If you borrow $450.00 at a 5% interest, how much will you pay in one month?

kati45 [8]

Look at the fractions!

9/10 is less than 1.

10/9 is more than 1. So it must be more than 9/10.

5 0

3 years ago

On a typical long distance call you talk for 30 minutes. On a typical local call you talk for 10 minutes. Your phone company off

klio [65]

The number of long distance calls to be made can be obtained by linear programming

To minimize the phone bill, the number of long distance call to be made is none

Reason:

Maximum number of number of long distance calls = 15

Maximum number of local calls = 30

Number of minutes of calls to make each month = 240 minutes

Duration of each long distance call = 30 minutes

Duration of each local call = 10 minutes

Cost of long distance call = $0.08

Cost of local call = $0.03

Solution:

Let X, represent the number of long distance calls, and let Y represent the number local calls

The objective function is given as follows;

P = 0.08·(30)·X + 0.03·(10)·Y

P = 2.4·X + 0.3·Y

The constraints are;

X ≥ 0

Y ≥ 0

X ≤ 15

Y ≤ 30

30·X + 10·Y ≥ 240

Solving we have;

Y ≥ 24 - 3·X

The above inequality can be plotted with MS Excel

From the objective function, P = 2.4·X + 0.3·Y, the coefficient of the long

distance calls, X is larger than the coefficient of the local calls Y, therefore,

making the minimum possible number of long distance calls of 0, and 24

local calls will give a cost;

P = 2.4 × 0 + 0.3 × 24 = 7.2

Therefore, to minimize the phone bill, the number of long distance calls to

be made is zero long distance calls

Learn more about linear optimization here:

brainly.com/question/21491539

5 0

3 years ago

Can someone please help me?

kvasek [131]

now, let's recall the rational root test, check your textbook on it.

so p = 18 and q = 1

so all possible roots will come from the factors of ±p/q

now, to make it a bit short, the factors are loosely, ±3, ±2, ±9, ±1, ±6.

recall that, a root will give us a remainder of 0.

let us use +3.

$\bf x^4-7x^3+13x^2+3x-18 \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{r|rrrrr} 3&1&-7&13&3&18\\ &&3&-12&3&18\\ \cline{1-6} &1&-4&1&6&0 \end{array}\qquad \implies (x-3)(x^3-4x^2+x+6)$

well, that one worked... now, using the rational root test, our p = 6, q = 1.

so the factors from ±p/q are ±3, ±2, ±1

let's use 3 again

$\bf \begin{array}{r|rrrrr} 3&1&4&1&6\\ &&3&-3&-6\\ \cline{1-5} &1&-1&-2&0 \end{array}\qquad \implies (x-3)(x-3)(x^2-x-2)$

and of course, we can factor x²-x-2 to (x-2)(x+1).

(x-3)(x-3)(x-2)(x+1).

7 0

3 years ago

Ben made 1/4 bath of brownies in five minutes. How many minutes did it take to make 1 batch of brownies?

Rom4ik [11]

Answer:

It took 20 minutes to make 1 batch

Step-by-step explanation:

If 1/4 batch of brownies takes 5 minutes then you just multiply 5 times 4 and that equals 20.

8 0

3 years ago

Read 2 more answers

On which interval is the average rate of change for bicycle production the greatest? A) 1950 to 1960 B) 1960 to 1970 C) 1970 to

Alborosie

Answer:

D) 1980 to 2000

Step-by-step explanation:

Finding the average rate of change in each interval to determine the greatest one.

Production per interval

1950-1960 = $(21-11) \ million$ = 10 million

1960-1970 = $(41-21)\ million$ = 20 million

1970-1980 = $(71-41)\ million$ = 30 million

1980-2000 = $(161-71)\ million= 90$ million

Rate of change (1950-1960)= $\frac{10}{11} \times 100= 90.90\%$

Rate of change (1960-1970) = $\frac{20}{21} \times 100= 95.23\%$

Rate of change (1970-1980)= $\frac{30}{41} \times 100= 73.17\%$

Rate of change (1980-2000)= $\frac{90}{71} \times 100= 126.76\%$

∴ rate of change between 1980 to 2000 is 126.78%

7 0

3 years ago