<img src="https://tex.z-dn.net/?f=Factor%3A%20x%5E2-9" id="TexFormula1" title="Factor: x^2-9" alt="Factor: x^2-9" align="absmidd

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Arada [10]

2 years ago

le" class="latex-formula">

Mathematics

1 answer:

Bezzdna [24]2 years ago

7 0

Answer:

x²-9=x²-3²=(x+3)(x-3)

Step-by-step explanation:

use formula

a²-b²=(a+b)(a-b)

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F(r) = -(1 – 12)(r + 3)<br> What are the zeros of the function

nexus9112 [7]

Answer:

r = -3

Step-by-step explanation:

The zero of a function is any replacement for the variable that will produce an answer of zero. Graphically, the real zero of a function is where the graph of the function crosses the x‐axis; that is, the real zero of a function is the x‐intercept(s) of the graph of the function.

8 0

2 years ago

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

2 years ago

Which of the following is the domainof the given relation?{(-6, 3), (-4 ,5), (0, 0)}A {0, 4, -6}B {0, 3, 5)C {3, 4, 0}D {-6, -4,

Bas_tet [7]

The domain is the input, which is the x value in the relation (x, y)

Hence the domain in the given relation {(-6, 3), (-4 ,5), (0, 0) is :

{-6, -4, 0}

6 0

8 months ago

Pls ANSWERRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR

natta225 [31]

Hi there!

$\large\boxed{ 9.4m}$

Find the perimeter using the formula:

P = 2l + 2w

It is given that the dimensions are 2.8 × 1.9 so plug these number into the equation:

P = 2(2.8) + 2(1.9)

P = 5.6 + 3.8

P = 9.4m

4 0

2 years ago

Victor is making a recipe for smoothies. His first recipe requires 2 cups of strawberries for every 7 cups of other ingredients.

sweet-ann [11.9K]

Answer:

2/7 or 3/9

3/9 is greater so the recipe with 3 cups per 9 cup requires more

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers