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Zolol [24]
4 years ago
12

What is 14g-98+3g when g is 7

Mathematics
2 answers:
V125BC [204]4 years ago
8 0
14g-98+3g=14g+3g-98=17g-98
g=7
17*7-98=119-98=21


pochemuha4 years ago
6 0
So there are two potential ways you could do this. First just multiple the g=7 into the equation and then add. Or another possibility is to first add the 14g and 3g to get the 17g and then multiply it by the g value and subtract 98. So if we combine the g's and then multiply it by 7, the 17g becomes 17*7=119. So now subtract 98 from 119, 119-98=21. So the overall answer to the expression is 21 when g equals 7.
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68% of buyers paid between $147,700 and $152,300.

Step-by-step explanation:

We are given that prices of a certain model of a new home are normally distributed with a mean of $150,000.

Use the 68-95-99.7 rule to find the percentage of buyers who paid between $147,700 and $152,300 if the standard deviation is $2300.

<u><em>Let X = prices of a certain model of a new home</em></u>

SO, X ~ Normal(\mu=150,000 ,\sigma=2,300)

The z score probability distribution for normal distribution is given by;

                             Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean price = $150,000

            \sigma = standard deviation = $2,300

<u>Now, according to 68-95-99.7 rule;</u>

Around 68% of the values in a normal distribution lies between \mu-\sigma and \mu-\sigma.

Around 95% of the values occur between \mu-2\sigma and \mu+2\sigma .

Around 99.7% of the values occur between \mu-3\sigma and \mu+3\sigma.

So, firstly we will find the z scores for both the values given;

         Z  =  \frac{X-\mu}{\sigma}  =  \frac{147,700-150,000}{2,300}  = -1

         Z  =  \frac{X-\mu}{\sigma}  =  \frac{152,300-150,000}{2,300}  = 1

This indicates that we are in the category of between \mu-\sigma and \mu-\sigma.

SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.

7 0
4 years ago
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