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uranmaximum [27]
2 years ago
12

g According to government data, 25% of employed women have never been married. If 10 employed women are selected at random, what

is the probability that two or fewer of them have never been married
Mathematics
1 answer:
ollegr [7]2 years ago
6 0

Answer:

The probability is 0.5263

Step-by-step explanation:

Probability of employed women being unmarried = 25% = 0.25 ( let this be p)

Probability of employed women ever being married = 1-0.25 = 0.75 (let this be q)

Now, we want to calculate the probability that two or less have never been married

We can use the Bernoulli approximation of the Binomial distribution here;

In this case, we have three cases

We have cases of;

Exactly two + Exactly one + None

For exactly two;

10 C 2 0.25^2 0.75^8 = 0.282

For exactly one

10 C 1 0.25^1 0.75^9 = 0.188

For none

10C 0 0.25^0 0.75^10 = 0.0563

Adding these 3 to get the probability of two or fewer, we have;

0.0563 + 0.188 + 0.282 = 0.5263

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PLEASE HELP ASAP!!
Fed [463]

Answer: The missing statements are,

In first blank:  ∠2≅∠1

In second blank:  AC≅AC

In third blank:  Reflexive

Step-by-step explanation:

Since, The hypotenuse angle theorem states that if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of another right triangle, then the two triangles are congruent to each other.

Here, given:

∠D and ∠B are right angles.

DC ║ AB

Prove: Δ ADC ≅ Δ CBA

              Statement                                       Reason

1.∠D and ∠B are right angles       1. Given

2. ∠2 ≅ ∠1                                      2. If lines are parallel then interior angles

                                                        are equal

3. AC≅AC                                      3. Reflexive

4.Δ ADC ≅ Δ CBA                        4. Hypotenuse angle theorem

                                                                     




6 0
3 years ago
Solve with completing the square method:<br>6x²-7x+2=0 and  ax²+bx+c=0
Norma-Jean [14]
6x ^2-7x+2=0\\ \\ a=6 , \ b = -7 , \ c=2 \\ \\\Delta = b^{2}-4ac = (-7)^{2}-4*6*2=49-48=1 \\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{7-\sqrt{1}}{2*6}=\frac{7-1}{12} =\frac{6}{12}= \frac{1}{ 2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{7+\sqrt{1}}{2*6}=\frac{7+1}{12} =\frac{8}{12}= \frac{2}{ 3}


7 0
3 years ago
The admission fee to a video game arcade is $1.25 per person, and it costs $0.50 for each game played. Latoya and donnetta
dalvyx [7]

Answer:

15 games.

Step-by-step explanation:

Here is the complete question: The admission fee to a video game arcade is $1.25 per person, and it costs $0.50 for each game played. Latoya and donnetta have total $10.00 to spend. What is the greatest number of games they will be able to play?

Given: Admission fees= $1.25 per person.

           Cost of each game played= $0.50.

           Total Money, Latoya and Donnetta have is $10.

First lets find total admission fees both will pay.

∴ Total addmission fees= Fees\ per\ person \times number\ of\ person

Total addmission fees= \$1.25\times 2= \$ 2.50

Now, money remaining after paying admission fees.

Money remaining = Total\ money - Admission\ fees

∴ Money remaining = \$10-\$2.5= \$ 7.5

∴ Money remained to play video games is $7.5.

Next, finding the maximum number of games, both will be able to play.

As we know, each game will cost $0.50.

Total number of games played= \frac{Money\ remaining}{Cost\ of\ each\ game}

⇒ Total number of games played=\frac{\$ 7.5}{\$ 0.5} = 15\ games

∴ 15 games is the maximum number of games they will be able to play.

3 0
3 years ago
How do I apply the rule of four to this?​
crimeas [40]

Answer:

Step-by-step explanation:

7 0
2 years ago
Suppose a jar contains 8 red marbles and 14 blue marbles. If you reach in the jar and pull out 2 marbles at random, find the pro
NikAS [45]

it would most likely be blue because its more

it would be a  57.1429%. of blue

8 0
2 years ago
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