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sweet [91]
2 years ago
5

Bags of pretzels are sampled to ensure proper weight. The overall average for the samples is 9 ounces. Each sample contains 25 b

ags. The standard deviation is estimated to be 3 ounces. The upper control chart limit (for 99.7% confidence) for the average would be ________ ounces.
Mathematics
1 answer:
77julia77 [94]2 years ago
4 0

Answer:

The value is UCL  =  10.8  

Step-by-step explanation:

From the question we are told that

    The sample mean is  \= x  =  9 \  ounce

    The sample size is  n =  25

    The standard deviation is  \sigma =  3 \ ounce

Given that the sample size is not large enough i.e  n<  30  we will make use of the student t distribution table  

From the question we are told the confidence level is  99.7% , hence the level of significance is    

      \alpha = (100 -99.7 ) \%

=>   \alpha = 0.003

Generally the degree of freedom is  df =  n- 1

=>  df =  25 - 1

=>  df =  24  

Generally from the student t distribution table the critical value  of  \frac{\alpha }{2} at a degree of freedom of df =  24   is  

   t_{\frac{\alpha }{2} , 24 } = 3.0

Generally the margin of error is mathematically represented as  

      E = t_{\frac{\alpha }{2} , 24} *  \frac{\sigma }{\sqrt{n} }

=>     E = 3.0 *  \frac{3 }{\sqrt{25} }

=>     E =1.8

Gnerally the  upper control chart limit  for 99.7% confidence is mathematically represented as

         UCL  =  \= x  + E

=>      UCL  =  9 + 1.8  

=>      UCL  =  10.8  

     

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An arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second. Let up
Papessa [141]

Answer:

s(t)=-9.8t^2+49t+58.8

Step-by-step explanation:

We have been given that an arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second. Let up be the positive direction. Because gravity is the force pulling the arrow down, the initial acceleration of the arrow is −9.8 meters per second squared.

We know that equation of an object's height t seconds after the launch is in form s(t)=-gt^2+v_0t+h_0, where

g = Force of gravity,

v_0 = Initial velocity,

h_0 = Initial height.

For our given scenario g=-9.8, v_0=49 and h_0=58.8. Upon substituting these values in object's height function, we will get:

s(t)=-9.8t^2+49t+58.8

Therefore, the function for the height of the arrow would be s(t)=-9.8t^2+49t+58.8.

6 0
3 years ago
The value of sin^2 30° - cos^2 30° is​
natulia [17]

Answer:

-1/2

Step-by-step explanation:

Im not going to go into detail about how to compute sin30 and cos 30, but know that sine and cosine are just the ratios between the sides of a 30-60-90 triangle with a hypothenuse of 1.

With that being said:

sin^2 30 - cos^2 30

=  (1/2)^2 - ((sqrt3)/2)^2

= 1/4 - 3/4

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7 0
3 years ago
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Melanie is 2 years older than twice her brother Bill’s age. If Bill is 5 years old,how old is Melanie
arlik [135]

she would be 12 because

10 * 2 = 10

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hope this helps :)

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the graph shows the location of a bank in a library each Square on the graph represents one block which path can be taken to go
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Answer:

Step-by-step explanation:

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3 years ago
01-1-
klio [65]

Answer:

Step-by-step explanation:

50 dollars = ∉34

∉1 = Rs 150

Rs 382800 / 150 Rs per ∉  =  382800/150 ∉

=2552 ∉

50 dollars = ∉34          so 50/34 dollars = ∉

2552 ∉ × 50/34 dollar per ∉ =       in the units the ∉ cancel out

                                                         ∉ × dollar/∉    = dollars

(2552 × 50) / 34    = 3752.94 dollars     <------------------

Here is another way of looking at the answer...........

Rs 382800 × 1∉/150 Rs × 50 dollars/34 ∉ =              note how the units cancel

= 382800 × (1/150) × (50/34)                          Rs × ∉/Rs × dollars/∉

=  3,752.94 dollars                                     leaving only dollars in the numerator

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