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kirill [66]
3 years ago
14

Can you answer these 3 questions for me plz

Mathematics
2 answers:
finlep [7]3 years ago
4 0
A.
1hr \: 15min = 75min

c = 0.22t = 0.22 \times 75 = 16.5

cost is $16.50

B.

26.40 = 0.22 \times t

rearrange to solve for t:

t = \frac{26.40}{0.22} = 120min

C. This is the same as part B, only the rate changed

t = \frac{26.40}{0.25} = 105.6min

they ask for the greatest number of minutes, so we round down (take the floor of the value).

the answer is 105 minutes.
victus00 [196]3 years ago
4 0
The cost is a function of time in minutes, and t represents minutes.

Part A.

1 hour + 15 minutes = 60 minutes + 15 minutes = 75 minutes

c = 0.22t

For 75 minutes, t = 75

c = 0.22 * 75

c = 16.5

Answer for Part A.: The cost of a 1 hour 15 minute call is $16.50

Part B.

Now we are given the cost, c, and we need to find t, the time in minutes.

c = 0.22t

26.4 = 0.22t

Divide both sides by 0.22

120 = t

t = 120

Answer to Part B.: 120 minutes.

Part C.

The new cost is now 0.25t.
We want the cost to be at most $26.40.
That means the cost must be less than or equal to $26.40.

0.25t \le 26.4

t \le 105.6

The number of minutes must be less than or equal to 105.6 minutes.
Since the calls are charged by whole minutes, the longest call can be 105 minutes.

Answer to Part C.: 105 minutes
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cluponka [151]

Answer:

Account 1 yields greater return.

Step-by-step explanation:

Lets solve for the first formula:

A=P(1+\frac{r}{n} )^{nt}

First, change 6.5% into a decimal:

6.5% -> \frac{6.5}{100} -> 0.065

Now, plug in the values for Account 1:

A=3,000(1+\frac{0.065}{2})^{2(10)}

A=5,687.51

<u>Now solve for Account 2:</u>

Change 6% into a decimal:

6% -> \frac{6}{100} -> 0.06

Now plug in the values:

A=3,000e^{0.06(10)}

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Account 1 yields better money after 10 years.

3 0
3 years ago
What percent of 50 is 20?
Papessa [141]

Answer:

40%

Step-by-step explanation:

20/50

= 2/5

= 0/4

= 40%

3 0
2 years ago
I need this on khan academy.
Alja [10]

9514 1404 393

Answer:

  A.  (1 2/3, 4 2/3)

Step-by-step explanation:

If you graph the equations, you see the lines intersect at the solution point:

  (x, y) = (1 2/3, 4 2/3)

7 0
3 years ago
In 2002, there were 972 students enrolled at Oakview High School. Since then, the number of students has increased by 1.5% each
Rudiy27

Answer:

N(t) = 972(1.015)^{t}

Growth function.

The number of students enrolled in 2014 is 1162.

Step-by-step explanation:

The number of students in the school in t years after 2002 can be modeled by the following function:

N(t) = N(0)(1+r)^{t}

In which N(0) is the number of students in 2002 and r is the rate of change.

If 1+r>1, the function is a growth function.

If 1-r<1, the function is a decay function.

In 2002, there were 972 students enrolled at Oakview High School.

This means that N(0) = 972

Since then, the number of students has increased by 1.5% each year.

Increase, so r is positive. This means that r = 0.015

Then

N(t) = N(0)(1+r)^{t}

N(t) = 972(1+0.015)^{t}

N(t) = 972(1.015)^{t}

Growth function.

Find the number of students enrolled in 2014.

2014 is 2014-2002 = 12 years after 2002, so this is N(12).

N(t) = 972(1.015)^{t}

N(12) = 972(1.015)^{12}

N(12) = 1162

The number of students enrolled in 2014 is 1162.

7 0
3 years ago
The seventh grade band has 15 drummers and 12 trumpet players. The eighth grade band band has 10 drummers and 8 trumpet players.
defon
Yes 7 has fifty percent more
7 0
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