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nika2105 [10]
2 years ago
5

Soup ordinarily priced at 2 cans for .33 cents may be purchased in lots of one dozen for 1.74, what is the savings per a can whe

n it is purchased in this way
Mathematics
1 answer:
Sidana [21]2 years ago
5 0

Answer:

Savings per can would be 0.02 cents when purchased in lots.

Step-by-step explanation:

Given:

Ordinary price of soup 2 cans = 0.33 cents.

When purchases in lots 12 cans = 1.74

We need to find the saving per can when purchased in Lots.

Solution:

Ordinary price of soup 2 cans = 0.33 cents.

1 can  = Cost of 1 can when purchased ordinary.

By Using Unitary method we get;

Cost of 1 can when purchased ordinary = \frac{0.33}{2} = 0.165\ cents/can

Now we will find the Cost of 1 can when purchased in lot.

12 cans = 1.74

1 can = Cost of 1 can when purchase in lot.

Again by using Unitary method we get;

Cost of 1 can when purchase in lot = \frac{1.74}{12} = 0.145\ cents/can

Savings = Cost of 1 can when purchase in lot - Cost of 1 can when purchased ordinary

Savings = 0.165-0.145=0.02\ cents/can

Hence Savings per can would be 0.02 cents when purchased in lots.

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Answer:

The base is reducing at a rate of 3.4cm per min.

Step-by-step explanation:

Let The Height of the Triangle, h

Area of the triangle = A

Area of a Triangle, A= \frac{1}{2}bh

When A=190 cm^2, h=10cm

190= \frac{1}{2}*b*10\\b=38cm

\frac{dA}{dt}=\frac{1}{2}h\frac{db}{dt}+ \frac{1}{2}b\frac{dh}{dt}\\\frac{dh}{dt}=1 cm/min, \frac{dA}{dt}=2cm^2/min,

2=\frac{1}{2}*10*\frac{db}{dt}+ \frac{1}{2}*38*1\\2=5\frac{db}{dt}+19\\2-19=5\frac{db}{dt}\\-17=5\frac{db}{dt}\\\frac{db}{dt}=-\frac{17}{5} =-3.4 cm/min

The base is reducing at a rate of 3.4cm per min.

8 0
2 years ago
Find the greatest common factor of these three expressions.<br> 18x^2, 42x^4, and 30x^5
DanielleElmas [232]

Answer:I think the answer is 6x because

18= 6x3

42= 6x7

30= 6x5

Step-by-step explanation:

18= 6x3

42= 6x7

30= 6x5 And 6x(3x+7x^3+5x^4) it became 18x^2+42x^4+30x^5

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2 years ago
Samples of a cast aluminum part are classified on the basis of surface finish (in microinches) and edge finish. The results of 1
liubo4ka [24]

Answer:

a) P (A)= 88/102= 0.8627

(b) P (B)=  89/102= 0.8725

c) P (A`) =  14/102 = 0.1372

(d) P (A∩B) = 84/102 =0.8235

(e) P(AUB)= 93/102 = 0.9117

(f)P (A`UB) = 98/102= 0.96078

Step-by-step explanation:

                                                                   edge                  

                                                  finish excellent       good           Total

<u>(A )surface finish excellent            84                      4                  88</u>

good                                    ( B) ⇵    5                       9                  14

Total                                               89                      13                102

a) Upper P left-parenthesis Upper A right-parenthesis equals

P (A)= 88/102= 0.8627

All the elements of set A = 84+4= 88

(b) Upper P left-parenthesis Upper B right-parenthesis equals

P (B)=  89/102= 0.8725

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c) Upper P left-parenthesis Upper A prime right-parenthesis equal

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All the elements of Universal set U which are not elements of set A = 102- 88= 14

(d) Upper P left-parenthesis Upper A intersection Upper B right-parenthesis equals

P (A∩B) = 84/102 =0.8235

Only those elements of set A and set B which are common

(e) Upper P left-parenthesis Upper A union Upper B right-parenthesis equals

P(AUB)= 93/102 = 0.9117

Totalling elements of set A and B= 88+5= 93

(f) Upper P left-parenthesis Upper A prime union Upper B right-parenthesis equals

P (A`UB) = 98/102= 0.96078

All the elements of Universal set U which are not elements of set A and the elements of Set B = 5+9+ 84= 98

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