To solve this, you need to isolate/get the variable "c" by itself in the equation:
4c + 8c = -55 + 3c You can first combine like terms (4c and 8c)
12c = -55 + 3c Subtract 3c on both sides
9c = -55 Divide 9 on both sides
Answer:
5+29
Step-by-step explanation:
<h3>
Answer: C) 81.5%</h3>
This value is approximate.
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Explanation:
We have a normal distribution with these parameters
- mu = 128 = population mean
- sigma = 30 = population standard deviation
The goal is to find the area under the curve from x = 68 to x = 158, where x is the number of text messages sent per day. So effectively, we want to find P(68 < x < 158).
Let's convert the score x = 68 to its corresponding z score
z = (x-mu)/sigma
z = (68-128)/30
z = -60/30
z = -2
This tells us that the score x = 68 is exactly two standard deviations below the mean mu = 128.
Repeat for x = 158
z = (x-mu)/sigma
z = (158-128)/30
z = 30/30
z = 1
This value is exactly one standard deviation above the mean
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The problem of finding P(68 < x < 158) can be rephrased into P(-2 < z < 1)
We do this because we can then use the Empirical rule as shown in the diagram below.
We'll focus on the regions between z = -2 and z = 1. This consists of the blue 13.5% on the left, and the two pink 34% portions. So we will say 13.5% + 34% + 34% = 81.5%
Approximately 81.5% of the the population sends between 68 and 158 text messages per day. This value is approximate because the percentages listed in the Empirical rule below are approximate.
What are th minimum,first quartile,median,third quartile,and maximum of the data set? 63,98,40,32,20,80,102,65
IRISSAK [1]
Median: 64
Minimum: 20
Maximum: 102
First Q: 32
Third Q: 98
Answer:
b. 0.37
Step-by-step explanation:
it asks for everyone from 60-69 so you add the male and female together and put it into a fraction it should come out as .37/1 or 0.37