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3 years ago

Which is the best estimate for (8.9x10^8)/(3.3x10^4) written in scientific notation? A. 3x10^2 B. 6x10^2 C. 3x10^4 D. 6x10^4

Mathematics

2 answers:

xenn [34]3 years ago

7 0

Esitmate:
9 /3 = 3
and
8-4 = 4
so it would be 3x10^4
answer is C

son4ous [18]3 years ago

3 0

Answer:

Option C

Step-by-step explanation:

The given expression is $\frac{8.9\times 10^{8}}{3.3\times 10^{4}}$

Now we have to solve this.

$\frac{8.9\times 10^{8}}{3.3\times 10^{4}}$

$=\frac{8.9}{3.3}\times \frac{10^{8}}{10^{4}}$

= 2.70 × 10⁽⁸⁻⁴⁾

= 2.70 × 10⁴

≈ 3 × 10⁴

Therefore, Option C is the answer.

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Wewaii [24]

Answer:

y = 10-2x

Step-by-step explanation:

2x+y=10

y=10-2x

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2 years ago

8(b-2)-12b what is the answer ?

arlik [135]

4(-b-4)

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3 years ago

Read 2 more answers

Consider the probability that greater than 26 out of 124 software users will call technical support. Assume the probability that

Mekhanik [1.2K]

Answer:

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

100% probability that greater than 26 out of 124 software users will call technical support.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

$np \geq 10$ and $n(1-p) \geq 10$

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.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Out of 124 software users

This means that $n = 124$

Assume the probability that a given software user will call technical support is 97%.

This means that $p = 0.97$

Conditions:

$np = 124*0.97 = 120.28 \geq 10$

$n(1-p) = 124*0.03 = 3.72 < 10$

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

Consider the probability that greater than 26 out of 124 software users will call technical support.

The lowest possible probability of those is 27, so, if it is 0, since it is considerably below the mean, 100% probability of being greater. We have that:

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

$P(X = 27) = C_{124,27}.(0.97)^{27}.(0.03)^{97} = 0$

1 - 0 = 1

100% probability that greater than 26 out of 124 software users will call technical support.

3 0

3 years ago

O.012 in words from

Scrat [10]

Answer:

word form of 0.012 is

zero point zero one two

Step-by-step explanation:

.-point

0-zero

1-one

2-two

3-three

4-four

5-five

6-six

7-seven

8-eight

9-nine

10-ten

4 0

2 years ago

Please help with this problem :(

Lilit [14]

36 - 24

= 12 + 16

= 28 - 10 2/3

Best of Luck to you.

8 0

3 years ago