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3 years ago

Solve. Tell whether the equations have all real numbers or no solutions.

Mathematics

2 answers:

Mila [183]3 years ago

8 0

9514 1404 393

Answer:

one solution:: k = 0

Step-by-step explanation:

The equation simplifies to ...

6k +22 = -k +22

7k = 0 . . . . . . . . . . add k-22 to both sides; next, divide by 7

k = 0 . . . . . . . . one solution

GenaCL600 [577]3 years ago

4 0

The answer is zero, so no solution is your answer.

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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

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

Part C

$\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971$

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Step-by-step explanation:

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Answer:

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

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Solving systems of equations using substitution/elimination

Step-by-step explanation:

Step 1: Define systems

13x - 6y = 22

x = y + 6

Step 2: Solve for y

Substitution

Substitute in x: 13(y + 6) - 6y = 22
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Combine like terms: 7y + 78 = 22
Isolate y term: 7y = -56
Isolate y: y = -8

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Define original equation: x = y + 6
Substitute in y: x = -8 + 6
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