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harina [27]
1 year ago
7

Erika calculated that she would spend $115 on school supplies this year. She actually spent $82.50 on school supplies. What is E

rika’s percent of error?
a 28.3%
b 2.8%
c 39.4%
d 32.5%
Mathematics
2 answers:
Sophie [7]1 year ago
7 0

Answer:

The answer is actually C

Step-by-step explanation:

This is because you do 115 - 82.50 = 32.5 then divide by 82.5 = 0.394 multiply by 100 = 39.4%. A lot of people use the theoretical number to divide instead of the actual, because people confuse the number 115 the theoretical number with the actual one. This causes people to use the number 115 to divide when you should use 82.5 to divide.

DerKrebs [107]1 year ago
6 0
The equation to find percent error is:
percent error=(experimental-theoretical)/theoretical *100
So, (115-82.5)/82.5=0.393939*100
39.3939%, or 39.94%

Hope this helps!!
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A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 449 gram setting. It is
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We accetp  H₀

Step-by-step explanation:

Information:

Normal distribution  

Population mean      =   μ₀  = 449

Population standard deviation  σ   unknown

Sample size   n  =  23        n < 30    we use t-student test

so   n  =  23    degree of fredom   df = n  - 1  df  = 23- 1   df = 22

Sample mean    μ =  448

Sample standard deviation   s  =  20

Significance level  α  =  0,05  

1.-Hypothesis Test

Null hypothesis                               H₀     μ₀  =  449

Alternative hypothesis                    Hₐ     μ₀  ≠  449

Problem statement ask for determine decision rule for rejecting the null hypothesis. For rejecting the null hypothesis we have to  get an statistic parameter wich implies  that μ is bigger or smaller than μ₀

2.-Significance level   α  =  0,05  ;  as we have a two tail test

α/2    =  0,025

Then from t - student table for  df =  22   and 0,025 (two tail-test)

t(c)  =  ±  2.074

3.- Compute  t(s)

t(s)   =  (  μ  -  μ₀ )  /  s /√n

plugging in values

t(s)   =  (448  -  449) /  20 /√23    ⇒   t(s)   =  -  1*√23 /20

t(s)   =  - 0.2398

4.-Compare t(c)   and  t(s)

t(s)  <  t(c)         - 0.2398  <  - 2.074

Therefore  t(s)  in inside acceptance region.  We accept  H₀

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