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Sergio [31]
2 years ago
14

Use the formula for the amount, a = p(1 + rt), to find the indicated quantity.

Mathematics
1 answer:
Alexeev081 [22]2 years ago
5 0

a= p(1+rt)

a= $4,100(1+12%×2×0.25)

a= $4,100×(1+6%) 100%=1 -> 6%= 0,06

a= $4,100×1,06

a= $4,346

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Is there enough information to prove each quadrilateral is a parallelogram? Explain.
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1. yes

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4. yes

Step-by-step explanation:

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3 years ago
Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
  • n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

8 0
3 years ago
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