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3 years ago

Write an equation in slope-intercept form that passes through (2, -3) and slope of -5.

Mathematics

1 answer:

Ksivusya [100]3 years ago

8 0

Step-by-step explanation:

Hey there!

The given slope is "-5" and it passes through point (2,-3).

Use one-point formula to find out the equation.

$(y - y1) = m(x - x1)$

Put all values.

$(y + 3) = - 5(x - 2)$

Simplify it.

$(y + 3) = - 5x + 10$

$y = - 5x + 10 - 3$

$y = - 5x + 7$

Therefore, the equation is y= -5x+7.

Hope it helps...

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3 years ago

Find the first four terms of the infinite sequence defined explicitly by the following rule.

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The only thing that you should do is to put 1, 2, 3, and 4 instead of the n and then you will get the first four terms!

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3 years ago

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3 years ago

Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline

yawa3891 [41]

Answer:

B) 4.07

Step-by-step explanation:

First we need to calculate the mean of all the data, which is the same as the mean of the means of each grade of gasoline:

Regular BelowRegular Premium SuperPremium

39.31 36.69 38.99 40.04

39.87 40.00 40.02 39.89

39.87 41.01 39.99 39.93

X1⁻=39.68 X2⁻= 39.23 X3⁻= 39.66 X4⁻= 39.95

Xgrand⁻ = (39.68+39.23+39.66+39.95)/4 = 39.63

Next we need to calculate the sum of squares within the group (SSW) and the sum of squares between the groups (SSB), and the respective degrees of freedom):

SSW = [ (39.31-39.68)² + (39.87-39.68)² + (39.87-39.68)² ] + [ (36.69-39.23)² + (40.00-39.23)² + (41.01-39.23)² ] + [ (38.99-39.66)² + (40.02-39.66)² + (39.99-39.66)² ] + [ (40.04-39.95)² + (39.89-39.95)² + (39.93-39.95)² ] = [0.2091] + [10.2129] + [0.6874] + [0.0121] = 11.12

SSW = 11.12

Degrees of freedom in this case is calculated by m(n-1), with m being the number of grades of gasoline (4) and n being the number of trial results for each one (3), so we would have 4(3-1) = 8 degrees of freedom

SSB = [ (39.68-39.63)² + (39.68-39.63)² + (39.68-39.63)²] + [ (39.23-39.63)² + (39.23-39.63)² + (39.23-39.63)² ] + [ (39.66-39.63)² + (39.66-39.63)² + (39.66-39.63)² ] + [ (39.95-39.63)² + (39.95-39.63)² +(39.95-39.63)² ] = [0.0075] + [0.48] + [0.0027] + [0.3072] = 0.7974

SSB = 0.80

For this case, the degrees of freedom are m-1, so we would have 4-1 = 3 degrees of freedom

Now we can establish the hypothesis for the test:

H0: μ1 = μ2 = μ3 = μ4

The null hypothesis states that the means of miles per gallon for each fuel are the same, indicating that the drade of gasoline does not make a difference, therefore our alternative hypothesis will be:

H1: the grade of gasoline does makes a difference

We will use the F statistic to test the hypothesis, which is calculated like follows:

F - statistic = (SSB/m-1) / (SSW/m(n-1)) = (0.80/3) / (11.12/8) = 0.19

We know that the level of significance we are using is α = 0.05, so to find the critical value F we need to look at some table of critical values for the F distribution for the 0.05 significance level (like the attached image). Then we just need to look fot the value that is located in the intersection between the degrees of freedom we have in the numerator (horizontal) and the denominator (vertical) of the statistic (3 and 8). That critical value is:

Fc = 4.07

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3 years ago

(1/4)^-2 - (5^0) • 1^-1 =

katrin [286]

Answer:

Step-by-step explanation:

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