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3 years ago

5x + 4y = 20 on a graph

Mathematics

1 answer:

Yakvenalex [24]3 years ago

3 0

Answer:

5x + 4y = 20 on a graph

Step-by-step explanation:

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Select all triangle whose base to height ratio proportional to the triangles pytho plotted

Aleonysh [2.5K]

If there was a picture maybe I could help.

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3 years ago

Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi

RSB [31]

Answer: $(0.273,\ 0.783)$

Step-by-step explanation:

Given : Total number of students who are asked to complete a survey anonymously : n= 240

Number of students

The proportion report that they have experienced bullying =54

Then , the proportion that students have experienced bullying = $p=\dfrac{54}{240}=0.225$

Significance interval : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

The confidence interval for the true proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.225\pm (1.96)\sqrt{\dfrac{(0.225)(1-0.225)}{240}}\\\\\approx0.225\pm0.0528\\\\=(0.255-0.528,\ 0.255+0.528)\\\\=(0.273,\ 0.783)$

8 0

3 years ago

I need help with 4x + 5y = 11 i am confused

AlekseyPX

<span>Simplifying 4x + 5y = 11

Solving 4x + 5y = 11

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-5y' to each side of the equation. 4x + 5y + -5y = 11 + -5y

Combine like terms: 5y + -5y = 0 4x + 0 = 11 + -5y 4x = 11 + -5y

Divide each side by '4'. x = 2.75 + -1.25y

Simplifying x = 2.75 + -1.25y</span>

4 0

3 years ago

At an interest rate of 8% compounded annually, how long will it take to double the following investments?

Paladinen [302]

Let's see, if the first one has a Principal of $50, when it doubles the accumulated amount will then be $100,

recall your logarithm rules for an exponential,

$\bf \textit{Logarithm of exponentials}\\\\
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-------------------------------\\\\
\qquad \textit{Compound Interest Earned Amount}
\\\\
$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$100\\
P=\textit{original amount deposited}\to &\$50\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t
\\\\\\
\cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
$

$\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------\\\\
$

now, for the second amount, if the Principal is 500, the accumulated amount is 1000 when doubled,

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$1000\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t
\\\\\\
$

$\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------$

now, for the last, Principal is 1700, amount is then 3400,

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$3400\\
P=\textit{original amount deposited}\to &\$1700\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}$

$\bf 3400=1700\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 3400=1700(1.08)^t
\\\\\\
\cfrac{3400}{1700}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t$

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4 years ago

What is the volume of a cone below?

Nonamiya [84]

The answer is 33pi ^3

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3 years ago

Read 2 more answers