How to find the greatest common factor of 24 and 90?

A bucket of paint has spilled on a tile floor. The paint flow can be expressed with the function p(t) = 6t, where t represents t

Neko [114]

Answer:

A) A[p(t)] = 36πt²

B) 7234.56 square units

Step-by-step explanation:

Given functions:

$\begin{cases}p(t)=6t \\ A(p)=\pi p^2 \end{cases}$

Part A

To find the area of the circle of spilled paint as a function of time, substitute the function p(t) into the given function A(p):

$\begin{aligned}A(p) & = \pi p^2\\\\ \implies A[p(t)] & = \pi [p(t)]^2\\& = \pi (6t)^2\\& = \pi 6^2 t^2\\& = 36\pi t^2\end{aligned}$

Part B

Given

t = 8 minutes
π = 3.14

Substitute the given values into the equation for A[p(t)} found in part A:

$\begin{aligned}A[p(8)] & = 36\pi t^2\\& = 36 \cdot 3.14 \cdot 8^2\\& = 36 \cdot 3.14 \cdot 64\\& = 113.04 \cdot 64\\& = 7234.56\:\: \sf square\:units\end{aligned}$

Therefore, the area of spilled paint after 8 minutes if 7234.56 square units.

4 0

1 year ago

Read 2 more answers

I don't know this please help

Komok [63]

Answer:

2/1 I believe, I suck at slope tho, I'm sorry if it's wrong

4 0

2 years ago

Read 2 more answers

Use the given information to find the p-value. Also, use a 0.05 significance level and state the conclusion about the null hyp

Hatshy [7]

Answer:

We can assume that the statistic is $z_{calc}=0.78$

$p_v = 2* P(z>0.78) = 0.435$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of interest is not different from 3/5

Step-by-step explanation:

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is equal to 3/5 or not.:

Null hypothesis: $p=3/5$

Alternative hypothesis: $p \neq 3/5$

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

We can assume that the statistic is $z_{calc}=0.78$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v = 2* P(z>0.78) = 0.435$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of interest is not different from 3/5

7 0

2 years ago

Can someone please help me with this? thank you

dybincka [34]

Answer:

what should I change my profile picture to?...

Step-by-step explanation: