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2 years ago

What is the slope of the line represented by the equation y = y equals StartFraction 4 Over 5 EndFraction x minus 3.x – 3?

Mathematics

2 answers:

lisov135 [29]2 years ago

5 0

Answer:

The slope of the line $y=\frac{4}{5}x-3$ is $\frac{4}{5}$

Step-by-step explanation:

The given equation can be written as

$y=\frac{4}{5}x-3\hfill (1)$ represents a line equation.

The general equation of the straight line is $y=mx+c\hfill (2)$

where m is the slope and c is intercept

Comparing equations (1) and (2) we get slope $m=\frac{4}{5}$

Therefore slope of the line $y=\frac{4}{5}x-3$ is $\frac{4}{5}$

aleksandrvk [35]2 years ago

4 0

Answer:

4 Over 5 (4/5)

Step-by-step explanation:

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2 years ago

The expression y2 – 10y + 24 has a factor of y – 4. What is another factor of y2 – 10y + 24?

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1 year ago

4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

pochemuha

Answer:

No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for Tunnel is $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

$t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

$t = - 0.0899$

Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

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