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3 years ago

If f(x) = 2x - 6 and g(x) = 3x + 9, find (f + g)(x)

Mathematics

1 answer:

alex41 [277]3 years ago

5 0

Answer:

5x+3

Step-by-step explanation:

$(f + g)(x) = f(x) + g(x) \\ =2x - 6 + 3x + 9 \\ = 5x + 3$

consider marking as Brainliest if this helped

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Solve the equation. 3x + 6 = 24

JulsSmile [24]

Hello!

Question 1
3x + 6 = 24

Subtract 6 from both sides

3x = 18

Divide both sides by 3

x = 6

The answer is 6
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Question 2
x/4 + 1 = -5

Subtract 1 from both sides

x / 4 = -6

Multiply both sides by 4

x = -24

The answer is -24
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Hope this helps!

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2 years ago

What is the unit rate of change of d with respect to t? (That is, a change of 1 unit in t will correspond to a change of how man

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Step-by-step explanation:

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2 years ago

Sinθ+cotθ×cosθ=cscθ

Assoli18 [71]

Answer:

Step-by-step explanation:

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2 years ago

A farmer had a farm with animals.

Gnom [1K]

Answer: Part B

Step-by-step explanation:

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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

2 years ago