Kora has 2 graham crackers and its enough for 1 s'mores. How many s'mores would Kora make if she had 20 graham crackers?
2 answers:
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We have that
<span>f(x)=3x
g(x)=x</span>²<span>+5
</span>g(x)/f(x)=[x²+5]/[3x]
<span>that new function will exist for all real numbers except the value of zero, value for which it becomes indefinite
</span>
the domain is the interval (-∞,0) U (0,∞)
using a graph tool
see the attached figure
<span>f(x)=3x
g(x)=x</span>²<span>+5
</span>g(x)/f(x)=[x²+5]/[3x]
<span>that new function will exist for all real numbers except the value of zero, value for which it becomes indefinite
</span>
the domain is the interval (-∞,0) U (0,∞)
using a graph tool
see the attached figure
Answer: (0,1), (1,0)
Step-by-step explanation:
plug in a number for y.
ex) x=0 y=1
y=-x + 1
1= -0 + 1
1=0 + 1
Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.
<h3>What is the binomial distribution formula?</h3>The formula is:
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 0.54% of the calls receive a busy signal, hence p = 0.0054.
- A sample of 1300 callers is taken, hence n = 1300.
The probability that at least 5 received a busy signal is given by:
In which:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Then:
Then:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.
0.8295 = 82.95% probability that at least 5 received a busy signal.
More can be learned about the binomial distribution at brainly.com/question/24863377
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