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4 years ago

three-quarters of students are wearing the school's colors. if there are 300 students how many are wearing the colors?

Mathematics

1 answer:

soldi70 [24.7K]4 years ago

4 0

Answer:

225 students are wearing school colors.

Step-by-step explanation:

1. we know that 75% or .75 of the students are wearing school colors.

2. to find what this number is we need to know what 75% of 300 is.

("of" is a keyword meaning multiply)

3. We can do this by multiplying:

300(.75) = 225

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Fiona has $18 to spend. She spent $4.25, including tax, to buy a notebook. She needs to save $9.75, but she wants to buy a snack

Sergeeva-Olga [200]

Answer:

4.25+9.75+0.5x≤18

Step-by-step explanation:

4.25+9.75+0.5x≤18

14+0.5x≤18

0.5x≤4

x≤8

She can buy at most 8 packs of crackers.

Hope this helps :)

8 0

3 years ago

A swimming pool is represented on a coordinate grid with the following vertices: Vertex A is at (−2,−3) . Vertex B is at (−2,

Semmy [17]

Answer:

26 yards.

Step-by-step explanation:

We can find the distance walked by Jared by finding the distances between each of the given vertex of swimming pool. We will use distance formula to find the distances between each vertex.

$\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

$\text{Distance between vertex A and vertex B}=\sqrt{(-2--2)^{2}+(-3-5)^{2}}$

$\text{Distance between vertex A and vertex B}=\sqrt{(-2+2)^{2}+(-8)^{2}}$

$\text{Distance between vertex A and vertex B}=\sqrt{(0)^{2}+(-8)^{2}}$

$\text{Distance between vertex A and vertex B}=\sqrt{64}=8$

Now let us find distance between vertex B and vertex C.

$\text{Distance between vertex B and vertex C}=\sqrt{(-2-3)^{2}+(5-5)^{2}}$

$\text{Distance between vertex B and vertex C}=\sqrt{(-5)^{2}+(0)^{2}}$

$\text{Distance between vertex B and vertex C}=\sqrt{25}=5$

Now let us find distance between vertex C and vertex D.

$\text{Distance between vertex C and vertex D}=\sqrt{(3-3)^{2}+(5--3)^{2}}$

$\text{Distance between vertex C and vertex D}=\sqrt{(0)^{2}+(5+3)^{2}}$

$\text{Distance between vertex C and vertex D}=\sqrt{64}=8$

Now let us find distance between vertex D and vertex A.

$\text{Distance between vertex D and vertex A}=\sqrt{(3--2)^{2}+(-3--3)^{2}}$

$\text{Distance between vertex D and vertex A}=\sqrt{(3+2)^{2}+(-3+3)^{2}}$

$\text{Distance between vertex D and vertex A}=\sqrt{(5)^{2}+(0)^{2}}$

$\text{Distance between vertex D and vertex A}=\sqrt{25}=5$

Now let us add all these distances to find the total distance walked by Jared.

$\text{Total distance walked by Jared}=8+5+8+5$

$\text{Total distance walked by Jared}=26$

Since we have been given that one unit on the grid equals 1 yard, therefore, Jared walked a distance equal to 26 yards.

3 0

3 years ago

What is the absolute value of -4? PLEASE I WILL GIVE BRAINLYEST

Drupady [299]

The Absolute Value of -4 is 4.

5 0

3 years ago

Read 2 more answers

How many sixteenths are there in 5/8

nadya68 [22]

Answer:

10!

Step-by-step explanation:

Exuse me if im wrong :')

6 0

3 years ago

Darren wanted to see how contagious yawning can be. To better understand this, he conducted a social experiment in which he yawn

denis23 [38]

Answer:

10.5 minutes

Step-by-step explanation:

Thinking about the problem

The modeling function is of the form P(t)=A⋅Bf(t), where B=4B=4B, equals, 4 and f(t)=\dfrac{t}{10.5}f(t)=

10.5

f, left parenthesis, t, right parenthesis, equals, start fraction, t, divided by, 10, point, 5, end fraction.

Note that each time f(t)f(t)f, left parenthesis, t, right parenthesis increases by 111, the quantity is multiplied by B=4B=4B, equals, 4.

Therefore, we need to find the ttt-interval over which f(t)f(t)f, left parenthesis, t, right parenthesis increases by 111.

Hint #22 / 3

Finding the appropriate unit interval

fff is a linear function whose slope is \dfrac{1}{10.5}

10.5

start fraction, 1, divided by, 10, point, 5, end fraction.

This means that whenever ttt increases by \Delta tΔtdelta, t, f(t)f(t)f, left parenthesis, t, right parenthesis increases by \dfrac{\Delta t}{10.5}

10.5

Δt

start fraction, delta, t, divided by, 10, point, 5, end fraction.

Therefore, for f(t)f(t)f, left parenthesis, t, right parenthesis to increase by 111, we need \Delta t=10.5Δt=10.5delta, t, equals, 10, point, 5. In other words, the ttt-interval we are looking for is 10.510.510, point, 5 minutes.

Hint #33 / 3

Summary

The number of people who yawned quadruples every 10.510.510, point, 5 minutes.

7 0

4 years ago