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3 years ago

10

Write an equation that expresses the following relationship. d varies directly with the cube of w and inversely with p. In your

equation, use k as the constant of proportionality.

2 answers:

Delvig [45]3 years ago

8 0

The formula for the indicated variable is d = k w³ / p

<h3>Further explanation</h3>

Solving linear equation mean calculating the unknown variable from the equation.

Let the linear equation : y = mx + c

If we draw the above equation on Cartesian Coordinates , it will be a straight line with :

<em>m → gradient of the line</em>

<em>( 0 , c ) → y - intercept</em>

Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :

$\large {\boxed {m = \frac{y_2 - y_1}{x_2 - x_1} } }$

If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :

$\large { \boxed {y - y_1 = m ( x - x_1 ) } }$

Let us tackle the problem.

This problem is about directly and inversely proportional.

<u>Given</u> :

d varies directly with the cube of w → $\boxed {d \propto w^3}$

d varies inversely with p → $\boxed {d \propto p^{-1} }$

∴ $\large {\boxed {d \propto ( w^3 \times p^{-1} )} }$

From the above relationship, we can write the equation for the variable d i.e:

$\large { \boxed {d = k \frac{w^3}{p} } }$

<h3>Learn more</h3>

Infinite Number of Solutions : brainly.com/question/5450548
System of Equations : brainly.com/question/1995493
System of Linear equations : brainly.com/question/3291576

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Linear Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point

pantera1 [17]3 years ago

5 0

Here d varies directly with cube of w, and inversely with p. So w cube have to be in the numerator and p have to be in the denominator. And we need to use k for constant of proportionality. Therefore the required equation is

$d=\frac{kw^3}{p}$

In direct proportionality, the relatioship is linear that is as w increases, d increases and as w decreases d decreases and for inverse, it is opposite .

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After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni

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Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.64=64>10$

$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

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

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

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Step-by-step explanation:

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2 years ago

Read 2 more answers

An automobile manufacturer has given its car a 52.6 miles/gallon (MPG) rating. An independent testing firm has been contracted t

Arte-miy333 [17]

Answer:

Null hypothesis: $\mu = 52.6$

Alternative hypothesis: $\mu \neq 52.6$

$z=\frac{52.8-52.6}{\frac{1.6}{\sqrt{250}}}=1.976$

$p_v =2*P(z>1.976)=0.0482$

Since the p value is lower than the significance level wedon't have enough evidence to conclude that the true mean is significantly different from 52.6 MPG.

Step-by-step explanation:

Information provided

$\bar X=52.8$ represent the sample mean for the MPG of the cars

$\sigma=1.6$ represent the population standard deviation

$n=250$ sample size of cars

$\mu_o =52.6$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis

We need to conduct a hypothesis in order to check if the true mean of MPG is different from 52.6 MPG, the system of hypothesis would be:

Null hypothesis: $\mu = 52.6$

Alternative hypothesis: $\mu \neq 52.6$

Since we know the population deviation the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Calculate the statistic

Replacing we have this:

$z=\frac{52.8-52.6}{\frac{1.6}{\sqrt{250}}}=1.976$

Decision

Since is a two tailed test the p value would be:

$p_v =2*P(z>1.976)=0.0482$

Since the p value is lower than the significance level wedon't have enough evidence to conclude that the true mean is significantly different from 52.6 MPG.

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2 years ago