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3 years ago

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Gina’s mother bought 200 lollipops and 150 pencils for goody bags for Gina’s birthday. What is the largest number of goody bags

that Gina's mother can make so that each goody bag has the same number of lollipops and the same number of pencils?

1 answer:

gtnhenbr [62]3 years ago

7 0

The greatest common factor is 50.
So it's fifty bags with 4 lollipops and 3 pencils in each bag.

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Molly tried to evaluate 93\times5193×5193, times, 51 using partial products. Her work is shown below.

Mrac [35]

Since molly's solution tally's with the given solution, hence <em>Molly's solution is correct.</em>

Given the working on a partial product of 93 and 51 carried out by Molly as shown:

$\begin{array}{llrr} &&93 \\ &&\underline{{}\times51} \\ &\blueD{\text{Step 1}}&\blueD{3}& \blueD{1\times3\text{ ones}}\\ &\greenD{\text{Step 2}}&\greenD{90}& \greenD{1\times 9\text{ tens}}\\ &\maroonD{\text{Step 3}}&\maroonD{150}& \maroonD{50\times 3\text{ ones}}\\ &\goldE{\text{Step 4}}&\underline{{}+\goldE{ 4{,}500}}& \goldE{50\times 9\text{ tens}}\\ &\purpleD{\text{Step 5}}&\purpleD{4{,}743}& \end{array}$

This partial product can also be solved as shown below:

$93 \times 51 = (90+3)\times (50+1)$

Applying the distributive law:

$93 \times 51 = 90(50) + 90(1) + 3(50) + 3(1)\\93 \times 51 =4500 + 90 + 150 + 3\\93 \times 51 =4500+240+3\\93 \times 51 =4740+3\\93 \times 51 =4743$

Since molly's solution tally's with the given solution, hence <em>Molly solution is correct.</em>

Learn more about partial product at: brainly.com/question/24716925

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A bag contains red, green, and yellow marbles. The probability of selecting a red marble is 35%. What is the probability of not

KonstantinChe [14]

Probablity of selecting a red marble= 35%

Probablity of not selecting a red marble= 100%-35%

=65%

Hope it helps you. Please mark brainliest.

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