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3 years ago

Can someome help me find the mean of this numbers? 15,16,17,18,19,20,21,22 (RANDOM ANXWERS WILL BE REPORTED)

Mathematics

2 answers:

bezimeni [28]3 years ago

4 0

The mean of these numbers is 18.5

Andrej [43]3 years ago

3 0

Hello! You find the mean by adding up all of the numbers and divided that total with the total amount of numbers!

First, solve this:

15+16+17+18+19+20+21+22= ?

Next, use the answer from the equation above the "?" is the answer from the problem above:

?/8=mean

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Number 2 I need help on how to solve it

Hunter-Best [27]

Let's simplify the following expression:

$\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}$

We get,

$\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}$ $\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}$

4 0

1 year ago

Find the value of x please help

____ [38]

using angle sum property

we get

110+35+(x+12)=180

(x+12)=180-145

(x+12)= 35

x= 35-12

x= 23

3 0

2 years ago

Read 2 more answers

Help me find the answer!! 2-step equation

Vladimir79 [104]

$\looparrowright \sf3\leqslant \dfrac{y}{ - 5} + 7$

$\looparrowright \sf3 - 7\leqslant \dfrac{y}{ - 5}$

$\looparrowright \sf - 4\leqslant \dfrac{y}{ - 5}$

$\looparrowright \sf - 4 \times - 5\leqslant {y}$

$\looparrowright \sf 20\leqslant {y}$

8 0

2 years ago

Equation

jekas [21]

You can use math away it would help!

6 0

2 years ago

Imagine that you would like to purchase a $275,000 home. Using 20% as

vfiekz [6]

Answer:

The mortgage chosen is option A;

15-year mortgage term with a 3% interest rate because it has the lowest total amount paid over the loan term of $270,470

Step-by-step explanation:

The details of the home purchase are;

The price of the home = $275,000

The mode of purchase of the home = Mortgage

The percentage of the loan amount payed as down payment = 20%

The amount used as down payment for the loan = $55,000

The principal of the mortgage borrowed, P = The price of the house - The down payment

∴ P = $275,000 - 20/100 × $275,000 = $275,000 - $55,000 = $220,000

The principal of the mortgage, P = $220,000

The formula for the total amount paid which is the cost of the loan is given as follows;

$Outstanding \ Loan \ Balance = \dfrac{P \cdot \left[\left(1+\dfrac{r}{12} \right)^n - \left(1+\dfrac{r}{12} \right)^m \right] }{1 - \left(1+\dfrac{r}{12} \right)^n }$

The formula for monthly payment on a mortgage, 'M', is given as follows;

$M = \dfrac{P \cdot \left(\dfrac{r}{12} \right) \cdot \left(1+\dfrac{r}{12} \right)^n }{\left(1+\dfrac{r}{12} \right)^n - 1}$

A. When the mortgage term, t = 15-years,

The interest rate, r = 3%

The number of months over which the loan is payed, n = 12·t

∴ n = 12 months/year × 15 years = 180 months

n = 180 months

The monthly payment, 'M', is given as follows;

M =

The total amount paid over the loan term = Cost of the mortgage

Therefore, we have;

220,000*0.05/12*((1 + 0.05/12)^360/( (1 + 0.05/12)^(360) - 1)

$M = \dfrac{220,000 \cdot \left(\dfrac{0.03}{12} \right) \cdot \left(1+\dfrac{0.03}{12} \right)^{180} }{\left(1+\dfrac{0.03}{12} \right)^{180} - 1} \approx 1,519.28$

The minimum monthly payment for the loan, M ≈ $1,519.28

The total amount paid over loan term, A = n × M

∴ A ≈ 180 × $1,519.28 = $273,470

The total amount paid over loan term, A ≈ $270,470

B. When t = 20 year and r = 6%, we have;

n = 12 × 20 = 240

$\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.06}{12} \right) \cdot \left(1+\dfrac{0.06}{12} \right)^{240} }{\left(1+\dfrac{0.06}{12} \right)^{240} - 1} \approx 1,576.15$

The total amount paid over loan term, A = 240 × $1,576.15 ≈ $378.276

The monthly payment, M = $1,576.15

C. When t = 30 year and r = 5%, we have;

n = 12 × 30 = 360

$\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.05}{12} \right) \cdot \left(1+\dfrac{0.05}{12} \right)^{360} }{\left(1+\dfrac{0.05}{12} \right)^{360} - 1} \approx 1,181.01$

The total amount paid over loan term, A = 360 × $1,181.01 ≈ $425,163

The monthly payment, M ≈ $1,181.01

The mortgage to be chosen is the mortgage with the least total amount paid over the loan term so as to reduce the liability

Therefore;

The mortgage chosen is option A which is a 15-year mortgage term with a 3% interest rate;

The total amount paid over the loan term = $270,470

8 0

3 years ago