(3,0) and (0,6)
Y= -2x + 6
0= -2x + 6
-6/-2 = 3
X= 3
(3,0)
Y = -2(0) + 6
Y=6
(0,6)
Answer:
The proportion of children that have an index of at least 110 is 0.0478.
Step-by-step explanation:
The given distribution has a mean of 90 and a standard deviation of 12.
Therefore mean,
= 90 and standard deviation,
= 12.
It is given to find the proportion of children having an index of at least 110.
We can take the variable to be analysed to be x = 110.
Therefore we have to find p(x < 110), which is left tailed.
Using the formula for z which is p( Z <
) we get p(Z <
= 1.67).
So we have to find p(Z ≥ 1.67) = 1 - p(Z < 1.67)
Using the Z - table we can calculate p(Z < 1.67) = 0.9522.
Therefore p(Z ≥ 1.67) = 1 - 0.9522 = 0.0478
Therefore the proportion of children that have an index of at least 110 is 0.0478
Answer:
Step-by-step explanation:
400yd($2304/192yd)=$4800
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Y - 2x = 8 . . . . (1)
16 + 4x = 2y . . (2)
From (1), y = 2x + 8 . . . (3)
Putting (3) into (2) gives
16 + 4x = 2(2x + 8) = 4x + 16
0 = 0
Therefore, there are infinite number of solutions.