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3 years ago

Write the equation in point-slope form of the line that passes through the given points.

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

4 0

Answer:

y-6=-1/3(x+6)

Step-by-step explanation:

y-y1=m(x-x1)

m=(y2-y1)/(x2-x1)

m=(3-6)/(3-(-6))

m=-3/(3+6)

m=-3/9

simplify

m=-1/3

y-6=-1/3(x-(-6))

y-6=-1/3(x+6)

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Please help with this AP Calculus question !

melomori [17]

Answer:

C. $\displaystyle \frac{cos(x)}{x} - ln(x)sin(x)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Trig Derivatives

Logarithmic Derivatives

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = ln(x)cos(x)$

Step 2: Differentiate

Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[ln(x)]cos(x) + ln(x)\frac{d}{dx}[cos(x)]$
Logarithmic Derivative: $\displaystyle f'(x) = \frac{1}{x}cos(x) + ln(x)\frac{d}{dx}[cos(x)]$
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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

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3 years ago

Find the product -8x5y2.6x2y

UNO [17]

Answer:

Answer is given below with explanations

Step-by-step explanation:

$- 8x.5 {y}^{2} .6{x}^{2} 2y = - 48 {x}^{3} .10 {y}^{3}$

HAVE A NICE DAY 

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

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3 years ago

Read 2 more answers

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Use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come f

AlladinOne [14]

Answer:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

Step-by-step explanation:

Data given and notation

n=75 represent the random sample taken

$\hat p=0.64$ estimated proportion of interest

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true proportion is higher than 0.5:

Null hypothesis: $p =0.5$

Alternative hypothesis: $p > 0.5$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

7 0

3 years ago