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kifflom [539]
3 years ago
8

Graph the line with intercept 6 and slope

la1" title=" - \frac{3}{2} " alt=" - \frac{3}{2} " align="absmiddle" class="latex-formula">
​

Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0

Given:

The y-intercept of a line = 6

The slope of the line = -\dfrac{3}{2}

To find:

The graph of the given line.

Solution:

The slope intercept form of a line is:

y=mx+b

Where, m is the slope and b is the y-intercept.

Putting m=-\dfrac{3}{2} and b=6 in the above equation, we get

y=-\dfrac{3}{2}x+6

At x=0,

y=-\dfrac{3}{2}(0)+6

y=0+6

y=6

At x=2,

y=-\dfrac{3}{2}(2)+6

y=-3+6

y=3

Plot these two points (0,6) and (2,3) on a coordinate plane and connect them by a straight line to get the graph of the required line.

The required graph is shown below.

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Answer:

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}= \mu_D

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C,D

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have p=4 groups and on each group from j=1,\dots,p we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2  

And we have this property  

SST=SS_{between}+SS_{within}  

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.

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3 years ago
For f(x) = 2x+1 and g(x) = x2 - 7, find (f+ g)(x).
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Answer:

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5 0
2 years ago
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IRISSAK [1]

Step-by-step explanation:

cos A = 16/20 = 4/5

tan A = 12/16 = 3/4

sin A = 12/20 = 3/5

4 0
3 years ago
Please answer CORRECTLY for brainliest guarantee. If your answer is not correct or does not have a explanation I will report you
Scorpion4ik [409]

Answer:

c. 45.34ft

Step-by-step explanation:

When leg one of the triangle is 15ft and leg two is 17 ft you would take   17^2 + 15^2 and then you find the square root of that (514) to get the hypotenuse of the triangle. After that you multiply by two to account for the 2/2 sqaure foot per each unit

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Answer:

6x - 11y = -13 is the answer.

Step-by-step explanation:

Let's plug in the points to see what sticks.

Start with (-4, -1)

1) 11x - 6y = 11(-4) - 6(-1) = -44 + 6 = -38 \neq 13

2) 6x - 11y = 6(-4) - 11(-1) = -24 + 11 = -13

3) 6x - 7y = 6(-4) - 7(-1) = -24 + 7 = -17 \neq 17

4) 6x - 11y = 6(-4) - 11(-1) = -24 + 11 = -13 \neq 13

The only one that fits is #2.  Let's try the other point to be sure.

2) 6x - 11y = 6(1.5) - 11(2) = 9 - 22 = -13

3 0
3 years ago
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