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3 years ago

A jar contains 4 purple, 3 blue, and 1 gold ticket. Three tickets are selected without replacement.

Mathematics

1 answer:

sveta [45]3 years ago

3 0

Answer:

a) 1/14

b) 1/56

c) 1/28

d) 5/28

Step-by-step explanation:

a) Since tickets are being selected without replacement, this means that the pool of outcomes diminishes with each selection. For the first pick, there are a total of 8 tickets to choose from, and 4 of those are purple. Therefore, the probability that the first ticket is purple is 4/8=1/2. For the second ticket, the total number of remaining tickets is 7, and the number of purple tickets is 3, meaning that the probability is 3/7. For the final ticket, 6 remaining tickets and 2 of them being purple yields a probability of 2/6=1/3. Multiplying all of these together, you get 1/2*3/7*1/3=1/14.

b) Similarly to the previous problem, the first probability is 3/8, followed by 2/7, and then by 1/6. Multiplying these together, you get a probability of 1/56.

c) The probability that the first two tickets are purple are 1/2 and 3/7, as we have already calculated. For the final ticket, there are 6 possible remaining with 1 of them being gold. This means a probability of 1/6, which yields a total of 1/28.

d) Since there are 5 tickets originally that aren't blue, the probability for the first pick is 5/8, followed by 4/7 and 3/6=1/2. Multiplying these together, you get a final answer of 5/8*4/7*1/2=5/28.

Hope this helps!

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Triphasil-28 birth control tablets are taken sequentially, 1 tablet per day for 28 days, with the tablets containing the followi

bazaltina [42]

Answer:

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

Step-by-step explanation:

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3}$

In which $L_{1}, L_{2}$ and $L_{3}$ are the number of miligrams of levonorgestrel taken in each phase.

The total milligrams of ethinyl estradiol is:

$E = E_{1} + E_{2} + E_{3}$

In which $E_{1},E_{2}$ and $E_{3}$ are the number of miligrams of ethinyl estradiol taken in each phase.

We can solve this by phase.

Phase 1: 6 tablets, each containing 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$6*0.050mg = 0.30mg$ of levonorgestrel and $6*0.030mg = 0.18mg$ of ethinyl estradiol are taken.

So $L_{1} = 0.30$ and $E_{1} = 0.18$

Phase 2: 5 tablets, each containing 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol

In this phase,

$5*0.075mg = 0.375mg$ of levonorgestrel and $5*0.040mg = 0.20mg$ of ethinyl estradiol are taken.

So $L_{2} = 0.375$ and $E_{2} = 0.20$

Phase 3: 10 tablets, each containing 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$10*0.125mg = 1.25mg$ of levonorgestrel and $10*0.030mg = 0.30mg$ of ethinyl estradiol are taken.

So $L_{3} = 1.25$ and $E_{3} = 0.30$

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3} = 0.30 + 0.375 + 1.25 = 1.925$

The total milligrams of ethinyl estradiol is:

$E = E_{1} + E_{2} + E_{3} = 0.18 + 0.20 + 0.30 = 0.68$

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

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3 years ago

Factor 27b - 18.<br> Write your answer as a product with a whole number greater than 1.

ElenaW [278]

9(3b-2) Iv factorised it but I dont know the rest sorry

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3 years ago

PLEASE HELP ANSWER !!!

geniusboy [140]

Answer:

the answer is 81

Step-by-step explanation:

1. find 75

2. add 75 + 24

3. subtract from 180 to find x

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3 years ago

Which equation has a graph that is perpendicular to the graph of -x + 6y = -12?

serious [3.7K]

Answer:

c) 6x + y = -52 is required equation perpendicular to the given equation.

Step-by-step explanation:

If the equation is of the form : y = mx + C.

Here m = slope of the equation.

Two equations are said to be perpendicular if the product of their respective slopes is -1.

Here, equation 1 : -x + 6y = -12

or, 6y = -12 + x

or, y = (x/6) - 2

⇒Slope of line 1 = (1/6)

Now, for equation 2 to be perpendicular:

Check for each equation:

a. x + 6y = -67 ⇒ 6y = -67 - x

or, y = (-x/6) - (67/6) ⇒Slope of line 2 = (-1/6)

but $\frac{1}{6} \times \frac{-1}{6} \neq -1$

b. x - 6y = -52 ⇒ -6y = -52 - x

or, y = (x/6) + (52/6) ⇒Slope of line 2 = (1/6)

but $\frac{1}{6} \times \frac{1}{6} \neq -1$

c. 6x + y = -52

or, y =y = -52 - 6x ⇒Slope of line 2 = (-6)

$\frac{1}{6} \times (-6) = -1$

Hence, 6x + y = -52 is required equation 2.

d. 6x - y = 52 ⇒ -y = 52 - 6x

or, y = 6x - 52 ⇒Slope of line 2 = (6)

but $\frac{1}{6} \times 6 \neq -1$

Hence, 6x + y = -52 is the only required equation .

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Answer:

61st term in the sequence

Step-by-step explanation:

125 = 2n + 3

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n = 61

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3 years ago