Translate each inequality. ¨a minimum of 10¨

Mathematics

1 answer:

Alexandra [31]2 years ago

5 0

Answer:

X ≥ 10

Step-by-step explanation:

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The first- and second-year enrollment values for a technical school are shown in the table below: Enrollment at a Technical Scho

lyudmila [28]

Answer:

The solution to f(x) = s(x) is x = 2012.

Explanation:

Rewrite the table and the choices for better understanding:

Enrollment at a Technical School

Year (x) First Year f(x) Second Year s(x)

2009 785 756

2010 740 785

2011 690 710

2012 732 732

2013 781 755

Which of the following statements is true based on the data in the table?

The solution to f(x) = s(x) is x = 2012.

The solution to f(x) = s(x) is x = 732.

The solution to f(x) = s(x) is x = 2011.

The solution to f(x) = s(x) is x = 710.

<h2>Solution</h2>

The question requires to find which of the options represents the solution to f(x) = s(x).

That means that you must find the year (value of x) for which the two functions, the enrollment the first year, f(x), and the enrollment the second year s(x), are equal.

The table shows that the values of f(x) and s(x) are equal to 732 (students enrolled) in the year 2012, x = 2012.

Thus, the correct choice is the third one:

The solution to f(x) = s(x) is x = 2012.

5 0

3 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$