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3 years ago

Which inequality is represented by the graph

Mathematics

2 answers:

castortr0y [4]3 years ago

8 0

Well, accordingly to the graph, it has a positive slope so we can eliminate b and d. So it is either a or c. As x increases, so does y. Y is almost not quite equal to 3x-2. In fact, it is greater, so I'd go with A. I'm pretty confident.

snow_tiger [21]3 years ago

7 0

Answer:

C) Inequality y ≤ 3x -2

Step-by-step explanation:

Given : Graph

To find : Which inequality is represented by the graph.

Solution : We have given graph.

Slope intercept form of line : y = mx + b.

Where, m = slope , b = y intercept,

We can see from given graph line is solid and shaded down.

For shaded down and solid we use less than or equal to sing

So, the inequality would be y ≤.

Y intercept ( where the line cut at y axis ).

Now, slope = 3 and y intercept = -2

So , equation become y ≤ 3x - 2.

Therefore, C) Inequality y ≤ 3x - 2.

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3 years ago

Which is bigger 4/6 or .77

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First, let's convert 0.77 to a fraction.

$\frac{0.77}{1}$

Multiply by 100

$\frac{77}{100}$

Now make the denominators the same, to make it easier to find which fraction is greater. We can do this by finding the LCM of the denominators, then however many times the denominator goes into that number, we multiply the numerator by the same amount.

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4 years ago

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agasfer [191]

Answer:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

Step-by-step explanation:

Information given

$\bar X=147.3$ represent the sample mean for the amount of time spent at part time jobs

$s=50$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =120$ represent the value to check

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to analyze if the true mean for the amount of time spent at part time jobs is higher than 120, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115$

Now we can calculate the degrees of freedom

$df=n-1=15-1=14$

If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got $t_{critc}= 1.761$

Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs

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