SIMPLE MATH PROBLEM! PLEASE HELP! The equation needs to use c, t, and =. Thanks!

"Find an integer, x, such that 5, 10, and x represent the lengths of the sides of an acute triangle.

Contact [7]

A, b, c - the lengths of the sides of the triangle
and a ≤ b ≤ c
then:
a + b > c and if the triangle is an acute triangle then a² + b² > c².

$1^o\\5\leq10\leq x\\\\\left\{\begin{array}{ccc}5+10 \ \textgreater \ x\\5^2+10^2 \ \textgreater \ x^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \ 15\\ x^2 \ \textless \ 125\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \ 15\\ x \ \textless \ \sqrt{125}\approx11.1\end{array}\right\\\\\boxed{x=11}\\\\2^o\\5\leq x\leq10\\\\\left\{\begin{array}{ccc}5+x \ \textgreater \ 10\\ 5^2+x^2 \ \textgreater \ 10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x=9}$

$3^o\\x\leq5\leq10\\\\\left\{\begin{array}{ccc}x +5 \ \textgreater \ 10\\ x^2+5^2 \ \textgreater \ 10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x\in\O}\\\\Answer:\boxed{x=9\ or\ x=11}\to your\ answer:\boxed{\boxed{x=11}}$

8 0

3 years ago

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Susan earned $75 for watching her neighbor's dog for 5 days. How much did Susan make in one day

Damm [24]

You need to find how many times 5 fits into 75 5x12=60 5x13=65 5x14=70 5x15=75. so, 5 fits in to 75 fifteen times. that means Susan makes $15 a day for 5 days. 5x15=75

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3 years ago

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g Suppose a factory production line uses 3 machines, A, B, and C for making bolts. The total output from the line is distributed

Vladimir [108]

Answer:

The probability that it came from A, given that is defective is 0.362.

Step-by-step explanation:

Define the events:

A: The element comes from A.

B: The element comes from B.

C: The element comes from C.

D: The elemens is defective.

We are given that P(A) = 0.25, P(B) = 0.35, P(C) = 0.4. Recall that since the element comes from only one of the machines, if an element is defective, it comes either from A, B or C. Using the probability axioms, we can calculate that

$P(D) = P(A\cap D) + P(B\cap D) + P(C\cap D)$

Recall that given events E,F the conditional probability of E given F is defined as

$P(E|F) = \frac{P(E\cap F)}{P(F)}$ , from where we deduce that

$P(E\cap F) = P(E|F)P(F)$ .

We are given that given that the element is from A, the probability of being defective is 5%. That is P(D|A) =0.05. Using the previous analysis we get that

$P(D) = P(A)P(D|A)+P(B) P(D|B) + P(C)P(D|C) = 0.05\cdot 0.25+0.04\cdot 0.35+0.02\cdot 0.4 = 0.0345$