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Mashcka [7]
3 years ago
8

9. Sarah saves $200 in her bank account for 3 years at a 5% interest rate. How much interest does she earn?​

Mathematics
2 answers:
Verizon [17]3 years ago
6 0

Answer:

$30

Step-by-step explanation:

Assuming this is an annual compound with a simple interest rate. the formula we can use is A=P(rt).

P=200

r=0.05

t=3

0.05*3=0.15

200*0.15=30

She earned $30

love history [14]3 years ago
6 0

Answer:

The amount of interest she earns is: $30

Step-by-step explanation:

Given

  • Principle P = $200
  • Time period t = 3 years
  • Interest rate r = 5% or 5/100 = 0.05

To determine

Interst I = ?

Using the equation to determine the interest amount

I = Prt

substitute P = 200, r = 0.05 and r = 3

I\:=\:200\:\times \:0.05\:\times \:3

I = 30 $

Therefore, the amount of interest she earns is: $30

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A researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic
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Answer:

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Step-by-step explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean iron concentration = 0.802 cc/cubic meter

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             n = number of water samples = 27

             \mu = population mean

<em>Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 90% confidence interval for the population mean, \mu is ;

P(-1.706 < t_2_6 < 1.706) = 0.90  {As the critical value of t at 26 degree of

                                                 freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 1.706) = 0.90

P( -1.706 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 1.706 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X -1.706 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +1.706 \times {\frac{s}{\sqrt{n} } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X -1.706 \times {\frac{s}{\sqrt{n} } , \bar X +1.706 \times {\frac{s}{\sqrt{n} } ]

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Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

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