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Mamont248 [21]
2 years ago
10

Katrina buys a ​40-ft roll of fencing to make a rectangular play area for her dogs. Use ​2(​l+w)=40 to write a function for the​

length, given the width. Graph the function. What is a reasonable domain for the​ situation? Explain.
Mathematics
1 answer:
Anuta_ua [19.1K]2 years ago
8 0
Why you write hi this needs to anser!!
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The half-life of a certain radioactive material is 78 hours. An initial amount of the material has a mass of 790 kg.. Write an e
adoni [48]
<span> half life i= 78 hours 
 amount after 78 hours is 395 kg: 
395 = 790e^(k*78) 

Dividing by 790 and taking natural log
ln (395/790) = (k*78) 
-0.6931 = 78k 
-0.00888 = k  

lets calculate  how much is left after 18 hours: 
Amount(18)
= 790e^(-0.00888*18) 
Amount = 673.301 kg
hope this helps</span>
4 0
3 years ago
Read 2 more answers
What is 9(x2) I don’t understand this
djverab [1.8K]

Answer:

the answer is 9x2 the 2 is above the x though.

5 0
3 years ago
Find the second derivative y=1/5x^2+1/11x
Amiraneli [1.4K]

\frac{dy}{dx}=\frac{1}{5}.x^{2-1}.2 + \frac{1}{11}.x^{1-1}.1

\frac{dy}{dx}=\frac{2}{5}.x+\frac{1}{11}

\frac{d^2y}{dx^2}=\frac{2}{5}.x^{1-1}.1

\frac{d^2y}{dx^2}=\frac{2}{5}

8 0
3 years ago
According to a study conducted in one city, 39% of adults in the city have credit card debts of more than $2000. A simple random
Butoxors [25]

Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean \mu = 0.39 and standard deviation s = 0.0488

Step-by-step explanation:

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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.39, n = 100

Then

s = \sqrt{\frac{0.39*0.61}{100}} = 0.0488

By the Central Limit Theorem:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean \mu = 0.39 and standard deviation s = 0.0488

5 0
2 years ago
(4a^2b-3ab^2+2ab+5)+(2ab^2b+3ab^2-7ab)
creativ13 [48]
Answer : 2ab^3+4a^2b-5ab+5
4 0
3 years ago
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