Maggie has a box of 200 colored blocks. The box has an equal number of green and blue blocks and an equal number of red and yellow blocks. If Maggie arranged all of the green blocks in stacks of 10 and all of the blue blocks in stacks of 7, how many yellow blocks are in the box?
If g and b are the number of stacks, then
10g = 7b
LCM(7,10) = 70
So, the smallest number of blocks would be
70 green and 70 blue, leaving 60 red and yellow blocks.
So, 30 of them are yellow.
Could there be more green/blue? No, since that would require at least 280 blocks.
I hate furrys
thx dude
even tho u answerd this 3 years ago
damn
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A florist has 64 red balloons and 48 blue balloons. What is the greatest number of identical balloon bouquets that can be made without any balloons leftover?
Responses
A 1212
B 1616
C 88
D 1818
The greatest number of identical balloon bouquets that can be made without any balloons leftover is the greatest common factor (GCF) of 64 and 48.
To find the GCF, we can list the factors of each number:
Factors of 64: 1, 2, 4, 8, 16, 32, 64
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The common factors are 1, 2, 4, 8, and 16. The greatest common factor is 16.
Therefore, the answer is B) 16.
Pencils are sold 9 to a package, and pens are sold 6 to a package. If an equal number of pencils and pens are purchased, what is the minimum number of each item?
Responses
A 1818
B 2828
C 3636
D 54
For the minimum number of each item, we need to find the least common multiple (LCM) of 9 and 6, which is 18. Each package of pencils contains 9 pencils, which is twice the number of packages of pens, since each package of pens contains 6 pens.
So, if x is the number of packages of pens, then the number of packages of pencils is 2x. We want an equal number of pencils and pens, so we must have:
2x = x
which means:
x = 0
This is not possible, so we must increase the number of packages of pencils and pens by a common factor until we get to a valid solution.
Let's try:
2 packages of pencils (18 pencils)
1 package of pens (6 pens)
This gives us equal numbers of pencils and pens (18 pencils and 6 pens).
Therefore, the minimum number of each item is 2 packages of pencils and 1 package of pens, or a total of 18 pencils and 6 pens.
The answer is not listed in the choices given, so it may be a mistake or incomplete.
Pencils are sold 9 to a package, and pens are sold 6 to a package. If an equal number of pencils and pens are purchased, what is the minimum number of each item?
Responses
A 18
B 28
C 36
D 54
For the minimum number of each item, we need to find the least common multiple (LCM) of 9 and 6, which is 18. Each package of pencils contains 9 pencils, which is twice the number of packages of pens, since each package of pens contains 6 pens.
So, if x is the number of packages of pens, then the number of packages of pencils is 2x. We want an equal number of pencils and pens, so we must have:
2x = x
which means:
x = 0
This is not possible, so we must increase the number of packages of pencils and pens by a common factor until we get to a valid solution.
Let's try:
2 packages of pencils (18 pencils)
1 package of pens (6 pens)
This gives us equal numbers of pencils and pens (18 pencils and 6 pens).
Therefore, the minimum number of each item is 2 packages of pencils and 1 package of pens, or a total of 18 pencils and 6 pens.
The answer is A) 18.
Find the LCM of 10 and 16.
We can find the LCM (Least Common Multiple) of 10 and 16 using the prime factorization method or by finding the multiples of the two numbers until we find a common one.
Prime factorization method:
10 factors into 2 x 5, while 16 factors into 2 x 2 x 2 x 2.
Then, we take the highest power of each prime factor.
So the LCM of 10 and 16 is (2 x 2 x 2 x 2) x 5 = 80.
Multiples method:
Multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80, ...
Multiples of 16: 16, 32, 48, 64, 80, ...
The LCM of 10 and 16 is the first common multiple, which is 80.
Either way, we get that the LCM of 10 and 16 is 80.
Which number is NOT a multiple of 10?
Responses
A 30
B 22
C 70
D 10
A number is a multiple of 10 if it ends with a 0.
Option A (30) and option C (70) both end with 0 and are therefore multiples of 10. Option D (10) itself is 10 and is therefore a multiple of 10.
Only option B (22) does not end with 0, so it is NOT a multiple of 10.
Therefore, the answer is B) 22.
Which number is NOT a multiple of 10?
Responses
A 30
B 2
C 70
D 10
A number is a multiple of 10 if it ends with a 0.
Option A (30) and option C (70) both end with 0 and are therefore multiples of 10. Option D (10) itself is 10 and is therefore a multiple of 10.
However, option B (2) does not end with 0, so it is NOT a multiple of 10.
Therefore, the answer is B) 2.
Find the GCF of the two numbers, and rewrite the sum of the two numbers using the distributive property.
14 + 24
Responses
A 2(7 + 12)
B 2(7 × 12)
C 4(10 + 20)
D 4(10 × 20)
To find the GCF (Greatest Common Factor) of 14 and 24, we can list the factors of each number and find the highest one they have in common:
Factors of 14: 1, 2, 7, 14
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
The highest common factor is 2, so the GCF of 14 and 24 is 2.
Using the distributive property, we can rewrite the sum of the two numbers as follows:
14 + 24 = 2 × 7 + 2 × 12
= 2(7 + 12)
Therefore, the answer is A) 2(7 + 12).
Find the GCF of the two numbers, and rewrite the sum of the two numbers using the distributive property.
18 + 42
Responses
A 6(3 + 7)
B 6(3 × 7)
C 3(6 + 14)
D 3(6 × 14)
To find the GCF (Greatest Common Factor) of 18 and 42, we can list the factors of each number and find the highest one they have in common:
Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42
The highest common factor is 6, so the GCF of 18 and 42 is 6.
Using the distributive property, we can rewrite the sum of the two numbers as follows:
18 + 42 = 6 × 3 + 6 × 7
= 6(3 + 7)
Therefore, the answer is A) 6(3 + 7).
Find the GCF of the two numbers, and rewrite the sum of the two numbers using the distributive property.
20 + 16
Responses
A 4(5 + 4)
B 2(10 + 8)
C 2 + (10 × 8)
D 4 + (5 × 4)
To find the GCF (Greatest Common Factor) of 20 and 16, we can list the factors of each number and find the highest one they have in common:
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 16: 1, 2, 4, 8, 16
The highest common factor is 4, so the GCF of 20 and 16 is 4.
Using the distributive property, we can rewrite the sum of the two numbers as follows:
20 + 16 = 4 × 5 + 4 × 4
= 4(5 + 4)
Therefore, the answer is A) 4(5 + 4).
Choose all that are correct.
Responses
A 64 + 24 = 8(8 + 3)
B 24 + 18 = 6(4 + 9)
C 45 + 54 = 9(7 + 6)
D 40 + 28 = 4(10 + 7)
To check if each equation is correct, we can use the distributive property to rewrite the sum of the two numbers and see if it matches the expression on the right-hand side.
A) 64 + 24 = 8(8 + 3)
Let's use the distributive property to check:
8(8 + 3) = 8 × 8 + 8 × 3 = 64 + 24
So, the equation is TRUE.
B) 24 + 18 = 6(4 + 9)
Let's use the distributive property to check:
6(4 + 9) = 6 × 4 + 6 × 9 = 24 + 54
The equation is NOT TRUE. The correct expression for 6(4+9) is 78, not 42.
C) 45 + 54 = 9(7 + 6)
Let's use the distributive property to check:
9(7 + 6) = 9 × 7 + 9 × 6 = 63 + 54
So, the equation is TRUE.
D) 40 + 28 = 4(10 + 7)
Let's use the distributive property to check:
4(10 + 7) = 4 × 10 + 4 × 7 = 40 + 28
So, the equation is TRUE.
The correct options are:
A) 64 + 24 = 8(8 + 3)
C) 45 + 54 = 9(7 + 6)
D) 40 + 28 = 4(10 + 7)
Therefore, the answer is options A, C, and D.
{49, 50, 45, 23, 35, 66, 34, 74, 54, 50}
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The mean is
The median is
The mode is
The range is
Suppose the value 60 is added to the data set.
The mean increases by
The median increases by
The mode is
The range is
The mean is the sum of all the values divided by the total number of values.
So, for {49, 50, 45, 23, 35, 66, 34, 74, 54, 50}, the sum is:
49 + 50 + 45 + 23 + 35 + 66 + 34 + 74 + 54 + 50 = 480
And the total number of values is 10.
So, the mean is:
Mean = (49 + 50 + 45 + 23 + 35 + 66 + 34 + 74 + 54 + 50)/10
= 48
The median is the middle value when the data set is arranged from smallest to largest.
Arranging the values in order, we get:
{23, 34, 35, 45, 49, 50, 50, 54, 66, 74}.
So, the median is the average of the two middle values, 49 and 50:
Median = (49 + 50)/2
= 49.5
The mode is the value that appears most frequently in the data set.
In this case, the value 50 appears twice, which is more frequently than any other value, so the mode is:
Mode = 50
The range is the difference between the largest and smallest values in the data set.
In this case, the smallest value is 23 and the largest value is 74, so the range is:
Range = 74 - 23
= 51
Now, we add the value 60 to the data set:
{49, 50, 45, 23, 35, 66, 34, 74, 54, 50, 60}
The new sum is:
49 + 50 + 45 + 23 + 35 + 66 + 34 + 74 + 54 + 50 + 60 = 540
And the new total number of values is 11.
So, the new mean is:
Mean = (49 + 50 + 45 + 23 + 35 + 66 + 34 + 74 + 54 + 50 + 60)/11
= 49.1
Adding the value 60 does not affect the position of the median, so the median is still:
Median = (49 + 50)/2
= 49.5
Adding the value 60 does not change the mode either, since 50 still appears most frequently.
So, the mode is still:
Mode = 50
The new largest value is 74 and the new smallest value is 23, so the new range is:
Range = 74 - 23
= 51
Therefore, the answers are:
The mean is 48.
The median is 49.5.
The mode is 50.
The range is 51.
After adding the value 60, the mean increases by 0.1 (from 48 to 49.1).
The median remains the same (49.5).
The mode remains the same (50).
The range remains the same (51).
Rainfall Data
Month Rainfall (cm)
January 4.3
February 3.9
March 4.0
April 5.0
May 2.9
June 3.2
July 1.9
August 2.0
September 3.7
October 3.8
November 4.2
December 4.9
What is the mode of the data in the rainfall table?
The mode is the value that appears most frequently in the data set.
Looking at the rainfall data, no value appears more than once, so there is no single mode. In this case, we say that the data has no mode or that it is multimodal.
Therefore, the mode of the rainfall data is not applicable or "N/A".
6 of 106 of 10 Items
14:04
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Question
Rainfall Data
Month Rainfall (cm)
January 4.3
February 3.9
March 4.0
April 5.0
May 2.9
June 3.2
July 1.9
August 2.0
September 3.7
October 3.8
November 4.2
December 4.9
What is the mode of the data in the rainfall table?
Responses
A no mode
B 3.7
C 3.65
D 5.0
E 3.1
The mode is the value that appears most frequently in the data set.
Looking at the rainfall data, no value appears more than once, so there is no single mode. In this case, we say that the data has no mode or that it is multimodal.
Therefore, the mode of the rainfall data is not applicable or "no mode."
The answer is A) no mode.
9 of 109 of 10 Items
20:56
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Question
{20, 30, 40, 50, 60, 70, 80}
The mean of this set of numbers is
Responses
A 20
B 80
C 55
D 50
E 60
To find the mean of a set of numbers, we add up all the numbers and divide by the total number of numbers.
For {20, 30, 40, 50, 60