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Marizza181 [45]

2 years ago

9

7 gifts are to be distributed among 10 children. how many distinct results are possible if no child is to receive more than 1 gi

ft?

1 answer:

Aneli [31]2 years ago

4 0

This is the number of combinations of 7 in 10 which is the same as number of combinations of 3 in 10:
10C3 = 10*9*8 / 3*2*1

= 120

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A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and tes

Anvisha [2.4K]

Answer:

Probability will be 0.5779

Step-by-step explanation:

Number of tablets n = 19

Probability of defect rems that is p = 0.03

Here number of trials is fixed and probability of success = 0.03

Here p = 0.03

So q = 1 - p = 1 -0.03 = 0.97

By using binomial distribution

Probability is given by $^nC_rp^rq^{n-r}$

So P(0 or 1) = $^{19}C_0{0.03}^0{0.97}^{19-0}+^{19}C_1{0.03}^1{0.97}^{19-1}=0.5606+0.0173=0.5779$

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3 years ago

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6 0

3 years ago

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dangina [55]

Answer:

$1.75$ × $10^{-15}$

I hope this helps!

8 0

3 years ago

-28 = -7(3x + 4) +21x

Vlada [557]

Answer:

1 Expandir.

-28=-21x-28+21x

−28=−21x−28+21x

2 Simplifica -21x-28+21x−21x−28+21x a -28−28.

-28=-28

−28=−28

3 Ya que ambos lados son iguales, hay infinitas soluciones.

Soluciones Infinitas

Step-by-step explanation:

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3 years ago

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The answer for this question is Y=3x+1

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3 years ago