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4 years ago

Calculate 1.6 ∙ 103 times 3.0 ∙ 10-6. (Box after the "10" in answer is for the exponent.)

2 answers:

Nutka1998 [239]4 years ago

7 0

(1.6 * 10^3) * (3.0 * 10^-6) =
(1.6 * 3.0) * (10^3 * 10^-6)
4.8 * 10^(3 - 6) =
4.8 * 10^-3 =
4.8 * 1/10^3 =
4.8 / 10^3 =
4.8 / 1000 =
0.0048 <==

Fudgin [204]4 years ago

6 0

Answer:

$1.6\cdot 10^3\times 3.0\cdot 10^{-6}=4.8\times10^{-3}$

Step-by-step explanation:

Given : Expression $1.6\cdot 10^3\times 3.0\cdot 10^{-6}$

To find : Calculate the given expression?

Solution :

Step 1 - Write the expression

$=1.6\cdot 10^3\times 3.0\cdot 10^{-6}$

Step 2 - Taking 10 power together

$=(1.6\times 3)\times(10^3\times 10^{-6})$

Step 3 - Solve the multiplication and if base are same then in multiplication power get added.

$=4.8\times10^{3-6}$

$=4.8\times10^{-3}$

Therefore, The solution of expression is

$1.6\cdot 10^3\times 3.0\cdot 10^{-6}=4.8\times10^{-3}$

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3 years ago

The mean per capita income is 16,127 dollars per annum with a variance of 682,276. What is the probability that the sample mean

MakcuM [25]

Answer:

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The standard deviation is the square root of the variance. So

$\mu = 16127, \sigma = \sqrt{682276} = 826, n = 476, s = \frac{826}{\sqrt{476}} = 37.86$

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Either it differs by 104 or less dollars, or it differs by more than 104 dollars. The sum of the probabilities of these events is 100. I am going to find the probability that it differs by 104 or less dollars first.

Probability that it differs by 104 or less dollars first.

pvalue of Z when X = 16127 + 104 = 16231 subtracted by the pvalue of Z when X = 16127 - 104 = 16023. So

X = 16231

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

By the Central Limit Theorem

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

$Z = \frac{16231 - 16127}{37.86}$

$Z = 2.75$

$Z = 2.75$ has a pvalue of 0.9970

X = 16023

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

$Z = \frac{16023 - 16127}{37.86}$

$Z = -2.75$

$Z = -2.75$ has a pvalue of 0.0030

0.9970 - 0.0030 = 0.9940

99.40% probability that it differs by 104 or less.

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

p + 99.40 = 100

p = 0.60

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

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4 years ago

1/3 (12x-24) =16

liraira [26]

Answer:

Step-by-step explanation:

Multiplying both sides by 3, we have 12x - 24 = 48. Adding both sides by 24, we have 12x = 72. That means x = 6

Can I have brainliest

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