Evaluate the distance between each pair of integers
1 answer:
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Solution:
To save Some amount of money, Janine is considering buying a water filter and a reusable water bottle rather than buying bottled water.
To answer this question, we have to look at the thought process of Janine, that why she has done this?
The information that we need to answer this Question: is Written in Bold Letters
1. Amount of Water Drunk by Janine in a day= 4 × 16.9 Ounces= 67.6 Ounces= Approximately 2 Liter
2. Cost of 24-packs of 16.9 ounce bottles = $3.89
Cost of 4×6 packs of 16.9 ounce bottles =$3.89
Cost of 2 liter or 4 packs of 16.9 ounce bottles =$
3. Cost of Reusable water bottle = $ 10
4. Time for which a reusable water bottle last= Infinitely
5. Cost of a faucet-mounted filter= $ 28
Cost of Refilling for a 3 pack filter=$ 33
So, Cost of Refilling a 1 pack filter= $11
1 filter will filter approximately =100 gallons of water=378 Liters
6. Cost of a water filter pitcher= $ 22
Now, coming to second part ,whether buying a water filter and a reusable water bottle rather than buying bottled water will save her money?
→Total cost of water filter and a reusable water bottle = $28+$10+ $ (33÷3) ×(Number of times She refills the 3 filter pack)= $ 38 + $11 n
For, 1 water filter and a reusable water bottle total money spent = $ 49
→As the filter contains , 378 Liters of water, and Janine drinks Approx 2 liters of water daily, so 1 filter will last for 189 days.
→1 Liter = 33.81 Ounce
→Cost of 378 liters of water== $ 122.535
So, We can Surely say that , buying a water filter and a reusable water bottle rather than buying bottled water will save Janine money.
Answer:
A) See the picture
B) 14
C) 45%
Step-by-step explanation:
A) To create a histogram like the one on the picture you can use an online tool like socscistatistics where the number of classes is customizable
B) Because the question B and C have to be responded using a frequency table with 8 classes the answer is 14; the method of using cumulative frequency tables should only be considered as a way of estimation, that is because you obtain values that depend on your choice of class intervals. The way to get a better answer would be to use all the scores in the distribution
Pc1 = 100*(4/40) = 10
Pc2 = 100*(4/40) = 10
Pc3 = 100*(3/40) = 7.5
Pc4 = 100*(11/40) = 27.5
Pc5 = 100*(5/40) = 12.5
Pc6 = 100*(4/40) = 10
Pc7 = 100*(7/40) = 17.5
Pc8 = 100*(2/40) = 5
Pc8 + Pc7 + Pc6 + Pc5 + Pc4 + Pc3 + Pc2 = 90%
Therefore, From class 8 to class 2 is the top 90% of the applicants and the minimum score is 14.
C) Scores equal to or greater than 20 are from class 8 to class 5
Pc8 + Pc7 + Pc6 + Pc5 = 45%
