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Jobisdone [24]
3 years ago
15

What is the total amount that will be repaid if the loan described is paid in 2 years?

Mathematics
1 answer:
zlopas [31]3 years ago
4 0

Given:

Principal value of loan = $1,245

Rate of interest = 5%

To find:

The total amount that will be repaid if the loan described is paid in 2 years.

Solution:

The formula for amount is

A=P\left(1+\dfrac{r}{100}\right)^t

Where, P is the principal, r is the rate of interest, t is the number of years.

Putting P=1245, r=5 and t=2, we get

A=1245\left(1+\dfrac{5}{100}\right)^2

A=1245\left(1+0.05\right)^2

A=1245\left(1.05\right)^2

On further simplification, we get

A=1245(1.1025)

A=1372.6125

A\approx 1372.61

Therefore, the total amount that will be repaid if the loan described is paid in 2 years is $1372.61.

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Substitution method<br>4x + y = 14<br>6x - 3y = 3​
lakkis [162]

Answer:

  • x = 2.5
  • y = 4

Step-by-step explanation:

<u>We know that:</u>

  • 4x + y = 14
  • 6x - 3y = 3

<u>Let's first solve for y by choosing any equation.</u>

  • 4x + y = 14
  • => y = 14 - 4x

<u>Now, let's substitute the value of y into the second equation.</u>

  • 6x - 3y = 3
  • => 6x - 3(14 - 4x) = 3
  • => 6x - 42 + 12x = 3
  • => 18x - 42 = 3
  • => 18x = 42 + 3
  • => 18x = 45
  • => x = 45/18 = 5/2 = 2.5

<u>Now, let's substitute the value of x into the equation of y.</u>

  • => 14 - 4x = y
  • => 14 - 4(2.5) = y
  • => 14 - 10 = y
  • => y = 4

Hence, the value of x and y are 2.5 and 4 respectively.

3 0
3 years ago
A gardener has 72 feet of fencing to enclose a
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5 0
3 years ago
1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The avera
ioda

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, \mu is given by;

P(-2.032 < t_3_4 < 2.032) = 0.95

P(-2.032 < \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } < 2.032) = 0.95

P(-2.032 * {\frac{s}{\sqrt{n} } < {Xbar - \mu} < 2.032 * {\frac{s}{\sqrt{n} } ) = 0.95

P(X bar - 2.032 * {\frac{s}{\sqrt{n} } < \mu < X bar + 2.032 * {\frac{s}{\sqrt{n} } ) = 0.95

95% confidence interval for \mu = [ X bar - 2.032 * {\frac{s}{\sqrt{n} } , X bar + 2.032 * {\frac{s}{\sqrt{n} } ]

                                            = [ 1080 - 2.032 * {\frac{260}{\sqrt{35} } , 1080 + 2.032 * {\frac{260}{\sqrt{35} } ]

                                             = [ 990.70 , 1169.30 ]

<em>No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.</em>

Therefore, the store manager believe is not correct.

8 0
3 years ago
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