What is the product of 3/8 × 8/9

A high school mathematics teacher can grade a set of exams in 7 hours working alone. If their student teacher helps them, it wi

Stels [109]

It takes 9 hrs 19.8 minutes for the student teacher to grade the exams if they worked alone

<u>Solution:</u>

Let the efficiency of the school teacher be “x” and the efficiency of the student teacher be “y”

Here the word efficiency means the rate of doing work

Let the total work to be completed be W

According to the question,

The time taken by the school teacher to complete the work is 7 hrs.

So we can write it in equation form

$W = x \times 7$ ---- eqn 1

where W is the total work and x is the efficiency

Similarly for student teacher and school teacher doing the work simultaneously, we can write the equation as

$W =(x +y ) \times 4$

Equating eqn 1 and 2

$x \times 7=(x+y) \times 4$

7x = 4x + 4y

3x = 4y

So in ratio x : y = 4 : 3

Now putting value of x = 4 in equation (1) we get total work to be 28

Now, <em>total time taken by the student teacher = total work ÷ efficiency of student teacher</em>

Hence, total time taken by the student teacher = 28 ÷ 3 = 9.33 hrs

= 9 hrs 19.8 minutes

Hence it takes 9 hrs 19.8 minutes for the student teacher to grade the exams if they worked alone

4 0

2 years ago

Read the passage from "Improve the Zoo!"

tensa zangetsu [6.8K]

I think it would be to refute the counterclaim.
Hope this helps!

7 0

2 years ago

Read 2 more answers

A ballplayer catches a ball 3.4s after throwing it vertically upward. with what speed did he throw it, and what height did it re

Daniel [21]

From the equation of motion, we know,

$s=ut+\frac{1}{2}at^{2}$

Where s= displacement

u= initial velocity

a= gravitational force

t= time

Displacement is 0 since the ball comes back to the same point from where it was thrown.

A = $-9.8m/s^{2}$ since the ball is thrown upwards.

Plug the known values into the equation.

=> $0=u\left ( 3.4 \right )+\frac{1}{2}\left ( -9.81 \right )(3.4^{2})$

Solving for u gives :

u= 16.67 m/ sec ....... equation (1)

At maximum height, final velocity i.e v is 0

Time take to reach the top = $\frac{3.4}{2} = 1.7 sec$

$v^{2}=u^{2}+2as$

=> $0=(16.67)^{2}+2(-9.81)(s)$

Solving for s we get

s= 14.16 m

4 0

2 years ago

Find the length of the missing side. Leave your answer in simplest radical form.

lozanna [386]

Answer:

The other side is 7.48 yards.

Step-by-step explanation:

Given that,

Hypotenuse = 15 yards

One leg = 13 yards

We need to find the length of the other leg. We can use the Pythagoras theorem to find it such that,

$H^2=b^2+h^2$

Where

h is other leg

Put all the values,

$h=\sqrt{H^2-b^2} \\\\h=\sqrt{(15)^2-(13)^2} \\\\h=7.48\ yd$

So, the other side is 7.48 yards.

6 0

2 years ago

On Tuesday, Michael walked to a grocery store in the morning and decided to buy a banana for $9.12 Michael handed the salesperso

mixas84 [53]

Michael received $0.01 or a penny

8 0

3 years ago