25.8%
First, determine how many standard deviations from the norm that 3 tons are. So:
(3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%
So the probability that more than 3 tons will be dumped in a week is 25.8%
Answer:
34
Step-by-step explanation:
x=3+√8
y =3-√8
now,
1/x^2+1/y^2
=1/(3+√8)² + 1/(3-√8)²
= [(3-√8)²+(3+√8)²] / (3+√8)²(3-√8)² [L.C.M = (3+√8)²(3-√8)² ]
=[(3-√8+3+√8)²-2(3-√8)(3+√8) ] / [(3+√8)(3-√8)]²
=[6²-2.(3²-√8² )] / (3²-√8²)² [ a²+ b²=(a+b)²-2ab]
=[36-2(9-8) ]/ (9-8)²
=[36-2.1] / 1²
=34
Technically, none of the equations.
The equation you're looking for would be:
x²+(x+20)²=150²
Divide the total amount by the portion size.
= 3 1/2 ÷ 1/4
convert 3 1/2 to improper fraction
= (2*3+1)/2 ÷ 1/4
= 7/2 ÷ 1/4
to divide a fraction, we multiply by the reciprocal/inverse of 1/4
= 7/2 * 4/1
multiply numerators, then multiply denominators
= (7*4)/(2*1)
= 28/2
= 14
ANSWER: 14
Hope this helps! :)
Answer:
It is the left one.
Step-by-step explanation:
This (<) symbol shows us which number is bigger. In this case, 4.626 is larger than 4.606, so we put this "<"
