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vitfil [10]
3 years ago
9

Learning Task 1: Read and analyze each problem carefully. Write your final answer in your notebook based on the given questions

below.
1) There are 2 2/3 pizzas. How many people are sharing when each has 2/3 of pizza?
2) A plastic bottle of mineral holds 600 ml of water when it is 3/5 full. What is the capacity of the bottle
3) Joel spend 4/8 of his monthly salary for their groceries. After spending had Php 980 left. How much his monthly salary?

Grade: V/5
Sub: Math​
Mathematics
2 answers:
solmaris [256]3 years ago
8 0

Solving the given problems using the appropriate arithmetic operation such as multiplication, divison and addition, the answers are :

  • <em>4</em><em> </em><em>people</em><em> </em>
  • <em>1000</em><em> </em><em>ml</em><em> </em>
  • <em>$</em><em>1960</em>

Total size of pizza = 2 \frac{2}{3} = \frac{8}{3}

Number of pizzas gotten by each person = \frac{2}{3

<u>Number of people sharing pizza</u> :

\frac{8}{3} \div \frac{2}{3} = \frac{8}{3} \times \frac{3}{2} = 4 \: people

2.)

  • <em>Volume bottle = 600 ml</em>

  • Fraction of water = \frac{3}{5}

Volume of full bottle = \frac{3}{5} \times n =  600 = \frac{3n}{5} = 600

Volume of full bottle = \frac{3n}{3000}

Volume of full bottle = n = \frac{3000}{3} = 1000 ml

3.)

  • Amount left = Php. 980
  • Fraction spent = \frac{4}{8}

<em>Let the monthly salary = m</em>

Fraction left = 1 - \frac{4}{8} =  \frac{4}{8} = \frac{1}{2}

Hence,

\frac{1}{2} \times m = 980

\frac{m}{2} = 980

m = 980 \times 2 = 1960

Therefore, Joel's monthly salary is $1960

Learn more :brainly.com/question/18796573

harkovskaia [24]3 years ago
7 0

Answer:

1. 4

2. 1000

3. 1960

thank you

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The mean of the data is the average value of the given data. The mean of the data is the ratio of the sum of all the values of data to the total number of values of data.

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Mode of a data set is the value, which occurs most times for that data set. The value which has the highest frequency in the given set of data is known as the mode of that data set.

The stem-and-leaf plot shows the number of digs for the top 15 players at a volleyball tournament. The table is attached below.

The data according to the table is,

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\mu=\dfrac{41+41+43+43+45+50+52+53+54+62+63+63+67+75+97}{15}\\\mu=\dfrac{849}{15}\\\mu=56.5

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The arranged data according to the table is,

41,41,43,43,45,50,52,53,54,62,63,63,67,75,97

Here, the middle value is 53. Thus, the median is 53.

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Mode is the data which occur most offen. Here the order of modes from least to greatest is,

41   41   43   43   45   50   52   53   54   62   63   63   67   75   97  

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Lowest value in data is 41 and highest is 97. Thus, the range is,

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  • The interquartile range is.

The third quartile of data is 63 while first quartile is 43. Thus, the interquartile range is.

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Thus, the data of players of volleyball tournament, the mean is 56.5, median is 53, range is 56 and interquartile range is 20.

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