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babunello [35]
2 years ago
6

In a nursery class of 18 pupils, the average is 4

title=" \frac{1}{2} " alt=" \frac{1}{2} " align="absmiddle" class="latex-formula">
years. What is the sum of the ages of pupils?​
Mathematics
2 answers:
Alisiya [41]2 years ago
6 0

Answer:

The sum of ages of all 18 pupils in the class = 81

Step-by-step explanation:

Total number of pupil in the class =  18

Average of years in the class = 4\frac{1}{2}

Now, 4\frac{1}{2}  = 4 + \frac{1}{2}  = 4 + 0.5 = 4.5

⇒Average of sum of ages  in the class = 4.05

Let us assume the sum of ages of all students in the class = m

By the formula for AVERAGE:

\textrm{Average of n obseravtions}  = \frac{\textrm{Sum of n observations}}{\textrm{n}}

\implies 4.5 = \frac{m}{18}

or, m =  18 x 4.5 = 81

Hence, the sum of ages of all 18 pupils in the class = 81

nata0808 [166]2 years ago
3 0

Answer:

81

Step-by-step explanation:

Average = 4.5 years

Total = 18

Sum = 4.5 × 18 = 81

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sattari [20]

Answer:

50th term is -338

Step-by-step explanation:

Common difference d = -2-5 = -7

first term is a1 = 5

50th term

a50 = a1 + (n -1)d

a50 =  5 -7(50-1)

a50 = 5 -7(49)

a50 = -338

Hope this will helpful.

Thank you.

6 0
3 years ago
What does 0 mean in a number line
Ivenika [448]
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3 years ago
Imagine you have some workers and some handheld computers that you can use to take inventory at a warehouse. There are diminishi
Svet_ta [14]

Answer:

a. $1.03

b. $0.93

c. 0.98

d. 2 workers

Step-by-step explanation:

a. Given that:

  • 1 computer : 1 worker : inventory 150 items per hour
  • 1 computer : 2 workers : inventory 200 items per hour
  • 1 computer : 3 workers : inventory 220 items per hour
  • 1 computer : 4+ workers : fewer than 235 items per hour
  • Cost: $100 per computer ; $25 per worker

The fixed production factor in the warehouse is the computer used:

-One computer used, but the number of users is varied to inventory a specified number of items.

-The variable production factor is the number of workers assigned per one computer.

#The cost of inventorying a single item by one worker is:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_1=\frac{125+30}{150}\\\\\\=1.03

Hence, the cost of inventorying a single item is $1.03

b. Using the information provided above, the cost of inventorying a single item when two workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_2=\frac{125+2\times30}{200}\\\\\\=0.925

Hence, the cost of inventorying a single item is $0.93

c.Using the information provided above, the cost of inventorying a single item when three workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_3=\frac{125+3\times30}{220}\\\\\\=0.98

Hence, the cost of inventorying a single item is $0.98

d. To determine the most cost-effective job assignment, we calculate the cost of 4+ workers.

Take any number less than 235(say 234) as the inventory units:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_4=\frac{125+4\times30}{234}\\\\\\=1.05

From our calculations, it's clear that two workers per computer costs the least amount($0.93) per unit item. Hence, it is best to assign two workers per computer.

4 0
3 years ago
Simplify 3p^4(4p^4 7p^3 4p 1
inessss [21]
I hope this helps you




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252.p^12
3 0
3 years ago
Evaluate.
lys-0071 [83]

Answer:

  361/900

Step-by-step explanation:

  \left(-\dfrac{1}{6}+0.6\left(-\dfrac{1}{3}\right)+1\right)^2=\left(-\dfrac{1}{6}-0.2+1\right)^2\\\\=\left(1-\left(\dfrac{1}{6}+\dfrac{1}{5}\right)\right)^2=\left(1-\dfrac{5+6}{6\cdot5}\right)^2=\left(\dfrac{19}{30}\right)^2=\boxed{\dfrac{361}{900}}

8 0
1 year ago
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