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3 years ago

24 students in a class took an algebra test. if 18 students passed, what percent passed.

Mathematics

2 answers:

tia_tia [17]3 years ago

8 0

Answer:

75%

Step-by-step explanation:

18/24=0.75

Tamiku [17]3 years ago

7 0

Answer:

75%

Step-by-step explanation:

24 divided by 4 is 6, so 25% of 24 is 6. 18 divided by 6 is 3, so 18 out of 24 is 75%.

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A tank had 6 liters of water. The tank had a small hole in its base through which water leaked out at a constant rate. The tank

Mariana [72]

Based on the given problem above, if we analyze it, we are going to look for the graph which with coordinates (0,6) which represents starting out with 6 gallons, and ending at (4,0). So, the graph that best represents this situation would be the first graph on the top left. Hope this answers your question. Have a great day!

8 0

2 years ago

Read 2 more answers

Solve the equation for y. Then find the value of Y for each value of X.

Hunter-Best [27]

Answer:

y=9-7x

If x= -1, y=16

If x= 0, y=9

If x=2, y= -5

Step-by-step explanation:

Because you are solving the equation for y, you want to have y by itself on one side. The only way to do that in this problem is by subtracting the 7x on both sides.

y+7x=9

-7x -7x ---> y=9-7x

The next part of the question says to find the value of Y for each value of X. Using the given X values, we plug them into the equation we just got in order to find Y.

y=9-7(-1) ----> y=16

y=9-7(0) ----> y=9

y=9-7(2) ----> y=-5

4 0

3 years ago

Read 2 more answers

What is the gcf of 18 and 9?

Zanzabum

The gcf of 9 and 18 is 18

8 0

3 years ago

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You take a True/False quiz that consists of 5 questions. What is the probability that you will get at least a B (at least 4 ques

Anna [14]

Answer:

20%

Step-by-step explanation:

6 0

3 years ago

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On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

$A = P \left(1 + \dfrac{r}{n}\right)^{nt}$

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = <u> -1 000.00 </u>

P = 22 944.00

1 Jan 2016 A = £23 517.60

5 0

3 years ago