Which expression is equivalent to (16 - 10) + (-2 + 19)

Mathematics

1 answer:

balu736 [363]3 years ago

8 0

Answer: =23

Step-by-step explanation: 16−10−2+19 =6−2+19 =6+17 =23

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If there is no relationship (linear or otherwise) between two quantitative variables as observed on a scatterplot, the value of

Ainat [17]

Answer:

Option 3) Closer to 0

Step-by-step explanation:

Correlation:

Correlation is a technique that help us to find or define a relationship between two variables.

A positive correlation means that an increase in one quantity leads to an increase in another quantity

A negative correlation means with increase in one quantity the other quantity decreases.
Range of Correlation

Values between 0 and 0.3 tells about a weak positive linear relationship, values between 0.3 and 0.7 shows a moderate positive correlation and a correlation of 0.7 and 1.0 states a strong positive linear relationship.

Values between 0 and -0.3 tells about a weak negative linear relationship, values between -0.3 and -0.7 shows a moderate negative correlation and a correlation value of of -0.7 and -1.0 states a strong negative linear relationship.

A value of 0 tells that there is no correlation between the two variables.

Thus, for the given situation, if there is no relationship between two quantitative variables then the value of the correlation coefficient, r, is close to 0

4 0

4 years ago

Find the limit............

liubo4ka [24]

Answer: $-\frac{9}{x^2}$

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Work Shown:

$\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9}{x+\Delta x} - \frac{9}{x}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x}{x(x+\Delta x)} - \frac{9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x-9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{9x-9x-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\$

$\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9}{x(x+\Delta x)}\\\\\\\displaystyle \frac{-9}{x(x+0)}\\\\\\\displaystyle -\frac{9}{x^2}\\\\\\$

The trick here is to first simplify the expression so that the $\Delta x$ term in the very lower denominator cancels out. This is to avoid dividing by zero. Once this division by zero error is avoided, we can replace the delta x term with 0 and evaluate the limit as $\Delta x$ approaches 0 from the right or positive direction.