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Feliz [49]
3 years ago
9

9. Raymond and Rose were working with exponents. Part A: Raymond claims that 5^5 * 5^2 = 5^3. Rose argues that 5^5 * 5^2 = 5^7.

Which one of them is correct? Use the properties of exponents to justify your answer.​
Mathematics
1 answer:
tamaranim1 [39]3 years ago
6 0

Answer: Rose is correct 5^5*5^2=5^7

Step-by-step explanation:

when multiplying two powers that have the same base you can add the exponent s (product property’s)

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Given that f(x) = 5x − 10 and g(x) = x + 3, solve for f(g(x)) when x = −1.
Masteriza [31]
When x=-1 g(x)=2 so f(g(x))=f(2)= 10-10=0
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3 years ago
A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of
Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

<u>Hours of TV per week (X)</u>: 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

                                P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean number of hours of TV watched per week = \frac{\sum X}{n} = 9.65

            s = sample standard deviation = \sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }  = 4.61

            n = sample of people = 20

           \mu = true average number of hours of TV watched per week

<em>Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the true average, </u>\mu<u> is ;</u>

P(-1.33 < t_1_9 < 1.33) = 0.80  {As the critical value of t at 19 degrees of

                                               freedom are -1.33 & 1.33 with P = 10%}  

P(-1.33 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.33) = 0.80

P( -1.33 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.33 \times {\frac{s}{\sqrt{n} } } ) = 0.80

P( \bar X-1.33 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.33 \times {\frac{s}{\sqrt{n} } } ) = 0.80

<u>80% confidence interval for</u> \mu = [ \bar X-1.33 \times {\frac{s}{\sqrt{n} } } , \bar X+1.33 \times {\frac{s}{\sqrt{n} } } ]

                                         = [ 9.65-1.33 \times {\frac{4.61}{\sqrt{20} } } , 9.65+1.33 \times {\frac{4.61}{\sqrt{20} } } ]

                                         = [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

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What is the slope of th line 7x-9y=63
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