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2 years ago

9

Customers of a phone company can choose between two service plans for long distance calls. The first plan has no monthly fee but

charges
$0.11
for each minute of calls. The second plan has a
$24
monthly fee and charges an additional
$0.07
for each minute of calls. For how many minutes of calls will the costs of the two plans be equal?

1 answer:

fredd [130]2 years ago

8 0

I think it is 600 minutes

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Pachacha [2.7K]

Answer:

Step-by-step explanation:

a-number of advance tickets

s-number of same day tickets

a*40+s*25=2400

a +s =75

I solve with the substitution method

s=75-a

40a+25*(75-a)=2400

40a+1875-25a=2400

15a=2400-1875

15a=525

a=525/15

a=35

so s=75-35

s=40

5 0

2 years ago

Numbers that are multiplied together to find the product are called

Zinaida [17]

Answer:

Multiples, or terms.

7 0

2 years ago

Read 2 more answers

A vacant city lot is being turned into a neighborhood garden. the neighbors want to fence in a triangular section of the lot and

Alexxx [7]

Let x represent the length of the shortest side. Then the longest side is

... 2x -7 . . . . . . 7 ft shorter than twice the shortest side

and the 3rd side is

... x +2 . . . . . . 2 ft longer than the shortest side.

The perimeter is the sum of the side lengths,

... x + (2x -7) + (x +2) = 59 . . . . . the given perimeter length

... 4x -5 = 59 . . . . collect terms

... 4x = 64 . . . . . . add 5

... x = 16 . . . . . . . . divide by 4

Then the other sides are

... 2×16 - 7 = 25 . . . . . longest side

... 16 +2 = 18 . . . . . . . . third side

The side lengths are 16 ft, 18 ft, and 25 ft.

3 0

3 years ago

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

4 0

2 years ago

What is the conditional relative frequency of selling an item marked 50% off, provided the item sold is item 2?

sukhopar [10]

Answer:

Conditional relative frequency = 0.67

Step-by-step explanation:

50% off Buy 1, get 1 free Total

Item 1 12 8 20

Item 2 20 10 30

Total 32 18 50

Condition relative frequency table:

50% off Buy 1, get 1 free Total

Item 1 $\frac{12}{20}=0.6$ $\frac{8}{20}=0.4$ $\frac{20}{20}=1$

Item 2 $\frac{20}{30}=0.67$ $\frac{10}{30}=0.33$ $\frac{30}{30}=1$

Total $\frac{32}{50}=0.64$ $\frac{18}{50}=0.36$ $\frac{50}{50}=1$

Therefore, conditional relative frequency of selling an item marked 50% off, provided the item sold is item 2 = 0.67

8 0

2 years ago