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hram777 [196]
4 years ago
12

PLEASE SHOW WORK! Neil decides to go to a technical college for 18 months to get a degree in computer networking. There are 6 sc

hool quarters in the program with a tuition of $530 each. Books for the entire program cost $1800. How much will it cost Neil to get his degree?
Mathematics
1 answer:
sukhopar [10]4 years ago
5 0

Total cost for Neil degree is $ 4980

<em><u>Solution:</u></em>

Given that,

Neil decides to go to a technical college for 18 months to get a degree in computer networking

Number of months = 18 months

There are 6 school quarters in the program with a tuition of $530 each

Thus there are 6 quarters

For one quarter, tuition fee is $ 530

For 6 school quarters, tuition fee is given as:

\rightarrow 6 \times 530 = 3180

Books for the entire program cost $1800

Total cost for degree = tuition cost + books cost

Total cost for degree = 3180 + 1800

Total cost for degree = 4980

Thus total cost for Neil degree is $ 4980

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if p(x)=x*2-1 and q(x)=5(x-1),which expression is equivalent to (p-q)(x)?
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(p - q)(x) = (x -1) (x - 4)

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(p - q)(x)

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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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Answer:

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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