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
Step-by-step explanation:

The answer is 4/10*cos 01
There are 10000 bugs and you add 15 bugs to it.
10,000+15= 10,015
The total is 10,015 bugs
There were already 10,000 bugs and 15 bugs joined them.
Which makes it an addition statement
Answer:Gary-18 years brother-10 years
Step-by-step explanation:1st case: brother=1 yrs,gary=3 yrs=>after 6 after: brother =7 yrs gary=9 years
2nd case:......
3rd case:.....
4th case:brother:4 years gary:12 yrs=>after 6 yrs:brother=10 yrs gary=18 yrs=>18-10=8