// Numbas version: finer_feedback_settings {"name": "FY001 - 2021 Exam Extra time", "metadata": {"description": "

FY001 exam extra time

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This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve.

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a)

\\begin{align}
\\var{h}x+\\var{k}y&=\\var{m}\\text{,}\\\\
\\var{j}x-\\var{l}y&=\\var{n}\\text{.}\\\\
\\end{align}

To find the solution to these equations, we need to cancel one of the unknowns.

Notice that $\\var{h}x$ in the first equation can be multiplied by $\\var{j/h}$ to match $\\var{j}x$ in the second equation. This means that we will only have to manipulate the first equation and can leave the second equation as it is.

We have to multiply the entire first equation by $\\var{j/h}$, not just the $x$ term to ensure the equation still holds.

$\\var{h}x+\\var{k}y=\\var{m}$ multiplied by $\\var{j/h}$ gives $\\var{j}x+\\var{k*(j/h)}y=\\var{m*(j/h)}.$

We now have a common $x$ term and we can cancel this by subtracting one equation from the other to find the $y$ term.

\\begin{align}
&&\\var{j}x+\\var{k*{j/h}}y&=\\var{m*(j/h)}\\\\
-&&\\var{j}x-\\var{l}y&=\\var{n}\\\\
&&\\overline{\\qquad} & \\overline{\\qquad}\\\\
&&0x+\\var{k*(j/h)+l}y&=\\var{m*(j/h)-n}\\\\[1em]
&&y&=\\frac{\\var{m*j/h-n}}{\\var{k*j/h+l}}\\\\
&&y&=\\var{y1}
\\end{align}

We can find the corresponding value of $x$ by substituting this value for $y$ back into either of the original equations.

\\begin{align}
\\var{h}x+(\\var{k}\\times\\var{y1})&=\\var{m}\\text{,}\\\\
\\var{h}x+\\var{k*y1}&=\\var{m}\\text{,}\\\\
\\var{h}x&=\\var{m-(k*y1)}\\text{,}\\\\
x&=\\var{x1}\\text{.}\\\\
\\end{align}

Therefore, $x=\\var{x1}$ and $y=\\var{y1}$.

b)

\\begin{align}
\\var{a}x+\\var{b}y&=\\var{c}\\text{,}\\\\
\\var{d}x+\\var{f}y&=\\var{g}\\text{.}\\\\
\\end{align}

To be able to solve the equations, we need to cancel one of the unknowns by manipulating the two equations so that the variable we wish to cancel is of the same value in each equation.

Although we can choose to cancel either variable, $x$ or $y$, a good rule of thumb is to look at the lowest common multiples of the coefficients for each variable and cancel the variable with the lowest LCM.

The LCM of the coefficients of the $x$ terms is $\\var{lcm(a,d)}$.

The LCM of the coefficients of the $y$ terms is $\\var{lcm(b,f)}$.

Therefore, we will choose to cancel the $x$ terms.

We need to multiply the equations individually to achieve the lowest common multiple identified.

\\begin{align}
\\simplify{ {a}x + {b}y } &= \\var{c} &\\text{multiply by } \\var{lcm(a,d)/a} \\text { to obtain } && \\simplify{ {lcm(a,d)}x + {b*lcm(a,d)/a}y} &= \\var{c*lcm(a,d)/a} \\\\
\\simplify{ {d}x + {f}y } &= \\var{g} &\\text{multiply by } \\var{lcm(a,d)/d} \\text { to obtain } && \\simplify{ {lcm(a,d)}x + {b*lcm(a,d)/d}y} &= \\var{c*lcm(a,d)/d}
\\end{align}

We now have a common $x$ term, and can cancel this by subtracting one equation from the other.

\\begin{align}
&& \\simplify{ {lcm(a,d)}x+{b*lcm(a,d)/a}y } = \\var{c*lcm(a,d)/a} \\\\
- && \\simplify{ {lcm(a,d)}x + {f*lcm(a,d)/d}y } = \\var{g*lcm(a,d)/d} \\\\
&& \\overline{\\simplify[]{ 0x+{b*lcm(a,d)/a-f*lcm(a,d)/d}y} = \\var{c*lcm(a,d)/a-g*lcm(a,d)/d}}
\\end{align}

\\begin{align}
\\var{(b*lcm(a,d)/a)-(f*lcm(a,d)/d)}y &= \\var{(c*lcm(a,d)/a)-(g*lcm(a,d)/d)}\\text{,}\\\\
y &= \\var{y2}\\text{.}
\\end{align}

We can find the corresponding value of $x$ by substituting thsi value of $y$ value back into either of the original equations.

\\begin{align}
\\simplify[]{ {a}x + {b}{y2}} &= \\var{c} \\\\
\\simplify[]{ {a}x + {b*y2}} &= \\var{c} \\\\
\\var{a}x&=\\var{c-b*y2} \\\\
x &= \\var{x2} \\text{.}
\\end{align}

Therefore, $x=\\var{x2}$ and $y=\\var{y2}$.

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Solve this set of simultaneous equations and give your answers for $x$ and $y$ below.

\\begin{align}
\\simplify{{h}x+{k}y} &= \\var{m} \\text{,} \\\\
\\simplify{{j}x+{l}y} &= \\var{n} \\text{.}
\\end{align}

$x =$ [[0]]

$y =$ [[1]]

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Solve this set of simultaneous equations and give your answers for $x$ and $y$ below.

\\begin{align}
\\simplify{{a}x + {b}y} &= \\var{c} \\text{,} \\\\
\\simplify{{d}x + {f}y} &= \\var{g} \\text{.}
\\end{align}

$x =$ [[0]]

$y =$ [[1]]

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Constant part of the LHS of the second equation in part a

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RHS of the second equation in part a

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RHS of the first equation in part a

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Value of $x$ in part a

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Value of $y$ in part b

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Value of $x$ in part b

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Value of $x$ in part a

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Constant part of the LHS of the first equation in part a

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$x$ coefficient of the second equation in part a. An integer multiple of the $x$ coefficient of the second equation.

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Coefficient of $y$ in the first equation of part b.

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$y$ coefficient of the second equation in part b. Never an integer multiple of the $y$ coefficient in the first equation.

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$x$ coefficient of the first equation in part a

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$x$ coefficient in the second equation of part b. Never an integer multiple of the $x$ coefficient in the first equation.

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Given the real functions below, you should be able to determine their domains.

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Division is defined for all real numbers except zero. So when a function involves division (such as a rational function), any input that will result in a division by zero is not allowed. Rational functions are simply a fraction/division of polynomials, so the only real numbers that are not in the domain of a rational function are the roots of the polynomial in the denominator. Sometimes, (for example part c above) you will need to factorise the denominator to determine its roots.

d) Even though 'something divided by itself is 1' division by zero is still undefined. So the domain of $h$ is not all of $\\mathbb{R}$, the domain does not include the number $\\var{c[0]}$. In other words, $h(\\var{c[0]})$ is undefined but for all other $r$, $h(r)=1$.

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There is a single real number that is not in the domain of the function

\\[f(x)=\\frac{1}{x}.\\]

That number is [[0]].

There are [[0]] real numbers that are not in the domain of {rat} these are [[1]].

Note: If the numbers were $-2,1$ and $4$ you would enter set(-2,1,4)

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Which of the following represents the domain of \\[\\simplify{g(x)=(x^2-{c[0]+c[1]}x+{c[0]*c[1]})/(x^2-{a+b}x+{a*b})} \\ ?\\]

(There could be more than one correct choice).

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$\\mathbb{R}$

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$\\{x\\in \\mathbb{R}:\\,\\var{a+b}<x<\\var{a*b}\\}$

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$\\{x\\in \\mathbb{R}:\\,x\\ne \\var{c[0]},\\var{c[1]}\\}$

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$\\{x\\in \\mathbb{R}:\\,x\\ne \\var{a},\\var{b}\\}$

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$\\mathbb{R}\\setminus \\{\\var{a},\\var{b}\\}$

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$\\mathbb{R}\\setminus \\{\\var{c[0]},\\var{c[1]}\\}$

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$\\{x\\in \\mathbb{R}:\\,x\\ne \\var{-a},\\var{-b}\\}$

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Which of the following is the domain of

\\[h(r)=\\simplify{(r-{c[0]})/(r-{c[0]})}?\\]

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$\\{r\\in\\mathbb{R}:\\,r\\ne \\var{c[0]}\\}$

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$\\mathbb{R}$

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a)Domain of $f(x)$ is all real numbers except $\\frac{-\\var{d}}{\\var{c}}$.

Let $y = \\frac{\\var{a}x \\ +\\ \\var{b}}{\\var{c}x\\ +\\ \\var{d}}$. Then

$y\\var{c}x \\ +\\ y\\var{d} = \\var{a}x\\ +\\ \\var{b}$

$y\\var{c}x -\\var{a}x\\ = -y\\var{d} +\\var{b}$

$x(y\\var{c} -\\var{a}) = -y\\var{d} +\\var{b}$

$x = \\frac{-\\var{d}y +\\var{b}}{\\var{c}y -\\var{a}}$

Hence $f^{-1}(x) = \\frac{-\\var{d}x +\\var{b}}{\\var{c}x -\\var{a}}$

Domain of $f^{-1}(x)$ is all real numbers except $\\frac{\\var{a}}{\\var{c}}$

b) Domain of $f(x)$ is all real numbers except $\\frac{\\var{k}}{\\var{q}}$.

Let $y = \\frac{\\var{s}x \\ -\\ \\var{t}}{\\var{q}x\\ -\\ \\var{k}}$. Then

$y\\var{q}x \\ -\\ y\\var{k} = \\var{s}x\\ -\\ \\var{t}$

$y\\var{q}x -\\var{s}x\\ = y\\var{k} -\\var{t}$

$x(y\\var{q} -\\var{s}) = y\\var{k} -\\var{t}$

$x = \\frac{\\var{k}y -\\var{t}}{\\var{q}y -\\var{s}}$

Hence $f^{-1}(x) = \\frac{\\var{k}x -\\var{t}}{\\var{q}x -\\var{s}}$

Domain of $f^{-1}(x)$ is all real numbers except $\\frac{\\var{s}}{\\var{q}}$

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Let $f(x) = \\frac{\\var{a}x \\ +\\ \\var{b}}{\\var{c}x\\ +\\ \\var{d}}$. Find the natural domain of $f$, $f^{-1}$ and the natural domain of $f^{-1}$.

Domain of $f$ is all real numbers except [[1]]

$f^{-1}(x) =$ [[0]]

Domain of $f^{-1}$ is all real numbers except [[2]]

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Let $f(x) = \\frac{\\var{s}x \\ -\\ \\var{t}}{\\var{q}x\\ -\\ \\var{k}}$. Find the natural domain of $f$, $f^{-1}$ and the natural domain of $f^{-1}$.

Domain of $f$ is all real numbers except [[0]]

$f^{-1} =$ [[1]]

Domain of $f^{-1}$ is all real numbers except [[2]]

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Students are given an exponential equation and asked to evaluate it at two points.

The constants in the exponential are randomised.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

A population of bacteria has a population, $P$, that can be described by the function $P = \\var{k} \\times \\var{a}^t $, where $t$ is the time in {timeperiod}.

", "advice": "

a) When $t=0$,

$ P = \\var{k} \\times \\var{a}^t = \\var{k} \\times \\var{a}^0 = 1$

b) When $t = \\var{time}$,

$P = \\var{k} \\times \\var{a}^\\var{time} = \\var{answer_raw}$

When this is rounded to the nearest whole number, we get

$P = \\var{answer}$

c) We need to take $t = -2$,

$P = \\var{k} \\times \\var{a}^{-2} = \\var{ansc}$,

When this is rounded to the nearest whole number, we get

$P = \\var{ansc_r}$.

d) We need to solve the following equation:

\\[\\var{l} = \\var{k} \\times \\var{a}^t\\]

for $t$. Take the appropriate logarithm on both sides

\\[log_{\\var{a}}(\\var{l}/\\var{k}) = \\var{ansd}\\]

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What is the initial population, $P$, when $t$ = 0?

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What is the population, $P$, when $t = \\var{time}$ {timeperiod}

Round your answer to the nearest whole number.

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What was the population, $P$, 2 {timeperiod} ago

Round your answer to the nearest whole number.

[[0]]

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Is the value of the truck exponential growth or decay?

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What is the original value of the truck?

$\\$$[[0]]

What is the growth or decay rate as a percent?

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What is the value of the truck $\\var{tDepreciate}$ years after it was built?

$\\$$[[0]]

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The value, $y$, of a truck can be modeled by $ y = \\var{A}(\\var{dFactor})^t $, where $t$ is the number of years since the truck was built.

", "rulesets": {}, "variable_groups": [], "metadata": {"licence": "All rights reserved", "description": ""}, "type": "question"}, {"name": "Ugur's copy of Using the Logarithm Equivalence $\\log_ba=c \\Longleftrightarrow a=b^c$", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}, {"name": "Ugur Efem", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/14200/"}], "tags": [], "metadata": {"description": "

Rearrange some expressions involving logarithms by applying the relation $\\log_b(a) = c \\iff a = b^c$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

a)

We can rearrange logarithms using indices.

\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]

Using this equivalence we can rewrite $\\log_\\var{f}x=\\var{f1}$.

\\[\\begin{align}
x&= \\var{f}^\\var{f1} \\\\
&=\\var{f^f1}
\\end{align}\\]

b)

We can use the equivalence to rewrite our equation.

\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]

We can write out our values to makes it easier.

\\[\\begin{align}
a&=x \\\\
b&=\\var{g1}\\\\
c&=y+\\var{g2}
\\end{align}\\]

Then we can write out our equation in the required form.

\\[x=\\var{g1}^{y+\\var{g2}}\\]

c)

We can use the same equivalence as in part b).

\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]

We have

\\begin{align}
a&=y+\\var{h1} \\\\
b&=x\\\\
c&=\\var{h2}\\text{.} \\\\ \\\\
\\log_{x}(y+\\var{h1}) &= \\var{h2} \\\\
\\implies y+\\var{h1} &= x^{\\var{h2}} \\\\
x &= (y+\\var{h1})^{\\frac{1}{\\var{h2}}}
\\end{align}

d)

The two in this list that don't equal $x$ are $\\log_e(x)$ and $\\log_{10}(x)$.

\\[\\begin{align}
\\log_e(x)&=\\ln(x)\\\\
\\log_{10}(x)&=\\log(x)\\text{.}
\\end{align}\\]

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Rearrange the equation to find $x$.

$\\log_\\var{f}(x)=\\var{f1}$

$x=$ [[0]]

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Make $x$ the subject of the following equation.

$\\log_\\var{g1}(x)=y+\\var{g2}$

$x=$ [[0]]

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Make $x$ the subject of the equation, leaving your answer in the form $a^{\\frac{1}{b}}$.

$\\log_x(y+\\var{h1})=\\var{h2}$

$x=$ [[0]]

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Which of the following expressions are equivalent to $x$?

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$\\log_a(a^x)$

", "

$a^{\\log_a(x)}$

", "

$e^{\\ln(x)}$

", "

$\\log_{10}(x)$

", "

$\\log_e(x)$

", "

$\\ln(e^x)$

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Given a sum of logs, all numbers are integers,

\n \t\t

$\\log_b(a_1)+\\alpha\\log_b(a_2)+\\beta\\log_b(a_3)$ write as $\\log_b(a)$ for some fraction $a$.

\n \t\t

Also calculate to 3 decimal places $\\log_b(a)$.

\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Answer the following question on logarithms.

", "advice": "\n

The rules for combining logs are

\\[\\begin{eqnarray*}&1.& \\log_b(ac)&=&\\log_b(a)+\\log_b(c)\\\\ \\\\ &2.& \\log_b\\left(\\frac{a}{c}\\right)&=&\\log_b(a)-\\log_b(c)\\\\ \\\\ &3.& \\log_b(a^r)&=&r\\log_b(a) \\end{eqnarray*} \\]

We see that:

\\[\\begin{eqnarray*}\\log_{\\var{b}}(\\var{a_1})-\\var{r_1}\\log_{\\var{b}}(\\var{a_2})+\\var{r_2}\\log_{\\var{b}}(\\var{a_3})&=&\\log_{\\var{b}}(\\var{a_1})-\\log_{\\var{b}}(\\var{a_2}^{\\var{r_1}})+\\log_{\\var{b}}(\\var{a_3}^{\\var{r_2}})\\mbox{ using 3.}\\\\&=&\\log_{\\var{b}}(\\var{a_1})-\\log_{\\var{b}}(\\var{a_2^r_1})+\\log_{\\var{b}}(\\var{a_3^r_2})\\\\&=&\\log_{\\var{b}}(\\var{a_1}\\times \\var{a_3^r_2})-\\log_{\\var{b}}(\\var{a_2^r_1}) \\mbox{ using 1.}\\\\&=&\\log_{\\var{b}}\\left(\\frac{\\var{a_1}\\times \\var{a_3^r_2}}{\\var{a_2^r_1}}\\right) \\mbox{ using 2.}\\\\&=&\\log_{\\var{b}}\\left(\\frac{\\var{a_1*a_3^r_2}}{\\var{a_2^r_1}}\\right)\\\\&=&\\log_{\\var{b}}\\left(\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\\right)\\mbox{ on cancelling common factors}.\\end{eqnarray*}\\]

Hence $\\displaystyle c=\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}$.

To calculate $\\displaystyle \\log_{\\var{b}}\\left(\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\\right)$ to 4 decimal places we use the fact that for any positive base $b$:

\\[\\log_b(c)=\\frac{\\ln(c)}{\\ln(b)}=\\frac{\\log_{10}(c)}{\\log_{10}(b)}\\]

and we can use either of the log functions, $\\ln$ or $\\log_{10}$ on our calculators to find the value.

Using $\\ln$ we find:

\\[ \\log_{\\var{b}}\\left(\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\\right)=\\frac{\\ln\\left(\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\\right)}{\\ln(\\var{b})}=\\var{ans}\\] to 4 decimal places.

\n ", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"c": {"name": "c", "group": "Ungrouped variables", "definition": "a_1*a_3^r_2/a_2^r_1", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "templateType": "anything", "can_override": false}, "a_3": {"name": "a_3", "group": "Ungrouped variables", "definition": "random(2..8)", "description": "", "templateType": "anything", "can_override": false}, "a_2": {"name": "a_2", "group": "Ungrouped variables", "definition": "random(2,4,8)*random(3,9)", "description": "", "templateType": "anything", "can_override": false}, "a_1": {"name": "a_1", "group": "Ungrouped variables", "definition": "random(5..20)", "description": "", "templateType": "anything", "can_override": false}, "r_1": {"name": "r_1", "group": "Ungrouped variables", "definition": "random(2..3)", "description": "", "templateType": "anything", "can_override": false}, "r_2": {"name": "r_2", "group": "Ungrouped variables", "definition": "random(2..3)", "description": "", "templateType": "anything", "can_override": false}, "tol": {"name": "tol", "group": "Ungrouped variables", "definition": "0.0001", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Ungrouped variables", "definition": "precround(log(c)/log(b),4)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["c", "b", "a_3", "a_2", "a_1", "r_1", "r_2", "tol", "ans"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find a fraction or integer $c$ such that:

$\\log_{\\var{b}}(\\var{a_1})-\\var{r_1}\\log_{\\var{b}}(\\var{a_2})+\\var{r_2}\\log_{\\var{b}}(\\var{a_3})=\\log_{\\var{b}}(c)$

$c=\\;$[[0]].

Input $c$ as an integer or as a fraction and not as a decimal.

Now calculate $\\log_{\\var{b}}(c)$ to 4 decimal places:

$\\log_{\\var{b}}(c)=\\;$[[1]].

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Given the first few terms of an arithmetic sequence, write down its formula, then find a couple of particular terms.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

In this question, consider the sequence

\\[ a = \\var{a1}, \\; \\var{a1+d}, \\; \\var{a1+d*2}, \\; \\var{a1+d*3}, \\; \\ldots \\]

A helpful person has drawn out a table of the terms so far.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\boldsymbol{n}$	$1$	$2$	$3$	$4$	$\\ldots$
$\\boldsymbol{a_n}$	$\\var{a1}$	$\\var{a1+d}$	$\\var{a1+2d}$	$\\var{a1+3d}$	$\\ldots$

", "advice": "

The formula for the $n^\\text{th}$ term, $a_n$, of an arithmetic sequence is

\\[ a_n=a_1+(n-1)d \\text{.} \\]

$a_1$ is the first term, and $d$ is the common difference between adjacent terms.

a)

In the given sequence, the common difference is $\\var{a1+d} - \\var{a1} = \\var{d}$, and the first term is $\\var{a1}$.

So, the formula for this sequence is

\\[ a_n = \\var{a1} + (n-1) \\times \\var{d} \\text{.} \\]

b)

\\[ a_\\var{small} = \\var{a1} + (\\var{small}-1) \\times \\var{d} = \\var{a1+(small-1)*d} \\text{.} \\]

c)

\\[ a_\\var{large} = \\var{a1} + (\\var{large}-1) \\times \\var{d} = \\var{a1+(large-1)*d} \\text{.} \\]

d)

\\[S_n = \\frac{n(a_1 + a_n)}{2}\\]

Plug in the values to get $S_{\\var{medium}} = \\frac{\\var{medium}(\\var{a1} + \\var{amed})}{2} = \\var{ansd} $

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A large index to compute

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A small index to compute

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The first term in the sequence

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Write out an expression for $a_n$, the $n^{\\text{th}}$ term of the sequence, in terms of $n$.

$a_n =$ [[0]]

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Find the $\\var{small}^{\\text{th}}$ term

$a_{\\var{small}} = $ [[0]]

Find the $\\var{large}^{\\text{th}}$ term

$a_{\\var{large}} = $[[0]]

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Find the sum of first $\\var{medium}$ terms.

$S_{\\var{medium}} =$ [[0]]

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A bag contains balls of three different colours. You're told how many there are of each, and asked the probability of picking a ball of a particular colour.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

a)For equally likely outcomes, you can calculate the probability of a particular event occurring by using the formula

$\\text{Probability of an event} = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}$.

We are told that the bag contains $\\var{red}$ red balls, $\\var{blue}$ blue balls and $\\var{green}$ green balls and that one ball is removed from the bag at random.

The total number of balls in the bag before the chosen ball is removed is

\\[\\var{red}+\\var{blue}+\\var{green} = \\var{total}.\\]

As the ball is being removed randomly from the bag, there is an equal probability of selecting any one of the $\\var{total}$ balls.

Therefore, the probability of the chosen ball being blue is

\\[
P(\\text{blue}) = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}} = \\displaystyle\\frac{\\var{blue}}{\\var{total}}
\\]

b) In total there are $\\var{red1} + n$ balls. Probability choosing one green ball is therefore $\\frac{n}{\\var{red1} + n}$.

We are given that $\\frac{n}{\\var{red1} + n} = \\frac{\\var{a}}{\\var{b}}$. Cross multiplication gives $\\var{b}n = \\var{a}\\times\\var{red1} + \\var{a}n$.

Simplfying yields $n = \\var{ansb}$

c) In total there are $\\var{red2}+\\var{blue2} +\\var{green}$ balls. Probability of picking a red ball is $\\frac{\\var{red2}}{\\var{red2}+\\var{blue2} +\\var{green}}$. Probability of picking a blue ball is $\\frac{\\var{blue2}}{\\var{red2}+\\var{blue2} +\\var{green}}$. Total probability is tehir sum:

\\[\\frac{\\var{red2}+\\var{blue2}}{\\var{red2}+\\var{blue2} +\\var{green}}.\\]

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number of red balls in part c

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number of green balls in part c.

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total number of balls in part c

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number of blue balls in part c

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A bag contains $\\var{red}$ red balls, $\\var{blue}$ blue balls and $\\var{green}$ green balls. One ball is chosen from the bag at random. What is the probability that the chosen ball will be blue?

[[0]]

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$\\displaystyle\\frac{\\var{red+green}}{\\var{total}}$

", "

$\\displaystyle\\frac{\\var{blue}}{\\var{green+red}}$

", "

$\\displaystyle\\frac{\\var{blue}}{\\var{total}}$

", "

$\\displaystyle\\frac{1}{\\var{blue}}$

", "

$\\displaystyle\\frac{1}{\\var{total}}$

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A bag contains $\\var{red1}$ red balls and $n$ green balls. Find $n$ is the probability of choosing a green ball is $\\frac{\\var{a}}{\\var{b}}$.

$n =$ [[0]]

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A bag contains $\\var{red2}$ red balls, $\\var{blue2}$ blue balls and $\\var{green}$ green balls. One ball is chosen from the bag at random. What is the probability that the chosen ball will be blue or red? Enter your answer as a fraction

[[0]]

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$\\text{mean}=\\;\\;$[[0]] (correct to two decimal places)

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$\\text{median}=\\;\\;$[[0]]

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$\\text{mode}=\\;\\;$[[0]]

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The fuel emissions (in g/km of CO2) of a sample of 7 diesel cars of similar type have been recorded as follows:

$\\var{a2}, \\var{a7}, \\var{a1}, \\var{a5}, \\var{a3}, \\var{a6}$ and $\\var{a4}$.

Calculate the mean, median and mode of these emissions.

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Topics covered are calculating the mean, median, mode and standard deviation.

rebelmaths

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Mean: Add up all the numbers and divide by the number of numbers.

Median: middle value

Mode: most common value

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