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3 years ago

6

Helia invested 4000 AED for 3 years with a compound interest rate of 1.5%, Calculate the worth of her investment at the end of t

his time period. ( Show each step of your calculation)

1 answer:

sineoko [7]3 years ago

6 0

Answer:

4180 AED

Step-by-step explanation:

Given data

Principal= 4000

Time= 3 years

Rate= 1.5%

The compound interest expression is given as

A= P(1+r)^t

substitute

A=4000(1+0.015)^3

A= 4000(1.015)^3

A= 4000*1.045

A= 4180

Hence the worth will be 4180 AED

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Find the domain and range of the relation given to the right.

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Answer:

Step-by-step explanation:

D:(0,10)

R:(14, infinitly)

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The manufacturer's price for a cooler is $28.50 A camping store that buys the coolers sells them for $40.76 What is the approxim

Nutka1998 [239]

The percent of the increase is

$I=\frac{N-O}{O}\times100$

I is the increasing value

N is the new value

O is the original value

The price of the cooler is $28.50, then

O = 28.50

The selling price of the cooler is $40.76, then

N = 40.76

We will substitute them in the rule above to find I

$I=\frac{40.76-28.50}{28.50}\times100$

Let us calculate it

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The answer is D

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1 year ago

How much greater is 0.0543 than 0.002?

aleksley [76]

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0.0523

Step-by-step explanation:

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2 years ago

6) A climber is on a hike. After 2 hours she is at an altitude of 400 feet. After 6 hours, she is at an

Bas_tet [7]

Answer:

<em>Her rate of change is 75 ft/h</em>

Step-by-step explanation:

<u>Rate of Change</u>

It measures how one quantity varies with respect to another. The second variable usually is the time.

The climber is hiking. We are given two points of her progress: At 2 hours, she is at 400 ft. This corresponds to the point (2,400).

At 6 hours, she is at 700 ft. This is the point (6,700)

The rate of change is the slope between these two points or the quotient of the variations of each variable:

$\displaystyle m=\frac{700\ ft-400\ ft}{6\ h-2\ h}$

$\displaystyle m=\frac{300\ ft}{4\ h}$

$m=75\ ft/h$

Her rate of change is 75 ft/h

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2 years ago

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$

Next, calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}$

$\bar x = \frac{1845.73}{146}$

$\bar x = 12.64$

Next, the sample variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}$

$\sigma^2 = \frac{2393.6875}{145}$

$\sigma^2 = 16.51$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{16.51}$

$\sigma = 4.06$

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2 years ago