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BigorU [14]
3 years ago
9

Which describes the best relationship between the age of cheese and the amount of lactic acid present in the cheese

Mathematics
1 answer:
katrin2010 [14]3 years ago
6 0
Positive correlation
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A rectangular lap pool measures 80 feet long by 20 feet wide if it needs to be filled to 48 and each cubic foot hose 7.5 gallons
zmey [24]

Answer: it will take 576000 gallons to fill the lap pool.

Step-by-step explanation:

The formula for determining the volume of water in the rectangular pool is expressed as

Volume = length × width × height

The rectangular lap pool measures 80 feet long by 20 feet wide if it needs to be filled to 48. It means that the volume of water that would be pumped inside the pool is

Volume = 80 × 20 × 48 = 76800 cubic feet

1 cubic foot = 7.5 gallons

76800 cubic feet = 76800 × 7.5 = 576000 gallons

5 0
3 years ago
Evaluate: 4+8 divided by 2x (6-3)
Volgvan

Answer:

Step-by-step explanation:

(4+8)/(2x(6-3)) = 12/(2x(6-3))  = 12/6x = 2/x

3 0
3 years ago
HELP WILL MARK BRAINLIEST
vaieri [72.5K]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
7. In A.ABC. JB and KA are medians, JK = 10x - 12, AB = 9x + 18, JM = 21, KM = 23. AJ = 60. and
Juliette [100K]

Answer:

A) JK and AB =

[tex] \boxed{( \theta) = \f{12}{( \theta)} } So, =》 \dfrac{10}{5} = \dfrac{1}{ \( \theta) } =》 \( \theta) = \dfrac{5}{2}[/tex]

B) AABC

[tex] \boxed{( \theta) = {1}{( \theta)} } So, 23}{-18=\\Ans } \boxed

C) AABM:

[tex] e \fbox{: } The value of k for which \sin(jm) = \cos(x) is \dfrac{\p}{2} ans[/tex]

----------

3 0
3 years ago
Bags of pretzels are sampled to ensure proper weight. The overall average for the samples is 9 ounces. Each sample contains 25 b
77julia77 [94]

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  

     

4 0
2 years ago
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