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
Hey there!
12/14
• BOTH NUMBERS HAVE THE COMMON FACTOR OF 2, SO WE CAN DIVIDE BOTH THE NUMERATOR (TOP number) FROM THE DENOMINATOR (BOTTOM number)
= 12 ÷ 2 / 14 ÷ 2
= 6/7
= 18/21
Therefore, your answer is: 6/7 or 18/21
Good luck on your assignment and enjoy your day!
~Amphitrite1040:)
Answer:
Step-by-step explanation:
If 3⁄4 gallons of water come out of a hose in 15 minutes, to calculate the rate of water that comes out in an hour (60 minutes), we will use the expression;
3/4 gallons = 15 minutes
x gallons = 60minutes
cross multiply
15x = 3/4 * 60
15x = 180/4
15x = 45
x = 45/15
x = 3/1
x = 3:1
Hence the water rate in gallons per hour is 3 gallons per hour
Answer:
22.7604763 degrees
Step-by-step explanation:
Note that the answer is NOT 20 degrees.
She adjusted the ramp to half the original <em>incline</em>
Therefore we have tan 40 = x = 0.839099631.
We want an incline of y degrees where tan y = x/2 = 0.839099631/2 = 0.419549815
Taking the arctan of both sides, we get
y = arctan(0.419549815) = 22.7604763 degrees
Answer:
-28 = x
Step-by-step explanation
[5x + 20] + [4x - 8] = 180
9x + 12 = 180
- 12 -12
______________
9x = 168
18⅔ = x
m∠EBC = 5[18⅔] + 20 = 113⅓
93⅓
You will get a negative answer if you set them equal to each other, which is not allowed when you are talking about <em>length</em><em>.</em>
I hope this is correct, and as always, I am joyous to assist anyone at any time.
