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Elodia [21]
2 years ago
10

Lynne bought a bag of grapefruit ,1 5/8 of apples ,and 2 3/16 pounds of bananas.The total weight of her purchases was 7 1/2 .How

much did the bag of grapefruit weigh
Mathematics
1 answer:
Vilka [71]2 years ago
7 0

Answer:

The weight of bag of grapefruit  =3\frac{27}{100} pounds

Step-by-step explanation:

The total weight of apples purchased   = 1  5/8 pounds

Now, 1\frac{5}{8}   = 1+ \frac{5}{8} = 1 + 0.625  =1.625

So, the weight of apples brought = 1.625 pounds.

The total weight of bananas purchased   = 2  3/16 pounds

Now, 12\frac{3}{16}   = 2+ \frac{3}{16} = 2+ 0.1875  =2.1875

So, the weight of bananas brought = 2.1875 pounds.

Let us assume the total weight of grapefruit weight = m pounds

Now, total weight = 7   1/12 pounds =  7.0833

⇒ Weight of ( Apples + Banana + Grapefruit) = Total Weight

⇒1.625 pounds +  2.1875 pounds + m = 7.0833 pounds

or, m  =  7.0833 - 3.8125 = 3 .27

or, m = 3.27

Now, 3.27  = 3 + 0.27  = 3 + \frac{27}{100}   = 3\frac{27}{100}

Hence. the weight of bag of grapefruit  =3\frac{27}{100}

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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

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Answer:

A. 29

B. 11

C. 10

D. 12

Step-by-step explanation:

A. 180 - 122 = 58

58 divided by 2 = 29

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B. 180 - 37 = 143

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C. the small box in a triangle means right angle = 90

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(2x - 2) we subtract before finding out what x is

22 - 2 = 20

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(8x + 3) + (2x + 9) + 4x = 180

add up the numbers with x and the numbers without them

14x + 12 = 180

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<span>(y=mx+b) or (ax+by=c)  hope this helped

</span>
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