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3 years ago

What is the simple interest on 500 for 5 years at the rate of 5% per annum

Mathematics

2 answers:

shepuryov [24]3 years ago

3 0

Answer:

125

Step-by-step explanation:

Formula Simple interest:

P x r x t

jeka943 years ago

3 0

I = principal x rate x time
I = 500 x .05 x 5
I = 125

( time is always expressed in years and you convert rate to a decimal )

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An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in t

EleoNora [17]

Answer:

Rachel

Step-by-step explanation:

We need to measure how far (towards the left) are the students from the mean in “standard deviations units”.

That is to say, if t is the time the student ran the mile and s is the standard deviation of the class, we must find an x such that

mean - x*s = t

For Rachel we have

11 - x*3 = 8, so x = 1.

Rachel is 1 standard deviation far (to the left) from the mean of her class

For Kenji we have

9 - x*2 = 8.5, so x = 0.25

Kenji is 0.25 standard deviations far (to the left) from the mean of his class

For Nedda we have

7 - x*4 = 8, so x = 0.25

Nedda is also 0.25 standard deviations far (to the left) from the mean of his class.

As Rachel is the farthest from the mean of her class in term of standard deviations, Rachel is the fastest runner with respect to her class.

8 0

3 years ago

This is done in for parts (separated by letter)

lord [1]

Answer:

Step-by-step explanation:

3 0

3 years ago

Which store has the lowest price per pound for the cheese?

Anettt [7]

The answer is the letter A

6 0

3 years ago

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Aerobics classes class $153.86 for 14 sessions. What is the fee for six sessions?

Brrunno [24]

153.86/14=10.99

10.99*6=$65.94

4 0

3 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

3 years ago