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Volgvan
1 year ago
10

On the first day of a family road trip, Kai's family travels 365 miles.

Mathematics
1 answer:
Gekata [30.6K]1 year ago
7 0
They traveled 292 miles on day two.

Known: On the first day they traveled 365 and on the second they traveled 20% less.

Solution:

If they traveled 20% less on the second day, that means they traveled 80% of the distance they traveled the first day.

365 miles * .8 = 292.

You could also solve this as:
20% of 365 is 73 miles
365 * .2 = 73.
So they traveled 73 less miles on the second day.
365 miles on the first day - 73 miles less on the second day = 292 miles.

I hope this helps!
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For this case we can make the following rule of three:

25% ----------------> 2 $

100% --------------> x

From here, we clear the value of x.

We have then:

Therefore, the value of the Taco that Andrea bought, before the tip, is $ 8.

the price of Andrea's tacos, before tip is $ 8.

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Erik drives 19.2 miles in 16 minutes.
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there are two pizzas. conor ate 1⁄4 of a pizza, brandon 2⁄8, tyler 3⁄4, and audrey 4⁄8. who ate the most of the two pizzas? a. a
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4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln
pochemuha

Answer:

No these these result do not differ at 95% confidence level  

Step-by-step explanation:

From the question we are told that

  The first concentrations is  c _1= 30.0 \ g/m^3

      The second concentrations is  c _2 = 52.9 \ g/m^3

  The first sample size is  n_1 =  32

    The second sample size is  n_2 =  32

   The  first standard deviation is \sigma_1 =  30.0

     The  first standard deviation is \sigma_1 =  29.0

The mean for  Turnpike is  \= x _1 = \frac{c_1}{n}  =  \frac{31.4}{32} = 0.98125

The mean for   Tunnel is  \= x _2 = \frac{c_2}{n}  =  \frac{52.9}{32} = 1.6531

The  null hypothesis is  H_o  :  \mu _1 - \mu_2  =  0

The  alternative hypothesis is  H_a  :  \mu _1 - \mu_2  \ne  0

Generally the test statistics is mathematically represented as

              t =  \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1}  +\frac{\sigma_2^2}{n_2} }}

         t =  \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32}  +\frac{29^2}{32} }}

        t = - 0.0899

Generally the degree of freedom is mathematically represented as

     df =  32+ 32 - 2

     df =  62

The  significance \alpha  is  evaluated as

      \alpha  =  (C - 100 )\%

=>   \alpha  =  (95 - 100 )\%

=>   \alpha  =0.05

The  critical value  is evaluated as

      t_c  =  2  *  t_{0.05 ,  62}

From the student t- distribution table  

        t_{0.05, 62} =  1.67

So

     t_c  =  2 * 1.67

=>  t_c  = 3.34

given that

       t_c  >  t we fail to reject the null hypothesis so  this mean that the result do  not differ

       

6 0
2 years ago
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