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Lena [83]
4 years ago
13

Of the following data, 17, 20,14,5,30,13,6, and 5 find the mean median, and mode

Mathematics
1 answer:
Kazeer [188]4 years ago
6 0
Arrange in ascending order:-

5 5 6 13 14 17 20 30

Mode = 5

Median = mean of middle numbers = (13+14) / 2 = 13.5
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Help would be appreciated,i wasnt there for most of this unit and don't know how to solve this
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3 years ago
Wilfred is feeding his goats. He gives 56% of the bag of feed to the goats in field A and the rest to the goats in field B. If t
krok68 [10]

Answer:

Feeds = 11.76kg

Step-by-step explanation:

Given

Field\ A= 56\%

Field\ B= The\ rest

Bag = 21kg\ feed

Required

Determine how much goats in field A, get.

To do this, we simply multiply the percentage feed received by goats in field A by the actual bags of feeds

i.e.

Feeds = 56\% * 21kg

Convert percentage to decimal

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<em>Hence, goats in field A get 11.76kg of feeds</em>

6 0
3 years ago
What is the answer? I can’t find it
monitta

Answer:

B. 360

Step-by-step explanation:

5 0
3 years ago
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The cost of 5 dresses and 2 belts is $48. The cost of 3 dresses and 2 belts is $32. Find the cost of a dress and a belt. Let x d
sergejj [24]
I think the answer is as
5 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
3 years ago
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