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3 years ago

What's 2+2 ? I have a hard time understanding what's the answer

Mathematics

2 answers:

madam [21]3 years ago

7 0

Answer:

2+2=4

Step-by-step explanation:

When adding 2 plus 2 you must add because + means adding when you add you get the sum of 4

Anarel [89]3 years ago

3 0

Answer:

Step-by-step explanation:

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State the requirements that must be satisfied to use the one-way ANOVA procedure. Select all that apply. A. The k samples must

Sergeu [11.5K]

Answer:

The correct option are: (A), (B), (C) and (E).

Step-by-step explanation:

The One-Way ANOVA is a statistical test used to compare the three or more population means.

The hypothesis is:

<em>H</em>₀: The population means are equal, i.e. μ₁ = μ₂ = μ₃ =...= μₙ

<em>H</em>ₐ: At least one of the population mean differ.

The Assumptions of One-way ANOVA are:

The samples selected from each population are independent of each other.
The population distribution of the dependent variables follow Normal distribution.
The population variances are always equal for all the groups.

Thus, the correct option are: (A), (B), (C) and (E).

3 0

2 years ago

A sweater is on sale for $45.00, down from $75.00. What is the

Dominik [7]

Answer:

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

The temperature at 12 am was 44 degrees. The temperature dropped steadily for the next 8 hours to reach the daytime low of 20 de

antoniya [11.8K]

Answer:

y=-3x+80

Step-by-step explanation:

Points (12,44) and (20,20)

slope = -3

Equation of line: y - 44 = -3(x - 12)

7 0

3 years ago

Find the values of m and b that make the following function differentiable. the piecewise function f of x equals x cubed when x

aalyn [17]

$f(x)=\begin{cases}x^3&\text{for }x\le1\\mx+b&\text{for }x>1\end{cases}$

$f$ must be continuous in order to be differentiable, so we need to have

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$

By its definition, $f(1)=1^3=1$ , and

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^3=1$

$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(mx+b)=m+b$

so that $\boxed{m+b=1}$ .

We want the derivative to exist at $x=1$ , which requires that we pick an appropriate value for $f'(1)$ so that $f'(x)$ is also continuous. At the moment, we know

$f'(x)=\begin{cases}3x^2&\text{for }x1\end{cases}{/tex]so we need to pick [tex]f'(1)$ such that

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)$

We have

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}3x^2=3$

$\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}m=m$

so that $\boxed{m=3}$ (which means we need to pick $f'(1)=3$ ) and so $m+b=1\implies\boxed{b=-2}$

5 0

3 years ago

Determine e which statement is true about the zeros of the function graphed below

Eddi Din [679]

Try this option:
1. according to the graph no points of intersection between function and X-axis, it means, that
2. D<0 and no real solutions. In other words function has two complex solution.
Answer: 2.

5 0

3 years ago