I hope this helps you
-8÷4 <4n÷4 <12÷4
-2 <n <3
n= -1,0,1,2
Answer:
2.5% and 2.5 · 10^-3, 0.25, 2/5, √5
Step-by-step explanation:
0.25, 2/5, 2.5 · 10^-3, 2.5%, √5
Now let's list them all in the same form, why not decimals.
0.25 = 0.25
2/5 = 4/10 = 40/100 = 0.4
2.5 · 10^-3 = 2.5 · 0.01 = 0.025
2.5% = 0.025
√5 ≅ 2.236
Answer:
x = 2.5
Step-by-step explanation:
4x-1/2 = x+7
4x-x = 7+0.5
3x = 7.5
x = 2.5
Answer:
The minimum score required for admission is 21.9.
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 = 20.6, \sigma = 5.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%2020.6%2C%20%5Csigma%20%3D%205.2)
A university plans to admit students whose scores are in the top 40%. What is the minimum score required for admission?
Top 40%, so at least 100-40 = 60th percentile. The 60th percentile is the value of X when Z has a pvalue of 0.6. So it is X when Z = 0.255. 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)
![0.255 = \frac{X - 20.6}{5.2}](https://tex.z-dn.net/?f=0.255%20%3D%20%5Cfrac%7BX%20-%2020.6%7D%7B5.2%7D)
![X - 20.6 = 0.255*5.2](https://tex.z-dn.net/?f=X%20-%2020.6%20%3D%200.255%2A5.2)
![X = 21.9](https://tex.z-dn.net/?f=X%20%3D%2021.9)
The minimum score required for admission is 21.9.