Kiara has a ribbon that measures 1/4 of an inch and Jillian has a ribbon that measures 2/3 of an inch. Jillian says their ribbons are equal in length because her numerator is one more than kiara's and her denominator is one less, therefore balancing the two fractions. Is she correct? Why or why not?
Answers
Answer:
Jillian is Incorrect as fraction can never be balanced based on numerator or denominator.
Step-by-step explanation:
Given:
Measure of ribbon Kiara has =
Measure of ribbon Jillian has =
Now given:
Jillian says their ribbons are equal in length because her numerator is one more than kiara's and her denominator is one less, therefore balancing the two fractions.
we need to find whether statement is correct or not.
First we will find the decimal value of the given fraction's
can rewritten as
Measure of ribbon Kiara has = 0.25 inch
can be Rewritten as
Measure of ribbon Jillian has = 0.67 inch
Now we can see that;
0.67 inch is greater than 0.25 inch.
Hence we can say that ribbons are not in equal length.
Also the fraction can never be balanced by decreasing or increasing the numerator or denominator.
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Answers
Answers
Step-by-step explanation:
Length, width and height are three dimensional objects.
I hope this will help you.
2)in mathmatics and sciecnce but not in english
3)in mathematics only
more than one subject only
Answers
Given that we have deduced the number of students who passed in two subjects, we can now solve for the number of students who passed only one subject.
English = 15 - (4 + 2 + 0) = 9
Mathematics = 12 - (4 + 3 + 2) = 3
Science = 8 - (4 + 3 + 0) = 1
1. In English but not in Science,
9 + 2 = 11
2. In Mathematics and Science but not in English
3 + 3 + 1 = 7
3. In Mathematics only
= 3
3. More than one subject only
3 + 4 + 2 + 9 = 18
It will really be helpful if you draw yourself a Venn Diagram for this item.
Answer:
there are 4 students who passed in all subjects, we can say that 2 students passed in English and Mathematics, 3 students passed in Mathematics and Science only, and no one passed in English and Science only.
Step-by-step explanation:
Given that we have deduced the number of students who passed in only two subjects, we can now solve for the number of students who passed only one subject.
English = 15 - (4 + 2 + 0) = 9
Mathematics = 12 - (4 + 3 + 2) = 3
Science = 8 - (4 + 3 + 0) = 1
1. In English but not in Science,
9 + 2 = 11
2. In Mathematics and Science but not in English
3 + 3 + 1 = 7
3. In Mathematics only
= 3
3. More than one subject only
3 + 4 + 2 + 9 = 18
Answers
I'm sure you can think of at least 15 more:
-- Pull 5 multiples of 3 from the bag.
-- Pull 3 multiples of 5 from the bag.
-- Pull 7 odd numbers from the bag.
-- Pull 7 even numbers from the bag.
-- Pull two slips from the bag whose sum is more than 21 .
-- Pull three slips from the bag whose sum is more than 33 .
-- Pull two slips from the bag whose sum is less than 3 .
-- Pull three slips from the bag whose sum is less than 6 .
-- Pull two slips from the bag whose product is more than 132 .
-- Pull two slips from the bag whose product is less than 2 .
etc.
Answers
f(x) =
2 - 9x ≥ 0
- 9x ≥ - 2
x ≤
Interval Notation: (-∞, ]
b) 1/7
c) 24/25
d) 3/4