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dybincka [34]
3 years ago
8

~[~p/\(q\/~p)] true or false

Mathematics
1 answer:
iren [92.7K]3 years ago
6 0
I believe you're answer is false.
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16 is the answer jxhsjcjaks
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Please help asap!! Will give brainlist :)
bezimeni [28]
214.2 yd? I’m not 109% percent sure
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Please help I need it now please and thank you
Bingel [31]

Answer:

It would be the first one or second one

Step-by-step explanation:

the answer is up ahead

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Allied Corporation is trying to sell its new machines to Ajax. Allied claims that the machine will pay for itself since the time
kvasek [131]

Answer:

z(s) is in the acceptance region. We accept H₀  we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

Step-by-step explanation:

We must evaluate the differences of the means of the two machines, to do so, we will assume a CI  of 95%, and as the interest is to find out if the new machine has better performance ( machine has a bigger efficiency or the new machine produces more units per unit of time than the old one) the test will be a one tail-test (to the left).

New machine

Sample mean                  x₁ =    25

Sample variance               s₁  = 27

Sample size                       n₁  = 45

Old machine

Sample mean                    x₂ =  23  

Sample variance               s₂  = 7,56

Sample size                       n₂  = 36

Test Hypothesis:

Null hypothesis                         H₀             x₂  -  x₁  = d = 0

Alternative hypothesis             Hₐ            x₂  -  x₁  <  0

CI = 90 %  ⇒  α =  10 %     α = 0,1      z(c) = - 1,28

To calculate z(s)

z(s)  =  ( x₂  -  x₁ ) / √s₁² / n₁  +  s₂² / n₂

s₁  = 27     ⇒    s₁²  =  729

n₁  = 45    ⇒   s₁² / n₁    = 16,2

s₂  = 7,56   ⇒    s₂²  = 57,15

n₂  = 36     ⇒    s₂² / n₂  =  1,5876

√s₁² / n₁  +  s₂² / n₂  =  √ 16,2  + 1.5876    = 4,2175

z(s) = (23 - 25 )/4,2175

z(s)  =  - 0,4742

Comparing z(s) and  z(c)

|z(s)| < | z(c)|  

z(s) is in the acceptance region. We accept H₀  we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean

3 0
3 years ago
Can someone please help me with this?
madreJ [45]

Answer:

2.664 * 10^{7}

Step-by-step explanation:

6 0
3 years ago
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