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3 years ago

I need 33. B) find the exact time when the radius reaches 10 inches 100 points! Plz help

Mathematics

1 answer:

stealth61 [152]3 years ago

3 0

Step-by-step explanation:

You have found a function r(V(t)). We can see that this function is a one variable function. The variable is time.

So in this specific function we can call r(v(t)), r(t).

So:

$r(t) = \sqrt[3]{ \frac{3 \times (10 + 20t)}{4\pi} }$

If α is the moment that the radius is 10 inches and since the function above gives radius in inches we have to solve the equation:

$r( \alpha ) = 10$

Which is the same as:

$\sqrt[3]{ \frac{3 \times (10 + 20 \alpha )}{4\pi} } = 10 \\ \frac{3 \times (10 + 20 \alpha )}{4\pi} = 1000 \\ (10 + 20 \alpha ) = \frac{4000\pi}{3} \\ 20 \alpha = \frac{(4000\pi - 30)}{3} \\ \alpha = \frac{(4000\pi - 30)}{60}$

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How do you solve this?

puteri [66]

Answer:

Width: 7

Length: 14

Step-by-step explanation:

The area of a rectangle can be found my multiplying the length by the width. We do not know neither the length nor the width. However, we do know that the length is 7 feet longer than the width. Because we do not know the value of either side, let's let $x$ represent the width.

Since the length is 7 feet longer, we can represent the length by $x+7$ .

Now that we have our values for the length and the width we can multiply them together to find our area.

$(x)(x+7)=x^{2} +7x$

Now we have our equation, so we can address the changes made by the original question. The original question states that when 7 is added to both the length and width the area becomes 3 times larger. To do so simply increase each side by 7 by adding 7 to the original values.

$(x+7)(x+14)$

And multiply our area by 3.

$3(x^{2} +7x)$

set these equal to each other to find your new equation.

$(x+7)(x+14)=3(x^{2} +7x)$

Now you need to solve for $x$ . To do this first muliply $(x+7)$ and $(x+14)$ .

$(x+7)(x+14)=\\x^{2} +14x+7x+98=\\x^{2} +21x+98$

Then multiply $3(x^{2} +7x)$

$3(x^{2} +7x)=\\3x^{2} +21x$

Now you can begin solving for $x$ .

$x^{2} +21x+98=3x^{2} +21x$

Subtract $x^{2}$ from both sides.

$21x+98=2x^{2} +21x$

Subtract $21x$ from both sides.

$98=2x^{2}$

Divide by 2.

$49=x^{2}$

And finally take the sqaure root of both sides.

$\sqrt{x^{2} }=\sqrt{49}\\x=7$

Remember, because we are dealing with length, there cannot be negative. So while normally we would get both +7 <em>and</em> -7, in this case we <em>only </em>get +7.

Now that we have the value of $x$ , we can plug it into our original values.

For width we simply get 7.

For length we get 7+7=14

To check our answer we cna multiply 7 by 14 to get 98. Then we can add 7 to our length and width to get 14 and 21. Multiply these together and we get 294. 294 divided by 3 is 98, proving our answer correct.

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Answer:

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Step-by-step explanation:

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The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.

Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If $Z \leq -2$ or $Z \geq 2$ , the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .