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3 years ago

How much bigger is 3/8 than 1/3 answer as a fraction

1 answer:

7 0

You would first need to find a common denominator for 3/8 and 1/3 the common denominator would be 24 because 8 times three is 24 and 3 times 8 is 24. After you find your common denominator your fraction becomes 9/24 + 8/24. This gives you 17/24.

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Simplify leaving your answer with positive exponents.

vaieri [72.5K]

$\bf \left( \cfrac{5m^{\frac{4}{3}}}{n^{\frac{2}{3}}} \right)^{-\frac{2}{3}}\left(\cfrac{m^4}{8n^5} \right)^{-\frac{4}{3}}\implies 
\left( \cfrac{n^{\frac{2}{3}}}{5m^{\frac{4}{3}}} \right)^{\frac{2}{3}}\left(\cfrac{8n^5}{m^4} \right)^{\frac{4}{3}}\impliedby 
\begin{array}{llll}
\textit{and now we}\\
\textit{distribute}\\
\textit{the exponent}
\end{array}$

$\bf \left( \cfrac{n^{\frac{2}{3}\cdot \frac{2}{3}}}{5^{\frac{2}{3}}m^{\frac{4}{3}\cdot \frac{2}{3}}} \right)\left( \cfrac{8^{\frac{4}{3}}n^{5\cdot \frac{4}{3}}}{m^{4\cdot \frac{4}{3}}} \right)\implies \cfrac{n^{\frac{4}{9}}}{5^{\frac{2}{3}}m^{\frac{8}{9}}}\cdot \cfrac{8^{\frac{4}{3}}n^{\frac{20}{3}}}{m^{\frac{16}{3}}}\implies \cfrac{8^{\frac{4}{3}}n^{\frac{4}{9}+\frac{20}{3}}}{5^{\frac{2}{3}}m^{\frac{8}{9}+\frac{16}{3}}}$

$\bf \cfrac{8^{\frac{4}{3}}n^{\frac{4+60}{9}}}{5^{\frac{2}{3}}m^{\frac{8+48}{9}}}\implies \cfrac{8^{\frac{4}{3}}n^{\frac{64}{9}}}{5^{\frac{2}{3}}m^{\frac{56}{9}}}$

4 0

3 years ago

What multiplied by itself 3 times gives you -216 please answer asap!!!

AlexFokin [52]

Answer:

-6

Step-by-step explanation:

given

x³ = -216

x = (-216)^⅓

factorize

x = [(-6)×(-6)×(-6)]^⅓

x = -6

4 0

3 years ago

Read 2 more answers

Can someone solve this for me and tell me the answer

Zolol [24]

Z (>/=) -17
start by subracting 26 then multiplying by 2

5 0

2 years ago

Read 2 more answers

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

7 0

2 years ago

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.

8 0

3 years ago