0.9 as a fraction is:

What place is the '9' in? It's in the tenth place.
0.9 can be said as

Nine

tenths

=
Answer: 77% of the apples will contain at least 2.36 ounces of juice.
Step-by-step explanation:
Since the amount of juice squeezed from each of these apples is approximately normally distributed,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = the amount of juice squeezed in ounces.
µ = mean
σ = standard deviation
From the information given,
µ = 2.25 ounces
σ = 0.15 ounce
Looking at the normal distribution table, the z score corresponding 77%(77/100 = 0.77) is 0.74
Therefore, the number of ounces of juice, x will be
0.74 = (x - 2.25)/0.15
Cross multiplying, it becomes
x - 2.25 = 0.74 × 0.15
x - 2.25 = 0.111
x = 0.111 + 2.25
x = 2.36
Answer:
The value is c = 21.1445.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The weight distribution of parcels sent in a certain manner is normal with mean value 12lb and standard deviation 3.5 lb.
This means that 
What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight?
This 1 added to the value of X for the 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.




1 + 20.1445 = 21.1445
The value is c = 21.1445.