<span>78.6 is a decimal number, you must name it with the whole part first and then the decimal part as part of tenths, hundredths or what it corresponds. In this case the whole part is 78 and the decimal part is 6/10 which is 6 tenths. Then, the number in words is sevety eight and six tenths.</span>
Answer:
2.28% of tests has scores over 90.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 80, \sigma = 5](https://tex.z-dn.net/?f=%5Cmu%20%3D%2080%2C%20%5Csigma%20%3D%205)
What proportion of tests has scores over 90?
This proportion is 1 subtracted by the pvalue of Z when X = 90. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{90 - 80}{5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B90%20-%2080%7D%7B5%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a pvalue of 0.9772.
So 1-0.9772 = 0.0228 = 2.28% of tests has scores over 90.
436 hamburgers and cheeseburgers sold Wednesday.
64 fewer cheeseburgers.
436/2= 218
218-64= 154
154 cheeseburgers were sold Wednesday.
Answer:
14/100 and 7/50
Step-by-step explanation:
Basically you convert the decimal into a fraction, 14 percent is the same as 14/100 and if you simplify that then you get 7/50