Use intercepts to graph 3y = -5x -30

George charged a motorized bicycle on his credit card for $2,000 on June 12. His balance before that purchase on that credit car

Gnesinka [82]

The calculation uses the accumulated daily balance method (ADB).
We assume the statement is based on calendar month (rare!).

George owes $500 from beginning to end of June, so 30 days out of 30.
Interest accrued is 500*0.013*30/30=$6.50.
He also owes $2000 from June 12 to June 30, so 19 days inclusively.
Interest accrued is $2000*.013*(19/30)=16.47
Total interest at the end of the month=$6.50+$16.47=$22.97

3 0

3 years ago

If Jay gets paid $125 for working 10 hours, how much does he get paid for one hour?

Wewaii [24]

125 dollars divided by ten = 12,5
He gets paid 12,5$ for one hour

8 0

3 years ago

Read 2 more answers

After the booster club sold 40 hotdogs at a football game, it had $90 in profit. After the next game, it had sold a total of 80

Ainat [17]

We can model the equation:y = m x + b, where y is the total profit and x is the number of hotdogs sold.The system of equations is:
90 = 40 m + b210 = 80 m + b---------------------b = 90 - 40 m210 = 80 m + 90 - 40 m210 - 90 = 40 m120 = 40 mm = 120 : 40m = 3b = 90 - 40*3b = 90 - 120b = - 30Answer: The equation is y = 3 x - 30

5 0

3 years ago

The distribution of the annual incomes of a group of middle management employees approximated a normal distribution with a mean

kondor19780726 [428]

Answer:

68% of the incomes lie between $36,400 and $38,000.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $37,200

Standard Deviation, σ = $800

We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.

Empirical rule:

Almost all the data lies within three standard deviation of mean for a normally distributed data.
About 68% of data lies within one standard deviation of mean.
About 95% of data lies within two standard deviation of mean.
About 99.7% of data lies within three standard deviation of mean.

Thus, 68% of data lies within one standard deviation.

$\mu \pm \sigma\\=37200 \pm 800\\=(36400,38000)$