the answer is 58.8 so i think these the answer
Answer:
The price that is two standard deviations above the mean price is 4.90.
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.
Mean of 3.22 and a standard deviation of 0.84.
This means that 
Find the price that is two standard deviations above the mean price.
This is X when Z = 2. So




The price that is two standard deviations above the mean price is 4.90.
Answer:
70
Step-by-step explanation:
63/9 = 7
7 * 10 = 70
we are given

where
price p( in dollars) of a new smartphone
t is the number of months since January
now, we can solve for t
step-1:
Add both sides by 4t^2


step-2:
Subtract both sides by p


step-3:
divide both sides by 4



step-4:
Take sqrt both sides


so, we get
..............Answer

Let's solve your problem:
The answer would be 280%.
We know the following information:
2pm - 15 toilet paper rolls
5pm - 57 toilet paper rolls
We have to find the <u>percent rate of change.</u>
Here is how we will do this problem:
<h3><u>
First, Set up a ratio:</u></h3>
Change in the quantity [15 - 57]
-------------------------------------
Original quantity [15]
57 - 15 = 42
<h3><u>
Next, we will divide. the ratio.</u></h3>
Given quantity - 42
Number of packs sold at 2pm - 15
Fraction - 42/15
<h3><u>
Convert the fraction to a decimal.</u></h3>
Fraction - 42/15
Decimal - 2.8
<h3>
<u>Make the decimal to a percent.</u></h3>
Decimal - 2.8
Percent - 28%
<u>Add a zero to the 28.</u>
<h2><u>
Our answer is 280%</u></h2>
<h3>
</h3>