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Lerok [7]
3 years ago
11

Determine the correct scientific notation form of the number. 0.000005304 A) 5.304 x 10−5 Eliminate B) 5.304 x 10−6 C) 53.04 x 1

0−7 D) 5304 x 10−9
Mathematics
2 answers:
Anuta_ua [19.1K]3 years ago
8 0

Answer: B) 5.304 x 10−6

Step-by-step explanation:

We have the number 0.000005304, now we can see that we have 5 zeros after the decimal point, in this case, we can write the number as the first digits different than zero (this will be the number before the decimal point, usually you only want one), multiplicated by 10 at the power of -(n + 1) where n is the number of zeros after the decimal point, so the correct way of writing it is:

5.305x10^-6

The correct option is B

Ksju [112]3 years ago
4 0

The answer is B) 5.304 × 10 to the power of -6

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zavuch27 [327]

Answer:

f(- 10) = 74

Step-by-step explanation:

To find f(- 10) , substitute n = - 10 into f(n) , that is

f(- 10) = - 7(- 10) + 4 = 70 + 4 = 74

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Tay–Sachs Disease Tay–Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carri
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Answer:

a) 0.0156 = 1.56% probability that all children will develop the disease.

b) 0.4219 = 42.19% probability that only one child will develop the disease.

c) 0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability that their offspring will develop the disease is approximately .25.

This means that p = 0.25

Three children:

This means that n = 3

Question a:

This is P(X = 3). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156

0.0156 = 1.56% probability that all children will develop the disease.

Question b:

This is P(X = 1). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.25)^{1}.(0.75)^{2} = 0.4219

0.4219 = 42.19% probability that only one child will develop the disease.

c. The third child will develop Tay–Sachs disease, given that the first two did not.

Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

Third develops, which 0.25 probability. So

p = 0.75*0.75*0.25 = 0.1406

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

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Answer:

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gift 2:

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We have to calculate the surface area of these gifts in order to know the amount of wrapping paper.

The formula for the surface area is:

A = 2*ab + 2*ac + 2*bc

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A = 2*6*10 + 2*6*9.5 + 2*10*9.5

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That means that gift 1 needs 424 cm^2 of wrapping paper

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That means that gift 2 needs 1420.5 cm^2 of wrapping paper

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This means that gift 1 needs 570 cm^3 of packaging peanuts

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