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rodikova [14]
3 years ago
11

A charitable organization packed 72 bags of rice, 48 blankets, and 24 cases of water equally in the boxes. Find the greatest pos

sible number of boxes that the items can be packed into so that there are no leftovers.
Mathematics
1 answer:
AysviL [449]3 years ago
6 0

Answer:

The greatest possible number of boxes that is needed to pack the items with no leftovers is 24.

Step-by-step explanation:

Given:

Number of rice bags = 72

Number of blankets =48

Number of water cases = 24

To Find :

The greatest possible number of boxes that the items can be packed into so that there are no leftover = ?

Solution:

we can find the greatest possible number of boxes that is required by finding the HCF(Highest common factor) of(24,48,72)

Finding the HCF (Highest common factor):

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

24 is the greatest common factor

Hence 24 will the number boxes required

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The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation 5 pounds. If 50,000 part
nalin [4]

Answer:

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

7933 parts

b) How many would have a tensile strength in excess of 48 pounds?

2739.95 parts

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

z = (x-μ)/σ

x = 35 μ = 40 , σ = 5

z = 35 - 40/5

= -5/-5

= -1

Determining the Probability value from Z-Table:

P(x<35) = 0.15866

Converting to percentage = 15.866%

We are asked how many will fail to meet this specification

We have 50,000 parts

Hence,

15.866% of 50,000 parts will fail to meet the specification

= 15.866% of 50,000

= 7933 parts

Therefore, 7933 parts will fail to meet the specifications.

b) How many would have a tensile strength in excess of 48 pounds?

z = (x-μ)/σ

x = 48 μ = 40 , σ = 5

z = 48 - 40/5

z = 8/5

z = 1.6

P-value from Z-Table:

P(x<48) = 0.9452

P(x>48) = 1 - P(x<48)

1 - 0.9452

= 0.054799

Converting to percentage

= 5.4799%

Therefore, 5.4799% will have an excess of (or will be greater than) 48 pounds

We are asked, how many would have a tensile strength in excess of 48 pounds?

This would be 5.4799% of 50,000 parts

= 5.4799% × 50,000

= 2739.95

Therefore, 2739.95 parts will have a tensile strength excess of 48 pounds

4 0
3 years ago
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