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sergeinik [125]
3 years ago
11

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>
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3 years ago
At an interest rate of 8% compounded annually, how long will it take to double the following investments?
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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}\\\\&#10;log_{{  a}}\left( x^{{  b}} \right)\implies {{  b}}\cdot  log_{{  a}}(x)\\\\&#10;-------------------------------\\\\&#10;\qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;

\bf A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &\$100\\&#10;P=\textit{original amount deposited}\to &\$50\\&#10;r=rate\to 8\%\to \frac{8}{100}\to &0.08\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annnually, thus once}&#10;\end{array}\to &1\\&#10;t=years&#10;\end{cases}&#10;\\\\\\&#10;100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t&#10;\\\\\\&#10;\cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)&#10;\\\\\\&#10;

\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\&#10;-------------------------------\\\\&#10;

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

\bf \qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &\$1000\\&#10;P=\textit{original amount deposited}\to &\$500\\&#10;r=rate\to 8\%\to \frac{8}{100}\to &0.08\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annnually, thus once}&#10;\end{array}\to &1\\&#10;t=years&#10;\end{cases}&#10;\\\\\\&#10;1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t&#10;\\\\\\&#10;

\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)&#10;\\\\\\&#10;log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\&#10;-------------------------------

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

\bf \qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &\$3400\\&#10;P=\textit{original amount deposited}\to &\$1700\\&#10;r=rate\to 8\%\to \frac{8}{100}\to &0.08\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annnually, thus once}&#10;\end{array}\to &1\\&#10;t=years&#10;\end{cases}

\bf 3400=1700\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 3400=1700(1.08)^t&#10;\\\\\\&#10;\cfrac{3400}{1700}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)&#10;\\\\\\&#10;log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t
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