All categories

2 years ago

There are 12 boys and 16 girls in a classroom. which represents the simplified ratio of girls to students in the classroom?

Mathematics

2 answers:

Nikolay [14]2 years ago

6 0

Answer:

Ratio of girls to STUDENTS is 4 to 7

Step-by-step explanation:

The information given is that there are 12 boys and 16 girls in the classroom. The question is not about the ratio of girls to boys but about the ratio of girls to STUDENTS. You need to add the amount of girls plus the amount of boys to get the total amount of students.

Girls to Students

16 to 28 (12 +16)

Simplify 16:28 by dividing each number by 4

Answer is 4:7

Papessa [141]2 years ago

3 0

The unsimplified version is 12:16
you have to find the highest common factor in them
in this case it would be 4
divide both numbers by 4
3:4

hope this helps you

You might be interested in

Which answers show a correct plan to solve this problem?

uranmaximum [27]

A. & B. Is the correct answer.

3 0

2 years ago

Can anyone explain me?

Alika [10]

Answer:

Step-by-step explanation:

a) (a + b)² = (a + b) * (a +b)

(a + b)³ = (a + b) * (a +b) * (a +b)

a²- b² = (a +b) (a - b)

Here (a + b) is common in all the three expressions

HCF = (a + b)

b) (x - 1) = (x - 1)

x² - 1 = (x - 1) * (x + 1)

(x³ - 1) = (x - 1) (x² + x + 1)

HCF = (x -1)

7 0

2 years ago

PLEASE HELP!!! EASY POINTS

SOVA2 [1]

Answer: 50 x

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

A car is moving with a velocity (3.0 m/s) x̂ + (1.0 m/s) ŷ and 3.0 seconds later its velocity is (6.0 m/s) x̂ - (3.0 m/s) ŷ. Wha

Mars2501 [29]

Here we know that the initial velocity of the car is given by:

$V_i=(3.0\hat x+1.0\hat y)m/s$

And the final velocity of the car is given by:

$V_f=(6.0\hat x-3.0\hat y)m/s$

It took 3 seconds to attain the final velocity, so we have $t=3 s$

Therefore, the acceleration can be obtained by:

$V_f-V_i=a\times t$

$a=\frac{V_f-V_i}{t}$

Plugging the values of the initial, final velocity and the time, we get:

$a= \frac{(6\hat x-3\hat y)-(3 \hat x+1 \hat y)}{3} =\frac{6\hat x-3 \hat y -3 \hat x-1 \hat y}{3} =\frac{3 \hat x-4 \hat y}{3}$

So the acceleration of the car is given by:

$a=\frac{3 \hat x}{3} -\frac{4 \hat y}{3} =1 \hat x- \frac{4}{3} \hat y$

Now we need to find the direction of the average acceleration of the car:

$\theta=tan^{-1} \frac{y}{x}$

Here, x and y are the coefficients of the 'x' and 'y' components of the vector:

$\theta=tan^{-1}(\frac{-\frac{4}{3} }{1} )=tan^{-1}(\frac{-4}{3}) =-53.13^\circ$

Therefore, the direction of the average acceleration of the car is $-53.13^\circ$ .

8 0

3 years ago

A baseball is thrown with a vertical velocity of 50 ft/s from an initial height of 6 ft. The height h in feet of the baseball ca

Bad White [126]

Vertex is highest point

ok so the t value of the vertex of a function in the form
f(x)=at^2+bt+c is
-b/2a

find t value
h(x)=-16t^2+50t+6
a=-16
b=50
-50/(2*-16)=1.5625
round
1.6 seconds
max height
input 1.6 for t and evaluate (for more precice, input 1.5625 for t)

max height is 45 feet

seems to be
b=initial velocity
c=intial height

a=-16
eqation is
h(t)=-16t^2+70t+5
vertex is -70/(2*-16)=2.1875
input for t to get max height
82 feet

answers are

1.6 seconds
45 feet
82 feet

4 0

2 years ago