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3 years ago

a family has five children. the probability of having a girl is 1/2. whats probability of having at leasr 4 girls g'

Mathematics

1 answer:

Dima020 [189]3 years ago

7 0

Answer:

Probability of having at least 4 Girls

= 0.6875

Step-by-step explanation:

Probability of having at least 4 Girls is 1-probability of having exactly 3 girls

Total number of children= 5 = N

Probability of having a girl p = 0.5

Probability of not having a girl q= 0.5

X= 3

Probability of at least 4 girls is given by

Probability= NCX(p)^x(q)^(N-x)

Probability = 5C3(0.5)^3(0.5)^(5-3)

Probability = 5C3(0.5)^3(0.5)^2

Probability= 5!/3!2!(0.5)^3(0.5)^2

Probability= 10(0.125)(0.25)

Probability= 0.3125

Probability of having at least 4 Girls

= 1- 0.3125

= 0.6875

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W(t)W, left parenthesis, t, right parenthesis models the daily water level (in \text{cm}cmstart text, c, m, end text) at a pond

erastovalidia [21]

Answer:

t=152 days (in radians)

Step-by-step explanation:

W(t)models the daily water level at a pond in Arizona, t days after the hottest day of the year. (t is entered in radian)

$W(t) = 15\cos\left(\dfrac{2\pi}{365}t\right) + 43$

We want to determine the first time,t at which the water level is 30cm.

When W(t)=30

$30 = 15\cos\left(\dfrac{2\pi}{365}t\right) + 43\\30-43=15\cos\left(\dfrac{2\pi}{365}t\right)\\-13=15\cos\left(\dfrac{2\pi}{365}t\right)\\-\dfrac{13}{15} =\cos\left(\dfrac{2\pi}{365}t\right)\\cos^{-1}(-\dfrac{13}{15})=\dfrac{2\pi}{365}t\right)\\cos^{-1}(\dfrac{13}{15})=\dfrac{2\pi}{365}t\right)$

$cos^{-1}(\dfrac{13}{15})=\dfrac{2\pi}{365}t\right)\\t=\dfrac{365}{2\pi}\cdot cos^{-1}(\dfrac{13}{15})\\t=152.16\\\approx 152 days \text{ (to the nearest whole day)}$

6 0

3 years ago

Read 2 more answers

Positive integers greater than -6 but less than +21

Nina [5.8K]

Step-by-step explanation:

Inequality to describe problem:

$- 6 < x < 21$

x represents the unknown integer(s).

Integers are positive or negative numbers that don't have a fraction, or a decimal.

Positive integers that work in the inequality:

$- 6 < 1< 21$

$- 6 < 2 < 21$

$- 6 < 3 < 21$

$- 6 < 4 < 21$

$- 6 < 5 < 21$

$- 6 < 6 < 21$

$- 6 < 7 < 21$

$- 6 < 8 < 21$

$- 6 < 9 < 21$

$- 6 < 10< 21$

$- 6 < 11 < 21$

$- 6 < 12 < 21$

$- 6 < 13 < 21$

$- 6 < 14 < 21$

$- 6 < 15 < 21$

$- 6 < 16 < 21$

$- 6 < 17 < 21$

$- 6 < 18 < 21$

$- 6 < 19 < 21$

$- 6 < 20 < 21$

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 work out as positive integers.

7 0

2 years ago

A bag contains 40 coins, all of them are either 2 cent or 5 cent coins. If the value of the money in the bag is $1.55, find the

dlinn [17]

Answer: Explanation:First, let's call the number of 2 cent coins: tNext, let's call the number of 5 cent coins: fWe can then write to equations from the information in the problem.Equation 1: t+f=40Equation 2: 0.02t+0.05f=1.55Step 1) Solve the first equation for t:t+f=40t+f−f=40−ft+0=40−ft=40−fStep 2) Substitute (40−f) for t in the second equation and solve for f:0.02t+0.05f=1.55 becomes:0.02(40−f)+0.05f=1.55(0.02×40)−(0.02×f)+0.05f=1.550.80−0.02f+0.05f=1.550.80+(−0.02+0.05)f=1.550.80+0.03f=1.550.80−0.80+0.03f=1.55−0.800+0.03f=0.750.03f=0.750.03f0.03=0.750.030.03f0.03=25f=25Step 3) Substitute 25 for f in the solution to the first equation at the end of Step 1 and calculate t:t=40−f becomes:t=40−25t=15The Solution Is:There are:15 two cent coins25 five cent coins

Step-by-step explanation:

5 0

2 years ago

1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$