All categories

3 years ago

Please help me out with this thingyyy!!!!!

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

5 0

Answer:

1. 18

2. 8

Step-by-step explanation:

the mean is calculated by:

total / number of values

the current total is

10 + 14 + 8 + 16 + 12 = 60

now when we add another number there will be 6 numbers and we want the mean to be 13 so:

total / 6 = 13

total = 13 × 6

total = 78

but our current total is 60

so the missing number is 78 - 60 = 18

for part 2:

if we remove a number there will be 5 values and we want the mean to be 14 so:

total / 5 = 14

total = 14 × 5

total = 70

but we have the total of 6 numbers as 78 so we have to remove 78 - 70 = 8

You might be interested in

Which chocolate chip cookies have the greater spread in the number of chips per cookie?

Arisa [49]

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

What’s #1?????? Plz help

liq [111]

Answer:

C) $400\pi$

Step-by-step explanation:

The formula for area is

A=pi * radius ^2

If the diameter is 40, the radius is 20

Therefore

$A=\pi 20^{2} \\A=\pi 400$

7 0

2 years ago

The price of an iPad Air is initially $500. iPads owned by a college student depreciate at a rate of 3% per

34kurt

Waddawadda waddawadda

5 0

3 years ago

Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .