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eduard
3 years ago
14

A zookeeper took a random sample of 30 days and observed how much food an elephant ate on each of those days. The sample mean wa

s 350 kg and the sample standard deviation was 25 kg. Find the 95% confidence interval for how much food the elephant eats.
Mathematics
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

341.95≤x≤358.95

Step-by-step explanation:

The confidence interval formula is expressed as;

CI = xbar±(z×s/√n)

xbar is the mean sample

s is the standard deviation

z is the z score at 95% confidence interval

n is the sample size

xbar = 350kg

z = 1.96

s = 25kg

n = 30

Substitute

CI = 350±(1.96×25/√30)

CI = 350±(1.96×4.564)

CI = 350±(8.946)

CI = (350-8.946, 350+8.946)

CI = (341.05, 358.95)

Hence the 95% confidence interval for how much food the elephant eats us 341.95≤x≤358.95

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Use Newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x3 + x +
nevsk [136]

Answer:

x_2 \approx -1.769

Step-by-step explanation:

Let f(x)=x^3+x+7

So f'(x)=3x^2+1

x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

Let x_1=-2

We are going to find x_2

So we are evaluating -2-\frac{f(-2)}{f'(-2)}

First step find f(-2)

Second step find f'(-2)

Third step plug in those values and apply PEMDAS!

f(-2)=(-2)^3+(-2)+7=-8-2+7=-10+7=-3

f'(-2)=3(-2)^2+1=3(4)+1=12+1=13

So

x_2=-2-\frac{-3}{13} \\\\ x_2=\frac{-26+3}{13} \\\\ x_2=\frac{-23}{13} \\\\ x_2 \approx -1.769

5 0
3 years ago
A) what is average rate of change of f(x) over the interval from x=5 to x=9? (table shows values of f(x). graph shows function o
SVETLANKA909090 [29]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

The average rate of change of a function over the interval [a,b] is equal to \frac{f(b)-f(a)}{b-a}, hence:

\frac{f(b)-f(a)}{b-a}\\\\\frac{f(9)-f(5)}{9-5}\\\\\frac{14-(-4)}{9-5}\\ \\\frac{14+4}{4}\\ \\\frac{18}{4}\\ \\\frac{9}{2}

Therefore, the average rate of change of f(x) over the interval [5,9] is \frac{9}{2}.

<u>Part B</u>

Do the same thing as in Part A:

\frac{f(b)-f(a)}{b-a}\\ \\\frac{f(1)-f(0.25)}{1-0.25}\\ \\\frac{2-5}{0.75}\\ \\\frac{-3}{0.75}\\ \\-4

Therefore, the average rate of change of g(x) over the interval [0.25,1] is -4.

<u>Part C</u>

To interpret our answer from Part B in terms of the real world it represents, we say that between 0.25 seconds and 1 second, the ball falls at a rate of 4 feet per second (since our average rate of change is negative).

5 0
2 years ago
I need help please.
Norma-Jean [14]

Answer:

No

Step-by-step explanation:

7 0
3 years ago
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I’m not sure sorrry but i hope someone else can help u
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What is h(-8)=-2(-8+5)^2 +4
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I got a fraction but, H= 7/4

3 0
3 years ago
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