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3 years ago

Write the pair of fractions as a part of fractions with a common denominator 1/4 and 2/3

Mathematics

1 answer:

sladkih [1.3K]3 years ago

4 0

Answer:

$\frac{3}{12} and \frac{8}{12}$ are the fractions with common denominator

Step-by-step explanation:

fractions are 1/4 and 2/3

$\frac{1}{4} and \frac{2}{3}$

To get common denominator we take LCD

LCD of 4 and 3 is 12

To get 12 in the denominator we multiply first fraction by 3 and second fraction by 4

$\frac{1*3}{4*3} and \frac{2*4}{3*4}$

$\frac{3}{12} and \frac{8}{12}$

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Is 0.4 over 0.8 a rational or irrational number

Rudik [331]

It is a rational number and it equals 0.5 or 1/2

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3 years ago

Which of the following expressions is equal to 5^6/5^2

Natali5045456 [20]

Answer:

$\dfrac{5^6}{5^2}$ = $5{\cdot}5{\cdot}5{\cdot}5$

Step-by-step explanation:

The given expression is :

$\dfrac{5^6}{5^2}$

We need to find this expression is equal to what.

$\dfrac{5^6}{5^2}=\dfrac{5^4\times 5^2}{5^2}\\\\=5^4\\\\=5\times 5\times 5\times 5\\\\\text{or}\\\\=5{\cdot}5{\cdot}5{\cdot}5$

Hence, $\dfrac{5^6}{5^2}$ is equal to $5{\cdot}5{\cdot}5{\cdot}5$ . Hence, the correct option is (c).

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2 years ago

Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbic

jolli1 [7]

Answer:

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

The randomly sampled 400 dogs from homes where an herbicide was used on a regular basis, diagnosing lymphoma in 230 of them.

This means that:

$p_h = \frac{230}{400} = 0.575, s_h = \sqrt{\frac{0.575*0.425}{400}} = 0.0247$