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Troyanec [42]
3 years ago
12

WILL MARK BRAINLIEST A number cube is rolled and a coin is flipped. Predict how many times you would get heads and a number grea

ter than 5 in 240 trials.
20
60
40
30
Mathematics
2 answers:
Over [174]3 years ago
8 0

Answer:

coin flip is 1/2

2 numbers are less than 3 so 2/6 = 1/3

1/2 x 1/3 = 1/6 probability total

240 x 1/6 = 40

answer is C) 40

Step-by-step explanation:

antoniya [11.8K]3 years ago
8 0

There is only one number greater than five on the die.

\frac{1}{2}\left(\frac{1}{6}\right)=\frac{1}{12}\\ \\ \left(\frac{1}{12}\right)240=20

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Solve x^2 - 3x - 10 = 0
Pachacha [2.7K]

Answer:

5 and -2

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Solve the following using Substitution method<br> 2x – 5y = -13<br><br> 3x + 4y = 15
Digiron [165]

\huge \boxed{\mathfrak{Question} \downarrow}

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\left. \begin{array}  { l  }  { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.

  • To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

2x-5y=-13, \: 3x+4y=15

  • Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

2x-5y=-13

  • Add 5y to both sides of the equation.

2x=5y-13

  • Divide both sides by 2.

x=\frac{1}{2}\left(5y-13\right)  \\

  • Multiply \frac{1}{2}\\ times 5y - 13.

x=\frac{5}{2}y-\frac{13}{2}  \\

  • Substitute \frac{5y-13}{2}\\ for x in the other equation, 3x + 4y = 15.

3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15  \\

  • Multiply 3 times \frac{5y-13}{2}\\.

\frac{15}{2}y-\frac{39}{2}+4y=15  \\

  • Add \frac{15y}{2} \\ to 4y.

\frac{23}{2}y-\frac{39}{2}=15  \\

  • Add \frac{39}{2}\\ to both sides of the equation.

\frac{23}{2}y=\frac{69}{2}  \\

  • Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

\large \underline{ \underline{ \sf \: y=3 }}

  • Substitute 3 for y in x=\frac{5}{2}y-\frac{13}{2}\\. Because the resulting equation contains only one variable, you can solve for x directly.

x=\frac{5}{2}\times 3-\frac{13}{2}  \\

  • Multiply 5/2 times 3.

x=\frac{15-13}{2}  \\

  • Add -\frac{13}{2}\\ to \frac{15}{2}\\ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

\large\underline{ \underline{ \sf \: x=1 }}

  • The system is now solved. The value of x & y will be 1 & 3 respectively.

\huge\boxed{  \boxed{\bf \: x=1, \: y=3 }}

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