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castortr0y [4]
2 years ago
10

Must m in the equation y=mx+b always be a positive number

Mathematics
2 answers:
Sindrei [870]2 years ago
7 0

Not at all.

'm' turns out to be the slope of the graph of the equation.

If the graph of the equation slopes upward as it goes
from left to right, then 'm' is positive.

If the graph of the equation slopes downward as it goes
from left to right, then 'm' is negative.

Nookie1986 [14]2 years ago
4 0
'm' is the slope. 

The answer is no. There could be a negative number, which means that the slope would be decreasing, or as known as a negative slope (negative rate of change). <span />
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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B2y-1%7D%7B15%7D%20%3D%5Cfrac%7By%7D%7B10%7D" id="TexFormula1" title="\frac{2y-1}{15}
bonufazy [111]

Answer:

y = 2

Step-by-step explanation:

Given

\frac{2y-1}{15} = \frac{y}{10} ( cross- multiply )

10(2y - 1) = 15y ← distribute left side

20y - 10 = 15y ( subtract 15y from both sides )

5y - 10 = 0 ( add 10 to both sides )

5y = 10 ( divide both sides by 5 )

y = 2

3 0
2 years ago
Need help on #12 and # 13 please and thank you
Paladinen [302]
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5 0
3 years ago
Write 5 1/3% as a fraction in simplest form
goblinko [34]
5+1/3=                                                                                                              15/3+1/3=                                                                                                        16/3
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2 years ago
Read 2 more answers
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Juli2301 [7.4K]
45/45/90 triangle has side ratios s/s/sroot2
leg = root 5
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6 0
2 years ago
10# A consumer advocacy group is doing a large study on car rental practices. Among other things, the consumer group would like
ella [17]

ANSWER:

104

STEP-BY-STEP EXPLANATION:

Given:

Standard deviation (σ) = 700

Confidence level = 95%

Mean error (Eμ) = 135

We have for a confidence level of 95%:

\begin{gathered} \alpha=100\%-95\%=5\%=0.05 \\  \\ \alpha\text{/2}=\frac{0.05}{2}=0.025 \\  \\ \text{ The corresponding Z value for \alpha/2 = 0.025 is as follows:} \\  \\ Z_{\alpha\text{/2}}=1.96 \end{gathered}

Now, we calculate the minimum value of the sample size as follows:

\begin{gathered} n=\left(\frac{Z_{\alpha\text{/2}}\cdot\sigma}{E}\right)^2 \\  \\ \text{ We replacing:} \\  \\ \:n=\left(\frac{1.96\cdot 700}{135}\right)^2 \\  \\ \:n=103.2858\cong104 \end{gathered}

The minimum sample size needed is 104

3 0
1 year ago
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