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3 years ago

Helllppp! please don’t give me something fake:(

Mathematics

1 answer:

Afina-wow [57]3 years ago

3 0

Answer:

B. -5

Step-by-step explanation:

Ok so g(0) is the output for when x is 0. On the table it shows that when x is 0 g(x) or g(0) is -5. So B -5 is your answer.

You might be interested in

What is the slope of the tangent line to the parametric curve x = t2 + 2t, y = t2 + 1 at t = 1?

notsponge [240]

Answer:

1/2

Step-by-step explanation:

x = t² + 2t

dx/dt = 2t + 2

y = t² + 1

dy/dt = 2t

dy/dx = (dy/dt) / (dx/dt)

dy/dx = 2t / (2t + 2)

dy/dx = t / (t + 1)

At t = 1, dy/dx = 1/2.

6 0

3 years ago

Investment firm claims that the mean increase in price per share companies invest in is $1.95 per share every year and national

djyliett [7]

Answer:

The Margin of Error in percentage is 5.88%

Step-by-step explanation:

The values of the factors in the question are;

The mean increase in price per share, = $1.95

The number of companies in the sample, n = 100

The mean increase in price per share, $\overline x$ = $1.82

The standard deviation of the population, σ = $0.30

The margin of error, MOE, is given by the following formula;

$Margin \ of \ Error= z \times \dfrac{\sigma}{\sqrt{n} }$

Where;

z = z-score, critical value = 1.96 at 95%

σ = The standard deviation of the population = $0.30

n = The sample size = 100

Therefore, we have;

$Margin \ of \ Error= 1.96 \times \dfrac{0.3}{\sqrt{100} } = 0.0588$

The Margin of Error in percentage = 0.0588 × 100 = 58.8%

The Margin of Error in percentage = 5.88%

7 0

3 years ago

Assume that the matrices below are partitioned conformably for block multiplication. Compute the product.

Masteriza [31]

Multiplying block matrices works just like multiplication between regular matrices, provided that component matrices have the right sizes.

$\begin{bmatrix}\mathbf I&\mathbf 0\\\mathbf K&\mathbf I\end{bmatrix}\begin{bmatrix}\mathbf W&\mathbf X\\\mathbf Y&\mathbf Z\end{bmatrix} = \begin{bmatrix}\mathbf{IW}+\mathbf{0Y}&\mathbf{IX}+\mathbf{0Z}\\\mathbf{KW}+\mathbf{IY}&\mathbf{KX}+\mathbf{IZ}\end{bmatrix}$

$\begin{bmatrix}\mathbf I&\mathbf 0\\\mathbf K&\mathbf I\end{bmatrix}\begin{bmatrix}\mathbf W&\mathbf X\\\mathbf Y&\mathbf Z\end{bmatrix} = \begin{bmatrix}\mathbf W+\mathbf 0&\mathbf X+\mathbf 0\\\mathbf{KW}+\mathbf Y&\mathbf{KX}+\mathbf Z\end{bmatrix}$

$\begin{bmatrix}\mathbf I&\mathbf 0\\\mathbf K&\mathbf I\end{bmatrix}\begin{bmatrix}\mathbf W&\mathbf X\\\mathbf Y&\mathbf Z\end{bmatrix} = \begin{bmatrix}\mathbf W&\mathbf X\\\mathbf{KW}+\mathbf Y&\mathbf{KX}+\mathbf Z\end{bmatrix}$