All categories

2 years ago

A $120,000, 5%, 200-day note dated June 6 is discounted on October 8. The discount period is how many days?

1 answer:

5 0

June: 30-6=24
July: 31
August:31
September:30
October: 8
Total=24+31+31+30+8=124 days

Discount period:
The period of time between the date a bill of exchange is drawn and the date on which payment is due.

So discount period = 200-124=76 days

You might be interested in

Todd’s average score for six tests was 92. If the sum of the scores of two of her tests was 188, then what was her average score

natita [175]

Answer:

Step-by-step explanation:

Todd’s average score for six tests = 92.

The sum of two of her test = 188

First, we need to find the total score for the six test. This given below:

Average = sum of all test / number of test

sum of all the test = average x number of test

average score for six tests = 92.

Number of test = 6

Sum of all the Tests = 92 x 6 = 552

Sum of four test = sum of all the test — sum of two test

Sum of four test = 552 — 188 = 364

Now we can solve for the average of the other four test as shown below:

Average of four test = 364/4= 91

3 0

2 years ago

Read 2 more answers

What is 54% of 450 on carnegie

Nadusha1986 [10]

54:450*100 =

(54*100):450 =

5400:450 = 12

7 0

2 years ago

Solve 0 = (x – 4)2 – 1 by graphing the related function.

fomenos

Answer:

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached.

Step-by-step explanation:

The solution of the graph could be obtained by finding the x-intercept.

$y=\left(x-\:4\right)^2-1$

Finding the x-intercept by substituting the value y = 0

$y=\left(x-\:4\right)^2-1$

$\:0\:=\:\left(x\:-\:4\right)^2\:-\:1$ ∵ y = 0

$\left(x-4\right)^2-1=0$

$\left(x-4\right)^2-1+1=0+1$

$\left(x-4\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$\mathrm{Solve\:}\:x-4=\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=1$

$x=5$

$\mathrm{Solve\:}\:x-4=-\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=-1$

$x=3$

So, when y = 0, then x values are 3, and 5.

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.