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3 years ago

I have two children. One is a boy born on a Tuesday. What is the probability that I have two boys?

2 answers:

8 0

Let's handle this case by case.

Clearly, there's no way both children can be girls. There are then two cases:

Case 1: Two boys. In this case, we have 13 possibilities: the first is born on a Tuesday and the second is not (that's 6 possibilities, six ways to choose the day for the second boy), the first is not born on a Tuesday and the second is (6 more possibilities, same logic), and both are born on a Tuesday (1 final possibility), for a total of 13 possibilities with this case.

Case 2: A boy and a girl. In this case, there are 14 possibilities: The first is a boy born on a Tuesday and the second is a girl born on any day (7 possibilities, again choosing the day of the week. We are counting possibilities by days of the week, so we must be consistent here.), or the first is a girl born any day and the second is a boy born on a Tuesday (7 possibilities).

We're trying to find the probability of case 1 occurring given that case 1 or case 2 occurs. As there's 13+14=27 ways for either case to occur, we have a 13/27 probability that case 1 is the one that occurred.

White raven [17]3 years ago

7 0

Let B1T denote the first boy being born on Tuesday, B2N denote the second boy being born any day of the week and similarly B1N,B2T.

Then by Baye's formula, $P=\frac{P((B1T\cap B2N)\cup(B1N\cap B2T))_{B1\cap B2}P(B1\cap B2)}{P((B1T\cap B2N)\cup(B1N\cap B2T)}=\frac{(\frac{1}{7}+\frac{1}{7}-\frac{1}{7^2})\cdot\frac{1}4}{\frac{1}{14}+\frac{1}{14}-\frac{1}{14^2}}=\boxed{\frac{13}{27}}$

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Atrain travels at a speed of 58.6 km'hi Express this speed a.ms b. m/min

marishachu [46]

Answer:

16.27

Step-by-step explanation:

if 1000m rep 1km

therefore :

58.6km/h in ms = 58.6km×1000/3600

=16.277 ms

8 0

2 years ago

A girl scout troop sold cookies. If the girls sold 5 more boxes the second week than they did the first, and if they doubled the

mihalych1998 [28]

We can let x be the number of boxes sold for the first week. We can as well express the number of boxes for the second and third week through x using the statements provided.

Since the girl scout sold 5 more boxes on the second week, we have (x + 5) number of boxes sold for the second week. Now, for the third week, since it's double that of the second week, we have 2(x + 5). Thus, we have the following:

first week: x
second week: x + 5
third week: 2(x + 5)

Given that the total number of boxes sold for the three weeks is 431. We have

x + (x + 5) + 2(x + 5) = 431
x + x + 5 + 2x + 10 = 431
4x + 15 = 431
4x = 431 - 15
4x = 416
x = 104

We have now the value of x. Using this, we can find the values for the second and third week.
x + 5 = `104 + 5 = 109
2(x + 5) = 2(109) = 218

Answer:
first week: 104
second week: 109
third week: 218

5 0

3 years ago

Solve the following using Substitution method 2x – 5y = -13 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.