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Delvig [45]
3 years ago
13

Which of the following are incorrect expressions for slope? Check all that apply.. . A. Change in the dependent variable relativ

e to change in the independent variable. B. x2-x1/y2-y1. C. run/rise. D. delta y/delta x. .
Mathematics
2 answers:
n200080 [17]3 years ago
7 0
A. B. and C. are incorrect expressions for the slope.

m (slope) = rise/run = y2-y1/x2-x1 = change in the independent variable relative to the change in dependent variable 

y is the independent variable and x is the dependent variable
Oxana [17]3 years ago
7 0

Answer:

(B) and (C)

Step-by-step explanation:

The slope or gradient of a line is a number that describes both the direction and the steepness of the line. Slope can be  calculated by finding the ratio of the "vertical change" to the "horizontal change" between two different points on a line.

Thus, from the definition of the slope, we have

(A) Slope=Change in the dependent variable relative to change in the independent variable.

which is correct expression for the slope.

(B) S=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

which is the formula for the slope.

Now, the given option is S=\frac{x_{2}-x_{1}}{y_{2}-y_{1}}

which is incorrect.

(C) The slope of is given as rise over run that is \frac{rise}{run} but the given option is \frac{run}{rise} which is incorrect.

(D) The slope formula is given as:

S=\frac{{\delta}y}{{\delta}x}

which is the correct expression.

Therefore, the options are (B) and (C) which are incorrect expressions for slope

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Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

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Step-by-step explanation:

Given:

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Sample mean, x' = 4.2

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Required:

Find a 90% CI for true mean time

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Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

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t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

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