3+(-7)<br> What is the answer

What is the mean of this data set? {14, 7, 2, 8, 3} Enter your answer as a decimal in the box.

Assoli18 [71]

The mean of the data set is 6.8.

Mean is the average value of the given set of observations. In statistics, we also come across different types of mean such as Arithmetic, Geometric and Harmonic mean.

Given

The data set is 14, 7, 2, 8, 3.

The mean of the data set is given by;

$\rm Mean = \dfrac{Sum \ of \ all \ data}{Number \ of \ data}\\\\Mean =\dfrac{14+7+2+8+3}{5}\\\\Mean = \dfrac{34}{5}\\\\Mean=6.8$

Hence, the mean of the data set is 6.8.

To know more about mean value click the link given below.

brainly.com/question/4209766

3 0

2 years ago

Jerry multiplied 3.04 x 7.091 and got the product 2.155664. Is his answer reasonable? Why or why not?

Readme [11.4K]

First we multiply the numbers ignoring the decimal and then place the decimal in the position

3.04 *7.091 = 2155664

Now the first number has two digits after the decimal and the second number has three digits after the decimal.

Therefore in the final result we have to move the decimal (2+3) =5 place from the left,

So the answer would be 21.55664

So, the correct answer is 21.55664

Jerry calculated the answer correctly , but made an error in the placement of the decimal point. The answer is not reasonable because 2.155664 and 21.55664 mean two completely different answer far apart from each other

6 0

3 years ago

Solve for the given variable. Record your work and check the solution, if possible.

nikdorinn [45]

Answer:

10

Step-by-step explanation:

75c-300=25c+200

-25c -25c

50c-300=200

+300 +300

50c=500

c=10

4 0

3 years ago

Read 2 more answers

_ 5x^20-7x^10+15 in quadratic form

Tpy6a [65]

Answer:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

Step-by-step explanation:

Solve for x:

-5 x^20 - 7 x^10 + 15 = 0

Substitute y = x^10:

-5 y^2 - 7 y + 15 = 0

Divide both sides by -5:

y^2 + (7 y)/5 - 3 = 0

Add 3 to both sides:

y^2 + (7 y)/5 = 3

Add 49/100 to both sides:

y^2 + (7 y)/5 + 49/100 = 349/100

Write the left hand side as a square:

(y + 7/10)^2 = 349/100

Take the square root of both sides:

y + 7/10 = sqrt(349)/10 or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

y = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Substitute back for y = x^10:

x^10 = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Taking 10^th roots gives (sqrt(349)/10 - 7/10)^(1/10) times the 10^th roots of unity:

x = -(1/10 (sqrt(349) - 7))^(1/10) or x = (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or y = -7/10 - sqrt(349)/10

Substitute back for y = x^10:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x^10 = -7/10 - sqrt(349)/10

Taking 10^th roots gives (-7/10 - sqrt(349)/10)^(1/10) times the 10^th roots of unity:

Answer: x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

8 0

3 years ago

A sundae at an ice-cream shop costs $3, and each extra ounce of toppings costs

anyanavicka [17]

Answer:

The number of ounce of topping that can be get is 4 .

Step-by-step explanation:

Given as :

The cost of a sundae = $3

The cost for each extra ounce of toppings = $0.50

Let The number of ounce of topping = n

The total spending cost on sundae with n extra topping = $5

Now, According to question

The total spending cost on sundae with n extra topping = The cost of a sundae + The number of ounce of topping × The cost for each extra ounce of toppings

Or , $3 + n × $0.50 = $5

or, 3 + 0.50 n = 5

or , 0.50 n = 5 - 3

or, 0.50 n = 2

∴ n = $\dfrac{2}{0.50}$

I.e n = 4

So, The number of ounce of topping = n = 4

Hence The number of ounce of topping that can be get is 4 . Answer

7 0

3 years ago

3+(-7) What is the answer