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3 years ago

Help me answer this question plz

Mathematics

1 answer:

Ostrovityanka [42]3 years ago

5 0

If f(x) = 2x -6+4
Then
f(-2) = 2(-2) -2
f(-2) = -6

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PLEASE HELP !! IT WOULD BE GREATLY APPREACIATED ! I DONT KNOW HOW TO DO THIS

serg [7]

(A) Bella's TV ratio of sides: 36"/27" = 1.333...

Lenny's: 52"/29.25" = 1.777...

Bella's and Lenny's TVs are NOT similar - their side ratios are different. Bella's is 4:3 while Lenny's is close to 16:9.

(B) Bella's TV has an exact ratio of 4:3 (36/27=4/3) and so is the older type.

Lenny's TV has a ratio of 52"/29.25". Divide both 52 and 29.25 by 3.25 to get 16/9, or 16:9. So Lenny's is the newer type.

4 0

3 years ago

Please answer!

Xelga [282]

Amswer

it is b Step-by-step explanation:

8 0

3 years ago

A) Find a particular solution to y" + 2y = e^3 + x^3. b) Find the general solution.

Reptile [31]

Answer:

a.P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x}$

Step-by-step explanation:

We are given that a linear differential equation

$y''+2y=e^{3x}+x^3$

We have to find the particular solution

P.I= $\frac{e^{3x}}{D^2+2}+\frac{x^3}{D^2+2}$

P.I= $\frac{e^{3x}}{3^2+2}+\frac{1}{2} x^3(1+\frac{D^2}{2})^{-2}$

P.I= $\frac{e^{3x}}{11}+\frac{1-2\frac{D^2}{4}+3\frac{D^4}{16}+...}{2}x^3$

P.I= $\frac{e^{3x}}{11}+\frac{x^3-2\frac{\cdot3\cdot 2x}{4}}{2}+0}$ (higher order terms can be neglected

P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.Characteristics equation

$D^2+2=0$

$D=\pm\sqrt2 i$

C.F= $C_1cos \sqrt2x+C_2 sin\sqrt2 x$

G.S=C.F+P.I

G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x)$

3 0

4 years ago

Solve this problem n-6/-4=6

babunello [35]

Answer:

N= 9/2

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home. Give the value of th

astra-53 [7]

Answer:

The value of the standard error for the point estimate is of 0.0392.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home.

This means that $n = 100, p = \frac{81}{100} = 0.81$

Give the value of the standard error for the point estimate.

This is s. So

$s = \sqrt{\frac{0.81*0.19}{100}} = 0.0392$

The value of the standard error for the point estimate is of 0.0392.

7 0

3 years ago