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3 years ago

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Marty was paying off a television that he bought on layaway, which means he has to pay it off over a period of time. Marty made

3 upfront payments over three weeks. One week he paid $20, next week $30, and the next week $40( remember he paid some money already). The total cost of the tv was $690. He will now make monthly payments to pay off the rest that he owes. If he has another 12 months to pay it off, how much will he have to pay per month?

1 answer:

ioda3 years ago

4 0

Answer:

he needs to be paying $60 per month

Step-by-step explanation:

please kindly check the attached files for explanation

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Charles has $87 in his checking account. He spent $14 and $17, then deposited $33. If he wants to buy a television that cost $18

tester [92]

Answer:

He need $99 more

Step-by-step explanation:

87-14-17=56+33=89

188-89=99

5 0

3 years ago

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Simplify the rational expression. State any restrictions on the variable.

ExtremeBDS [4]

we are given

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}$

Firstly, we will factor numerators and denominators

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-6)(n^2-4)}{(n^2-3)(n^2-6)}$

we can see that

n^2 -6 is factor on both numerator and denominator

so, it will get cancelled

and n^2 -6 can not be equal to 0

so, one of restriction is

$n^2-6\neq 0$

$n\neq -+ \sqrt{6}$

we can simplify it

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-4)}{(n^2-3)}$

we know that denominator can not be zero

$n^2-3\neq 0$

$n\neq -+ \sqrt{3}$

so, option-B.......Answer

3 0

3 years ago

A candy company claims that 20% of the candies in its bags are colored green. Steve buys 30 bags of 30 candies, randomly selects

Allisa [31]

Answer:

The probability of Steve agreeing with the company’s claim is 0.50502.

Step-by-step explanation:

Let <em>X</em> denote the number of green candies.

The probability of green candies is, <em>p</em> = 0.20.

Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies.

So, <em>n</em> = 30 candies are randomly selected.

All the candies are independent of each other.

The random variable <em>X</em> follows a binomial distribution with parameter <em>n</em> = 30 and <em>p</em> = 0.20.

It is provided that if there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct.

Compute the probability of 5, 6 and 7 green candies as follows:

$P(X=5)={30\choose 5}(0.20)^{5}(1-0.20)^{30-5}=0.17228\\\\P(X=6)={30\choose 6}(0.20)^{6}(1-0.20)^{30-6}=0.17946\\\\P(X=7)={30\choose 7}(0.20)^{7}(1-0.20)^{30-7}=0.15328$

Then the probability of Steve agreeing with the company’s claim is:

P (Accepting the claim) = P (X = 5) + P (X = 6) + P (X = 7)

= 0.17228 + 0.17946 + 0.15328

= 0.50502

Thus, the probability of Steve agreeing with the company’s claim is 0.50502.

7 0

3 years ago

The options are 1. is, is not

djverab [1.8K]

Answer:

1. is not

2. increase

3. remain the same

Step-by-step explanation:

8 0

3 years ago

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Use matrices and elementary row to solve the following system:

LiRa [457]

I assume the first equation is supposed to be

$5x-3y+2z=13$

and not

$5x-3x+2x=4x=13$

As an augmented matrix, this system is given by

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]$

Multiply through row 3 by 1/2:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]$

Add -1(row 2) to row 3:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]$

Multiply through row 3 by 1/5:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]$

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

$\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Add -3(row 2) to row 1:

$\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Multiply through row 1 by -1:

$\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Add -2(row 1) to row 2:

$\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]$

Multipy through row 2 by -1:

$\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]$

The solution to the system is then

$\boxed{x=1,y=-2,z=1}$

5 0

3 years ago