Okay lets create an eqn from that information
A first int, B second int, C third int, D fourth int.
B = A + 2
C = A + 4
D = A + 6
A is the smallest integer
B + D = 0.5 (A + C)
Now lets substitute
(A + 2) + (A + 6) = 0.5(A + (A + 4))
now lets dist
2a + 8 = 0.5(2a +4)
2a + 8 = a + 2
a + 8 = 2
a = -6
B = -6 +2
B = -4
C = -6 + 4
C = -2
D = -6 + 6
D = 0
Now using B + D = 0.5(A + C)
-4 + 0 = 0.5(-6 + (-2))
-4 = 0.5 (-8)
-4 = -4
Correct
Therefore, First integer is -6, second integer is -4, third integer is -2 and fourth integer is 0
Answer:
if -5n+7<57, then n>-10.
if 6n+2<8, then n<1.
Step-by-step explanation:
-5n+7<57
1. subtract 7 from both sides:
-5n+7 -7< 57 -7
-5n<50
2. divide both sides by -5 (remember to flip the sign whenever you divide by a negative number):
-5n ÷-5<50 ÷-5
n>-10
6n+2<8
1. subtract 2 from both sides:
6n+2 -2<8 -2
6n<6
2. divide both sides by 6:
6n ÷6<6 ÷6
n<1
We have the following:
therefore, the answer is 3.35 x 10^6
Answer:
The probability that the cost for someone's dog is higher than for the cat is 33.2%.
Step-by-step explanation:
With the data given, we know that the difference in cost of medical care of dogs and cats have a normal distribution with μ=-20 and σ=46.
To know the probability of the cost for the dog is higher than the cat, we have to calculate the probability of P(d>0).
Then we have to calculate z, and look up in a standarized normal distribution table.
Calculate z:
The probability of the difference being higher than 0, we have:
We can say that the probability that the cost for someone's dog is higher than for the cat is 33.2%.
