Answer:
50%
Step-by-step explanation:
Let the total number of trees be T and suppose that this is made up of eucalyptus trees E and other kinds of trees, O.
Then E+O=T, and since 99% of the trees are eucalyptus trees, EE+O=0.99.
Therefore
0.99E+0.99O=E
0.01E=0.99O
E=99O.
Suppose we reduce the number of eucalyptus trees by some amount x and that the remaining number of eucalyptus trees are now 98% of the total trees.
Then we would have E−x(E−x)+O=0.98.
Therefore,
0.02(E−x)=0.98O
0.02E−0.02x=0.98O
0.02x=0.02E−0.98O
x=E−50(0.98)O
x=E−49O
x=E−49(E99)
x=5099E≈0.505050...×E.
So we end up removing a little over half of the eucalyptus trees. EDIT: Forgot to mention - removing 50.505050...% of the eucalyptus trees means that we are removing 0.50505050....×0.99T which is 0.5T or half the trees in the neighborhood.
Answer:
(a) 0.119
(b) 0.1699
Solution:
As per the question:
Mean of the emission, 
million ponds/day
Standard deviation, 
million ponds/day
Now,
(a) The probability for the water pollution to be at least 15 million pounds/day:
= 1 - P(Z < 1.178)
Using the Z score table:
= 1 - 0.881 = 0.119
The required probability is 0.119
(b) The probability when the water pollution is in between 6.2 and 9.3 million pounds/day:
P(Z < - 0.86) - P(Z < - 1.96)
Now, using teh Z score table:
0.1949 - 0.025 = 0.1699
Answer:
14
Step-by-step explanation:
4x4 -2
It would be true no explanation needed
About 2.43 kg of pepper
Divide 17 by 7
