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Sergio [31]
2 years ago
13

Please i need help ASAP

Mathematics
1 answer:
Flauer [41]2 years ago
6 0
Sorry the reason I’m typing is I need points
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The price (in dollars) p and the quantity demanded q are related by the equation: p^2+2q^2=1100.If R is revenue, dR/dt can be ex
bagirrra123 [75]
FOR THE FIRST PROBLEM:

Differentiating p^2 + 2q^2 = 1100 with respect to time => 2p(dp/dt) + 4q(dq/dt) = 0 

<span>dp/dt = -2q/p(dq/dt) </span>

<span>R = pq </span>

<span>dR/dt = p(dq/dt) + q(dp/dt) = -p²/2q(dp/dt) + q(dp/dt) = (2q² - 1100)/2q * (dp/dt) + q(dp/dt) </span>

<span>A = (2q - 550/q)

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
3 0
3 years ago
Witch one of these is correct
Lady bird [3.3K]
42:105 is the answer
4 0
3 years ago
Read 2 more answers
The volumes of two similar square pyramids is 128 and 250 meters respectively. The side length of the smaller pyramid is 8 meter
Yuliya22 [10]

Answer:

Height of bigger square pyramid = 3.072 meter

Step-by-step explanation:

Using equivalent ratios to find the side of the bigger square prism

128:8 = 250:X

X = (250 x 8) ÷ 128 = 15.625

Volume of a square pyramid = a x a x h/3

Hence to evaluate the height

250 = 15.625 x 15.625 x h/3

H = 3.072 meters

6 0
3 years ago
These box plots show the prices for two different brands of shoes.
nignag [31]

Answer: Choice C

The interquartile range (IQR) for brand A, $10, is less than the IQR for brand B, $20.

===========================================================

Explanation:

Let's go through the answer choices one by one to see which is true and which is false. Also, we'll see which is helpful when it comes to the spread.

----------------------

Choice A

The IQR is found by subtracting the values of Q1 and Q3, which are the left and right edges of the box in that order.

For boxplot A, we have an IQR = Q3-Q1 = 80-70 = 10.

So we can immediately rule out choice A because the IQR for boxplot A is not 20, but instead it's 10.

----------------------

Choice B

The statement made in choice B is true, since the center vertical lines represent the median, but it's not useful to determine the spread. The median only tells us the center of the distribution. It doesn't tell us how spread out the distribution is.

Imagine you didn't have access to the graph. Now if you're told that the median for brand A is larger than brand B's median, then you'd probably think boxplot A is more spread out; however, the boxplot diagram shows the opposite is the case.

So again the median isn't used here. We can rule out choice B.

----------------------

Choice C

This statement is true. We subtract the box edge values to get the IQR values for each boxplot

For brand A we have IQR = Q3-Q1 = 80-70 = 10

For brand B we have IQR = Q3-Q1 = 70-50 = 20

This is the answer we're after.

The IQR is one tool we can use to measure the spread. The range is another option. We could also use the variance or the standard deviation (two slightly similar ideas) to measure the spread.

As you can see from the diagram, the larger the IQR, the more spread out the data will be.

----------------------

Choice D

This is false because earlier we've shown 10 is the IQR for brand A.

7 0
3 years ago
Problem PageQuestion The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate paramet
asambeis [7]

Answer:

<em>t = 1.51</em>

Step-by-step explanation:

<u>Exponential Model</u>

The exponential model is often used to simulate the behavior of a magnitude that either grow or decay in proportion to the existing amount of that magnitude.

The model can be expressed as

M=M_oe^{kt}

In this case, Mo is the initial mass of the radioactive substance and k is a constant which value is positive if the mass is growing or negative if the mass is decaying.

The value of k is not precisely given in the question, we are assuming k=-0.2

The model is now

M=M_oe^{-0.2t}

We are required to compute the time it takes the mass to reach one-half of its initial value:

\displaystyle \frac{M_o}{2}=M_oe^{-0.2t}

Simplifying

\displaystyle \frac{1}{2}=e^{-0.2t}

Taking logarithms

\displaystyle ln\frac{1}{2}=ln(e^{-0.2t})=-0.2t

Solving for t

\displaystyle t=-\frac{ln\frac{1}{2}}{0.2}=1.51

6 0
3 years ago
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