8
Calculating the payout w s.t:
![E[netpay]=0=(-5) \times 0.9544+w\times (1-0.9544)\\\\ w =(5 \times \frac{0.9544}{1-0.9544}) =\$ 104.65](https://tex.z-dn.net/?f=E%5Bnetpay%5D%3D0%3D%28-5%29%20%5Ctimes%200.9544%2Bw%5Ctimes%20%281-0.9544%29%5C%5C%5C%5C%20w%20%3D%285%20%5Ctimes%20%5Cfrac%7B0.9544%7D%7B1-0.9544%7D%29%20%3D%5C%24%20104.65)
therefore, we assume that the weight of the dog is a normal distribution with std. deviation that is 8.
