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Fynjy0 [20]
3 years ago
10

Please include work! :)

Mathematics
1 answer:
UNO [17]3 years ago
6 0

Answer:

m = 13.

Step-by-step explanation:

Recall that adding logarithmic equations, (ex: log a + log b) is the same as log (a·b). Therefore:

log 12 + log 5 = log (4m + 8)

log (12·5) = log (4m + 8)

log (60) = log (4m + 8)  **Ignore 'Log' to solve for 'm'

60 = 4m + 8  **Subtract both sides by 8

52 = 4m  **Divide both sides by 4

m = 13.

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Jeremy bought a 5-kilogram can of peanuts for $4.50. What is the unit price?
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disa [49]

Answer:

The  Decision Rule

Fail to reject the null hypothesis

The conclusion

 There is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

Step-by-step explanation:

From the question we are told that

   The data is  

      Car Ages 4 0 8 11 14 3 4 4 3 5 8 3 3 7 4 6 6 1 8 2 15 11 4 1 6 1 8

     Taxi Ages 8 8 0 3 8 4 3 3 6 11 7 7 6 9 5 10 8 4 3 4

      The  level of significance \alpha = 0.05

 Generally the null hypothesis  is  H_o  :  \mu_1 - \mu_2  = 0

                  the alternative hypothesis is   H_a  :  \mu_1 - \mu_2 >  0

Generally the sample mean for the age of  cars is mathematically represented as

        \= x_1 = \frac{\sum x_i }{n}

=>     \= x_1 = \frac{4+ 0+ 8 +11 + \cdots + 8
}{27}

=>     \= x_1 = 5.56

Generally the standard deviation of age of  cars

     \sigma _1  = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }

=>  \sigma _1  = \sqrt{\frac{(4 - 5.56)^2 + (0 - 5.56)^2+ (8 - 5.56)^2 + \cdots + 8}{ 27} }

=>  \sigma _1  =  3.88

Generally the sample mean for the age of taxi is mathematically represented as

        \= x_2 = \frac{\sum x_i }{n}

=>     \= x_2 = \frac{8 +8 +0  + \cdots + 4
}{20}

=>     \= x_2 = 5.85

Generally the standard deviation of age of  taxi

\sigma _2  = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }

=>  \sigma _2  = \sqrt{\frac{(8 - 5.85)^2 + (8 - 5.85)^2+ (0 - 5.85)^2 + \cdots + 8}{ 20} }

=>  \sigma _2  = 2.83

Generally the test statistics is mathematically represented as

   t = \frac{(\= x_ 1 - \= x_2 ) - 0}{\sqrt{\frac{\sigma^2_1}{n_1}  + \frac{\sigma^2_2}{n_2} }  }

=> t = \frac{(5.56 - 5.85 ) - 0}{\sqrt{\frac{(3.88)^2}{27}  + \frac{(2.83)}{20} }}  

=> t = -0.30  

Generally the degree of freedom is mathematically  represented as

   df =  n_1 + n_2 -2

    df =  27 +  20 -2

    df =  45

From the t distribution table  the P(t >  t ) at the obtained degree of freedom = 45 is  

   P(t >  -0.30 ) = 0.61722067

So  the  p-value  is

    p-value  =  P(t >  T) =  0.61722067

From the obtained values we see that the  p-value  >  \alpha hence we fail to reject the null hypothesis

Hence the there is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

5 0
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