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DerKrebs [107]
3 years ago
13

the price of one share of ABC company decreased a total of $45 in 5 days what was the average change of the price of one share p

er day
Mathematics
1 answer:
garri49 [273]3 years ago
6 0
9 because if u divide 45 by 5 you get 9
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At a local pet store guppies are on sell for $0.70 each and sword tails $1.80 each. If amanda buys 4 guppies and 3 sword tails h
Lesechka [4]

Answer:

$8.20

Step-by-step explanation:

0.70*4=2.80

1.80*3=5.40

5.40+2.80=8.20

$8.20

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The graph represents function 1, and the equation represents function 2:
inessss [21]

Answer:

Option D.

Step-by-step explanation:

The slope of a horizontal line is 0.

It is given that the function 1 is a horizontal line that passing through the y-axis at y = 4.

It means the rate of change of function 1 is 0.

The slope intercept form of a linear function is  1

where, m is slope and b is y-intercept.

The function 2 is  2

On comparing (1) and (2), we get

The rate of change of function 2 is 8.

The difference between rate of change is

The rate of change of function 2 is 8 more than the rate of change of function 1.

Therefore, the correct option is D.

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3 years ago
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The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the c
Kitty [74]

Answer:

The 90% confidence interval  -49.8

The null hypothesis is  H_o : \mu_1 = \mu_2

The alternative hypothesis H_a  :  \mu_1 <  \mu_2

The distribution test statistics is t = -3.222

The rejection region is  p-value < \alpha

The decision rule is reject the null hypothesis

The conclusion is

      There is sufficient evidence to conclude that there are more passengers riding the 8:30  train

The p-value  is  p-value  =0.000951

Step-by-step explanation:

From the question we are told that

    The first sample size n_1 = 30

    The first sample mean is  \= x _1 = 323

    The first standard deviation is s_1 = 41

    The second sample size is n_2 = 45

    The second sample mean is  \=x_2 = 356

    The second standard deviation is s_2 = 45

given that the confidence level is 90% then the level of significance is mathematically represented as

          \alpha  = (100 -90)\%

         \alpha  = 0.10

Generally the critical value of \frac{\alpha }{2} obtained from the normal distribution table is  

   Z_{\frac{\alpha }{2} } = 1.645

Generally the pooled variance is mathematically represented as

        s^2 = \frac{(n_1 - 1)s_1^2  + (n_2 -1)s_2^2 }{n_1 + n_2 -2}

      s^2 = \frac{(30 -1)(41^2) + (45-1)45^2}{30+45 -2}

     s^2 = 1888.34

Generally the standard error is mathematically represented as

     SE =  \sqrt{\frac{s^2}{n_1} + \frac{s^2}{n_2}  }

=>  SE =  \sqrt{\frac{1888.34}{30} + \frac{1888.34}{45}  }

=>   SE =  10.24

Generally the margin of error is mathematically evaluated as  

      E =  Z_{\frac{\alpha }{2} } * SE

       E =  1.645* 10.24

       E = 16.85

Generally the 90% confidence interval is mathematically represented as

     \=x_1 -\=x_2 -E < \mu_1 -\mu_2 < \=x_1 -\=x_2 +E

     323 -356 -16.84

     -49.8

The null hypothesis is  H_o : \mu_1 = \mu_2

The alternative hypothesis H_a  :  \mu_1 <  \mu_2

Generally the test statistics is mathematically represented as

     t =  \frac{\= x_1 - \=x_2 }{SE}

=>   t = \frac{323-356}{10.24}

=>   t = -3.222

Generally the degree of freedom is mathematically represented as

     df =  n_1+n_2 -2

      df = 30 + 45 -2

     df = 73

The p-value is obtained from the student t distribution table at degree of freedom of 73 at 0.05 level of significance

    The value is  p-value  =0.000951

Here the level of significance is  \alpha =  5\%  =  0.05

Given that the p-value < \alpha then we  reject the null hypothesis

Then the conclusion is  

  There is sufficient evidence to conclude that there are more passengers riding the 8:30  train

               

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