Ask question

Ask question

All categories

3 years ago

10

Solve -4(x+2)-2x+4 plzzzzzzzzzzzzzzzzzzzz

2 answers:

Ronch [10]3 years ago

8 0

Answer:

-6x-4

Step-by-step explanation:

nikitadnepr [17]3 years ago

6 0

Answer:

-6x-4

Step-by-step explanation:

You might be interested in

Complete the equation of the line through (3,-1)(3,−1)left parenthesis, 3, comma, minus, 1, right parenthesis and (4,7)(4,7)left

Mandarinka [93]

Answer: $y=8x-25$

Step-by-step explanation:

Given

Line passes through $(3,-1)\ \text{ and}\ (4,7)$

Equation of line when it passes through two points $(x_1,y_1),(x_2,y_2)$

$\Rightarrow \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{y-y_1}{x-x_1}$

Insert the values

$\Rightarrow \dfrac{7+1}{4-3}=\dfrac{y+1}{x-3}\\\\\Rightarrow \dfrac{8}{1}=\dfrac{y+1}{x-3}\\\\\Rightarrow 8x-24=y+1\\\Rightarrow y=8x-25$

8 0

3 years ago

Can someone please help me I tryed already

Anit [1.1K]

Answer:look at the picture below

Step-by-step explanation:

3 0

3 years ago

Help me out please????

frutty [35]

Answer:

fromfromfromfromfromfromfromfromfromfromfromfromfromfromfromfromfrom

Step-by-step explanation:

6 0

3 years ago

Please help.. omgg urgent

Inga [223]

The top one. they can't be expressed in a fraction or decimal

4 0

3 years ago

Read 2 more answers

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

6 0

3 years ago