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Hoochie [10]
4 years ago
5

Now consider the expression 4.0×103+4×102. determine the values of a and k when the value of this expression is written in scien

tific notation.
Mathematics
2 answers:
natta225 [31]4 years ago
6 0
Let`s say that the expression is written in the form:a * 10^k.4 * 10^3 + 4 * 10^2 = = 4 * 10 * 10^2 + 4 * 10^2 = = 40 * 10^2 + 4 * 10^2 = = 44 * 10^2 = 4.4 * 10^3.Answer:a = 4.4 and k = 3 and the expression in the scientific notation is:4.4 * 10^3.

Feliz [49]4 years ago
4 0

Answer:

The values of ''a'' and ''k'' are :

a=4.4

k=3

Step-by-step explanation:

If we want to write a number in scientific notation we can write it as :

a.(10)^{k}    (I)

Where ''a'' is a real number and :

1\leq a

Where ''k'' is an integer number.

The first step is to find the value of the expression.

The expression is :

4.(10)^{3}+4.(10)^{2} ⇒

4.(10)^{3}+4.(10)^{2}=4000+400=4400

We need to write the number ''4400'' in scientific notation.

We can write it as :

4400=(4.4).(10)^{3}

If we compare this final expression with (I) ⇒

a=4.4

k=3

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An automobile manufacturer finds that 1 in every 2500 automobiles produced has a particular manufacturing defect. ​(a) Use a bin
Advocard [28]

Answer:

a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

Step-by-step explanation:

Binomial 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.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

To use the Poisson approximation for the binomial, we have that:

\mu = np

1 in every 2500 automobiles produced has a particular manufacturing defect.

This means that p = \frac{1}{2500} = 0.0004

a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.

This is P(X = 4) when n = 7000. So

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

P(X = 4) = C_{7000,4}.(0.0004)^{4}.(0.9996)^{6996} = 0.1558

0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.

Using the approximation:

\mu = np = 7000*0.0004 = 2.8. So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 4) = \frac{e^{-2.8}*(2.8)^{4}}{(4)!} = 0.1557

0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

6 0
3 years ago
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