How are equivalent fractions used when adding unlike fractions

Parker Middle School's chior needs to raise $480 for an upcoming trip they will take. At their first fundraising project, they r

harkovskaia [24]

The money does the choir still need to raise is $180.

Step-by-step explanation:

The money needed to be raised by the choir for an upcoming trip = $480.

It is given that, the money needed was raised in two fundraising projects.

During the first fundraising project :

The choir raised 25% of the money needed for the trip.

⇒ 25% of 480

⇒ (25/100) × 480

⇒ (1/4) × 480

⇒ 120

Therefore, the first fundraising project raised $120.

During the second fundraising project :

The choir raised 3/8 of the money needed for the trip.

⇒ 3/8 of 480

⇒ (3/8) × 480

⇒ 3 × 60

⇒ 180

Therefore, the second fundraising project raised $180.

To find the money does the choir still need to raise :

You need to find the remaining amount from the 480 after raising money from the two fund raising projects.

For that, find out how much you raised from those fund raising projects and subtract it from the 480.

Money needed = Total money required for the trip - Raised money

⇒ 480 - (120+180)

⇒ 480 - 300

⇒ 180 dollars.

The money does the choir still need to raise is $180.

6 0

2 years ago

Find the sixth term of the sequence 1/2, -3/8, 9/32

sukhopar [10]

Answer:

The sixth term is -243/2048 ⇒ answer B

Step-by-step explanation:

* Lets explain the geometric sequence

- There is a constant ratio between each two consecutive numbers

- Ex:

# 5 , 10 , 20 , 40 , 80 , ………………………. (×2)

# 5000 , 1000 , 200 , 40 , …………………………(÷5)

* General term (nth term) of a Geometric sequence:

# U1 = a , U2 = ar , U3 = ar² , U4 = ar³ , U5 = ar^4

# Un = ar^(n-1), where a is the first term , r is the constant ratio

between each two consecutive terms and n is the position of the

number in the sequence

- Ex: U5 = ar^4 , U7 = ar^6 , U10 = ar^9 , U12 = ar^11

- Lets solve the problem

∵ The sequence is 1/2 , -3/8 , 9/32

- Lets find the constant ratio r

∵ The first term is a = 1/2

∵ The second term is U2 = ar

∵ The second term U2 = -3/8

∴ ar = -3/8

∴ 1/2 r = -3/8 ⇒ multiply both sides by 2

∴ r = -3/4

- Lets find the sixth term

∵ a = 1/2 and r = -3/4

∵ n = 6

∴ U6 = ar^5

∴ U6 = 1/2 (-3/4)^5 = 1/2 × -243/1024 = -243/2048

* The sixth term is -243/2048

8 0

2 years ago

Read 2 more answers

What is bigger 7 1/8 or 7.025?

Stolb23 [73]

7.025 is greater than 7 1/8

4 0

2 years ago

Read 2 more answers

A professor has recorded exam grades for 10 students in his class, but one of the grades is no longer readable. If the mean sco

Nostrana [21]

Answer:

unreadable score = 35

Step-by-step explanation:

We are trying to find the score of one exam that is no longer readable, let's give that score the name "x". we can also give the addition of the rest of 9 readable s scores the letter "R".

There are two things we know, and for which we are going to create equations containing the unknowns "x", and "R":

1) The mean score of ALL exams (including the unreadable one) is 80

so the equation to represent this statement is:

mean of ALL exams = 80

By writing the mean of ALL scores (as the total of all scores added including "x") we can re-write the equation as:

$\frac{R+x}{10} =80$

since the mean is the addition of all values divided the total number of exams.

in a similar way we can write what the mean of just the readable exams is:

$\frac{R}{9} = 85\\$ (notice that this time we don't include the grade x in the addition, and we divide by 9 instead of 10 because only 9 exams are being considered for this mean.