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gogolik [260]
3 years ago
15

Simplify the complex fraction.

Mathematics
1 answer:
lilavasa [31]3 years ago
5 0

Given:

$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}

To find:

The simplified fraction.

Solution:

Step 1: Simplify the numerator

$\frac{(4 r)^{3}}{15 t^{4}}=\frac{4^3 r^{3}}{15 t^{4}}=\frac{64 r^{3}}{15 t^{4}}

Step 2: Simplify the denominator

$\frac{16 r}{(3 t)^{2}}=\frac{16 r}{3^2 t^{2}}= \frac{16 r}{9 t^{2}}

Step 3: Using step 1 and step 2

$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}} \right)}

Step 4: Using fraction rule:

$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a \cdot d}{b \cdot c}

$\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}}\right)}=\frac{64r^3 \cdot 9t^2}{16 r \cdot 15 t^4}

Cancel the common factor r and t², we get

           $=\frac{64 r^{2} \cdot 9 }{16  \cdot 15 t^2 }

Cancel the common factors 16 and 3 on both numerator and denominator.

           $=\frac{4 r^{2} \cdot 3 }{  5 t^2 }

           $=\frac{12 r^{2}  }{  5 t^2 }

$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{12 r^{2}  }{  5 t^2 }

The simplified fraction is \frac{12 r^{2}  }{  5 t^2 }.

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well, 6.2% of 4,300 can't be over half. she has made a calculation error, and the real answer is (this is where the mistake was made) 4,300 * 0.062 (not 6.2, because percents are always divided by 100 to make them into a usable number), which is equal to $266.60.

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Please help me please I really need help
expeople1 [14]

Answer:

5%

Step-by-step explanation:

Multiply both numerator and denominator by 100. We do this to find an equivalent fraction having 100 as the denominator.

0.05 × 100/100

= (0.05 × 100) × 1/100 = 5/100

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2 years ago
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Round 564,938 to the nearest 1000 is it 565000
lord [1]

Answer:

YES!

Step-by-step explanation:

564,938 -> 565,000

BRAINLIEST?

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3 years ago
in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smok
grandymaker [24]

Answer:

z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869  

p_v =2*P(Z>1.869)=0.0616  

If we compare the p value obtained and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .  

Step-by-step explanation:

1) Data given and notation

n=232 represent the random sample taken

X represent the adults were no longer smoking after one month

\hat p=0.849 estimated proportion of adults were no longer smoking after one month

p_o=0.80 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.8.:  

Null hypothesis:p=0.8  

Alternative hypothesis:p \neq 0.8  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(Z>1.869)=0.0616  

If we compare the p value obtained and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .  

6 0
3 years ago
Can somebody help me please ​
frozen [14]

Answer:

lets use X as a number

(x×2)+(x)+(x-40)

4 0
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