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Amiraneli [1.4K]
2 years ago
8

Gobe weighs 53.6 kg. he wears shoes which weigh 450 gm and carries a school bag weighing 4.5 kg. if he stands on the scales in h

is shoes and carrying the school bag. what weight will be recorded?​
Mathematics
1 answer:
-Dominant- [34]2 years ago
6 0

Answer:

The key to this question is converting all the mass units to the same unit and adding them all up.

The only value that isn't consistent with the other units is 450 gm or grams.

Recall that 1 gram is equal to 0.001 kilo grams. So if you want to convert grams to kilograms you would multiply that amount by 0.001 or 1/1000

450gm * \frac{1}{1000} = 0.45kg

Now let's add up all the masses.

53.6kg + 0.45kg + 4.5kg = 58.55 kg\\

(Now a few footnotes here, that I considered but wasn't sure about. The answer asks for weight, but you're only given mass to work with. That would be fine, if the question asked you to convert it to SI weight units like Newtons, but there's no mention of that. And also I'm not sure if the question requires significant digits. But I'll continue the answer if you think so)

To convert this into weight in Newtons, multiply by 9.81 m/s^2

58.55kg*9.81m/s^2 = 574.3755 N\\\\= 5.7 * 10^2 N

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\displaystyle\\
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It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 122 feet. A transportat
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Complete Question

The complete question is shown on the first uploaded image

Answer:

the null hypothesis is  H_o  :  \mu =  122

the alternative hypothesis is H_a  :  \mu \ne  122

The test statistics is  t =  - 1.761

The p-value is  p =  P(Z  <  t ) = 0.039119

so

    p  \ge 0.01

Step-by-step explanation:

From the question we are told that

   The population mean is  \mu  = 122

     The sample size is  n=  38

    The sample mean is  \= x  =  116 \ feet

     The standard deviation is \sigma  =  21

     

Generally the null hypothesis is  H_o  :  \mu =  122

                the alternative hypothesis is H_a  :  \mu \ne  122

Generally the test statistics is mathematically evaluated as

         t =  \frac { \= x - \mu }{\frac{ \sigma }{ \sqrt{n} } }

substituting values

         t =  \frac { 116 -  122 }{\frac{ 21 }{ \sqrt{ 38} } }

         t =  - 1.761

The p-value is mathematically represented as

      p =  P(Z  <  t )

From the z- table  

     p =  P(Z  <  t ) = 0.039119

So  

     p  \ge 0.01

 

         

     

           

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