Answers
Given:-
The denominator of a fraction is one more than three times and if 8 is added to the neumerator and the demominator then the solution is 1/2.
To find the equation.
So now the given fraction is,
[tex]\frac{x}{1+3x}[/tex]So now we add. so we get,
[tex]\frac{x+8}{1+3x+8}=\frac{1}{2}[/tex]So now we get,
[tex]\frac{x+8}{3x+9}=\frac{1}{2}[/tex]So this is the required solution.
Related Questions
The teacher awards 128 points to each student at the begging of the quarter. Half of the bonus points are removed each time a student does not complete a homework assignment on time. Write a rule for the number of points left after x homework assignments not done on time. Y=
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If we represent the number of points left after x homework by x
Then if there are initially 128 points
Then for every question missed, we will obtain half of the initial points, which will be
[tex]\frac{1}{2}\text{ x 128 x X}[/tex]=> 64x
Then the
Graph the equation after rewriting it on in slope intercept form. 2y-4x=3
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The slope intercept form is defined as
[tex]\begin{gathered} y=mx+b \\ \text{where} \\ m\text{ is the slope} \\ b\text{ is the y-intercept} \end{gathered}[/tex]Rearrange the given equation by transferring the x term to the left side and dividing both sides by 2.
[tex]\begin{gathered} 2y-4x=3 \\ 2y=4x+3 \\ \frac{2y}{2}=\frac{4x+3}{2} \\ y=2x+\frac{3}{2} \\ \\ \text{Therefore, the slope intercept form of the equation }2y-4x=3\text{ is} \\ y=2x+\frac{3}{2} \end{gathered}[/tex]Now that we have the slope intercept form, get at least any of the two points of the equation by substituting any two values of x, for this case, we will use x = 0, and x = 1
[tex]\begin{gathered} \text{If }x=0 \\ y=2x+\frac{3}{2} \\ y=2(0)+\frac{3}{2} \\ y=\frac{3}{2} \\ y=1.5\rightarrow(0,1.5)\rightarrow\text{first point} \\ \\ \text{If }x=1 \\ y=2x+\frac{3}{2} \\ y=2(1)+\frac{3}{2} \\ y=2+\frac{3}{2} \\ y=\frac{7}{2} \\ y=3.5\rightarrow(1,3.5)\rightarrow\text{second point} \end{gathered}[/tex]Now that we have two coordinates (0, 1.5) and (1, 3.5), plot the two points in the graph, and then connect a line between them. This will result with the graph.
A polygon has a perimeter of 18 units. It is dilated with a scale factor of . What is theperimeter of its image?
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The new perimeter is as follows:
[tex]P=18\cdot\frac{3}{2}=9\cdot3=27[/tex]We multiply the former perimeter by the scale factor. Thus, it should be 27 units.
Find the value of Angle S and round it off to the nearest Tenth.
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From the figure, the only information given is the measure of its Hypotenuse and the Adjacent Side. The triangle also has 90 degrees interior angle, thus, this triangle is a right triangle and we can apply this Trigonometric Function:
[tex]\text{Cosine(}\Theta)\text{ = Cosine (}\angle\text{S) = }\frac{Adjacent\text{ Side}}{Hypotenuse}[/tex]Given the following information:
Hypotenuse = 7
Adjacent Side = 6
Let's now find the value of Angle S.
[tex]\text{Cosine(}\Theta)\text{ = Cosine(}\angle S)=\frac{Adjacent\text{ Side}}{Hypotenuse}\text{ }\rightarrow\text{ Cosine(}\angle S)\text{ = }\frac{6}{7}[/tex][tex]\text{ }\angle S\text{ = }\cos ^{-1}(\frac{6}{7})[/tex][tex]\angle S=31.002719^{\circ}[/tex]Rounding it to the nearest Tenth, we get:
[tex]\angle S=31.002719^{\circ}=31.0^{\circ}[/tex]1) The given line markings show how the roads in a town are related to each other. A) Name a pair of parallel lines. B) Name a pair of perpendicular lines. C) Is ST I NV?
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The given figure is;
Parallel Lines : parallel lines are lines in a point which do not meet
In the given figure parallel lines are;
WX and YZ
UV and QR
A) WX || YZ
B)
Perpendicular lines : lines that intersect at a right (90 degrees) angle.
In the given figure, perpendicualr lines are;
Line ST and UV
B) ST ⊥ UV
C)
Since it is given that SN is perpendicular to the line UV
Then sum of all angle at a point on a line is equal to 180 degree
Thus, Angle SNV + Angle SNU = 180
Angle SNV + 90 = 180
Angle SNV = 180- 90
Angle SNV = 90
Thus, ST ⊥ NV
...
I’m working on review questions to prepare for test and need help with this.
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SOLUTION
Step1: write out the expression
[tex]\sqrt[]{125x}[/tex]we have to write this expression in its simplest form to identify the value of A and B
Step2: Identify the perfect square and write has a product
[tex]125=25\times5[/tex]Step3: replace the product above with 125 in the expression
[tex]\sqrt[]{125x}=\sqrt[]{25\times5\times x}[/tex]Step4: simplify the last expression by applying the rational rule below
[tex]\sqrt[]{a\times b}=\sqrt[]{a}\times\sqrt[]{b}[/tex]Hence we have
[tex]\sqrt[]{25\times5\times x}=\sqrt[]{25}\times\sqrt[]{5}\times\sqrt[]{x}[/tex][tex]\sqrt[]{25}\times\sqrt[]{5}\times\sqrt[]{x}=5\times\sqrt[]{5}\times\sqrt[]{x}=5\sqrt[]{5x}[/tex]Hence
[tex]\sqrt[]{125x}=5\sqrt[]{5x}[/tex]Therefore
A=5 and B=5x
Find the degree measure of the angle or the intercepted arc indicated in each figure. I’m struggling with number 8 :(
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We are going to use the following theorem to solve the problem we have
The first thing we can say is that the angle of mGFH is equal to:
[tex]m\angle GFH=50[/tex]Using the theorem shown above we can determine the angle for c
[tex]m\angle GEH=25[/tex]Arc in part a
[tex]mEG=155[/tex]Arc in part b
[tex]mEHG=205[/tex]A. Unfortunately, the precise data used by Eratosthenes was lost long ago. However, if Eratosthenes used a meter stick for his experiment today, then the stick’s shadow in Alexandria would be 127 mm long. Determine the angle 0 that the sunrays made with the meter stick. Remember that a meter stick is 1000 millimetres long.B. Assuming that the sun’s rays are essentially parallel, determine the central angle of the circle if the angle passes through Alexandria and Syene. How did you find your answer?
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A
Answer:
Explanation:
The right triangle formed is shown below.
To find θ, we would apply the tangent trigonometric ratio which is expressed as
tan θ = opposite side /adjacent side
From the triangle,
opposite side = 127
adjacent side = 1000
tanθ = 127/1000 = 0.127
taking the tan inverse of 0.127
θ = tan^1(0.127)
θ = 7 degrees
Hello! High school student in Calculus here. I need help solving the problem attached in the image. I'd like help FINDING THE DERIVATIVE using: - PRODUCT RULE- QUOTIENT RULEIf someone could help break down the steps and explain how to solve the problem using BOTH rule methods (because my teacher advised us to learn how to solve both ways), I would greatly appreciate it!
Answers
The function we have is:
[tex]h(\alpha)=\alpha^2\tan \alpha[/tex]Let's find the derivative using the product rule and the quotient rule.
-Using the product rule.
This is the better method for finding the derivative of this function because the function is already a product between two functions, which we will call f(alpha) and g(alpha):
[tex]\begin{gathered} f(\alpha)=\alpha^2 \\ g(\alpha)=\tan \alpha \end{gathered}[/tex]The product rule we will use is as follows:
[tex]h^{\prime}(\alpha)=f(\alpha)\times g^{\prime}(\alpha)+f^{\prime}(\alpha)\times g(\alpha)[/tex]Where the ' means derivative.
Let's first calculate the derivative of function f and g:
[tex]f(\alpha)=a^2\longrightarrow f^{\prime}(\alpha)=2\alpha[/tex]here we have used the following rule of derivation:
[tex]f(x)=x^{n^{}}\longrightarrow f^{\prime}(x)=nx^{n-1}[/tex]Now we find the derivative of g:
[tex]g(x)=\tan \alpha\longrightarrow g^{\prime}(\alpha)=\sec ^2\alpha[/tex]This is a derivate that we can find tables of derivation.
Now, we apply the product rule mentioned earlier:
[tex]h^{\prime}(\alpha)=f(\alpha)\times g^{\prime}(\alpha)+f^{\prime}(\alpha)\times g(\alpha)[/tex]And substitute all of the known values:
[tex]h^{\prime}(\alpha)=\alpha^2\sec ^2\alpha+2a\tan \alpha[/tex]That is our final result.
We can also calculate this using the quotient rule.
As we mentioned at the beginning, since there is a multiplication it is better to use the product rule to find the derivative, but we can also find it by using the quotient rule.
To use the quotient rule there need to be a division in the function, so we modify the original function:
[tex]h(\alpha)=\alpha^2\tan \alpha[/tex]And express it as a division as follows:
[tex]h(\alpha)=\frac{\alpha^2\sin\alpha}{\cos\alpha}[/tex]This is because
[tex]\tan \alpha=\frac{\sin \alpha}{\cos \alpha}[/tex]To use the quotient rule, we need to take or function:
[tex]h(\alpha)=\frac{\alpha^2\sin\alpha}{\cos\alpha}[/tex]And define one function f to be the numerator, and another function g to be the denominator:
[tex]f(\alpha)=\alpha^2\sin \alpha[/tex][tex]g(\alpha)=\cos \alpha[/tex]And the quotient rule is:
[tex]h^{\prime}(\alpha)=\frac{g(\alpha)f^{\prime}(\alpha)-f(\alpha)g^{\prime}(\alpha)}{g(\alpha)^2}[/tex]Before we find this, we need the derivatives of g and f:
[tex]g(\alpha)=\cos \alpha\longrightarrow g^{\prime}(\alpha)=-\sin \alpha[/tex]And for the derivative of f we apply the product rule because there is a multiplication:
[tex]f(\alpha)=a^2\sin \alpha\longrightarrow f^{\prime}(\alpha)=a^2(\sin \alpha)^{\prime}+(a^2)^{\prime}\sin \alpha[/tex]Solving the derivatives of this product rule:
[tex]f^{\prime}(\alpha)=a^2\cos \alpha+2a\sin \alpha[/tex]And now we are ready to use the quotient rule:
[tex]h^{\prime}(\alpha)=\frac{g(\alpha)f^{\prime}(\alpha)-f(\alpha)g^{\prime}(\alpha)}{g(\alpha)^2}[/tex]Substituting the known values:
[tex]h^{\prime}(\alpha)=\frac{\cos \alpha(a^2\cos \alpha+2a\sin \alpha)-a^2\sin \alpha(-\sin \alpha)}{\cos ^2\alpha}[/tex]Simplifying the expression:
[tex]h^{\prime}(\alpha)=\frac{a^2\cos^2\alpha+2a\sin\alpha\cos\alpha+a^2\sin^2\alpha}{\cos^2\alpha}[/tex]We factor the terms that contain a^2:
[tex]h^{\prime}(\alpha)=\frac{a^2(\cos ^2\alpha+\sin ^2\alpha)+2a\sin \alpha\cos \alpha}{\cos ^2\alpha}[/tex]Here, we use the following property of the cosine and the sine:
[tex]\cos ^2\alpha+\sin ^2\alpha=1[/tex]And we simplify again:
[tex]\begin{gathered} h^{\prime}(\alpha)=\frac{a^2+2a\sin \alpha\cos \alpha}{\cos ^2\alpha} \\ h^{\prime}(\alpha)=\frac{a^2}{\cos^2\alpha}+\frac{2a\sin \alpha\cos \alpha}{\cos ^2\alpha} \end{gathered}[/tex]We simplify further using:
[tex]\frac{1}{\cos^2\alpha}=\sec ^2\alpha[/tex][tex]h^{\prime}(\alpha)=a^2\sec ^2\alpha+\frac{2a\sin\alpha}{\cos\alpha}[/tex]which is equal to:
[tex]h^{\prime}(\alpha)=a^2\sec ^2\alpha+2a\tan \alpha[/tex]Answer:
[tex]h^{\prime}(\alpha)=a^2\sec ^2\alpha+2a\tan \alpha[/tex]1. How would you write the logarithm of x base 3 equations 27 in a formula?2. How much money would you have in the bank if you deposited $300 at an interest rate of 4% after 1 yearcompounded continuously?Hint: P-initial amount, r=rate in decimal, t=number of yearsP(t) = Poe^rt
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Continuous Interest
When an investment is made at continuous interest, the amount of money available is calculated with the formula:
[tex]P(t)=P_o\cdot e^{r\cdot t}[/tex]Where Po is the initial amount, r=interest rate, t= number of years.
The data given in the problem is:
Po= $300
r = 4% = 0.04
t = 1 year
Substituting:
[tex]P(1)=300\cdot e^{0.04\cdot1}=300\cdot1.0408=312.24[/tex]You would have $312.24 in the bank
Statement: Using the figure, determine the following ratios. ∆RTS is a right triangle with a right angle at
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By definition:
[tex]\text{tan(angle)}=\frac{\text{opposite side}}{adjacent\text{ side}}[/tex]From the picture:
[tex]\begin{gathered} \tan (S)=\frac{80}{18}=4.4444 \\ \tan (R)=\frac{18}{80}=0.225 \end{gathered}[/tex]If f(x) = x^2 + 3x, find f(-3). *
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Answer:
The value of f(-3) is;
[tex]f(-3)=0[/tex]Explanation:
Given that;
[tex]f(x)=x^2+3x[/tex]To get f(-3), we will substitute -3 for x in the given function f(x);
[tex]\begin{gathered} f(x)=x^2+3x \\ f(-3)=(-3)^2+3(-3) \\ f(-3)=9^{}-9 \\ f\mleft(-3\mright)=0 \end{gathered}[/tex]Therefore, the value of f(-3) is;
[tex]f(-3)=0[/tex]The perimeter of a rectangle measuring (2x + 2)cm by (3x -3) cm is 58cm. Calculate its area. 13. The perimeter
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Given:
The length of the given rectangle is l =(2x+2) cm.
The width of the given rectangle is w =(3x-3) cm.
The perimeter of the given rectangle is P =58cm.
Required:
We need to find the area of the rectangle.
Explanation:
Consider the perimeter of the rectangle formula.
[tex]P=2(l+w)[/tex]Substitute l=2x+2, w =3x-3 and P=58 in the formula.
[tex]58=2(2x+2+3x-3)[/tex][tex]58=2(5x-1)[/tex][tex]58=10x-2[/tex]Add 2 to both sides of the equation.
[tex]58+2=10x-2+2[/tex][tex]60=10x[/tex]Divide both sides by 10.
[tex]\frac{60}{10}=\frac{10x}{10}[/tex][tex]x=6[/tex]Substitute x =6 in the equations l=2x+2.
[tex]l=2x+2=2(6)+2=12+2=14cm[/tex][tex]Substitute\text{ }x=6\text{ }in\text{ }the\text{ }equations\text{ }w=3x-3.[/tex][tex]w=3x-3=3(6)-3=18-3=15cm[/tex]The area of the rectangle is
[tex]A=lw[/tex]Substitute l=14cm and w =15cm in the formula.
[tex]A=14\times15[/tex][tex]A=210cm^2[/tex]Final answer:
The area of the given rectangle is 210 square cm.
Carlos is putting a rectangular swimming pool in his back yard. The length of the pool is `12` feet `6` inches. The width is `30` feet `4` inches. He wants to put in a sidewalk around the pool that is `2` feet wide.What is the length in feet only? In other words, convert `12` feet `6` inches to a decimal number, rounded to the nearest hundredth.What is the width in feet only? In other words, convert 30 feet 4 inches to a decimal number, rounded to the nearest hundredth.Use the feet dimensions to find the total area of the sidewalk. Round to the nearest hundredth.
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Remember that 1 foot is equal to 12 inches.
a)
To find the length in feet only, convert 6 inches to feet and add that amount to the length of 12 feet:
[tex]\begin{gathered} 6in=6in\times\frac{1ft}{12in}=0.5ft \\ \\ 12ft+6in=12ft+0.5ft=12.5ft \end{gathered}[/tex]Then, the length of the swimming pool in feet is 12.50.
b)
To find the width in feet only, convert 4 inches to feet and and that amount to the width of 30 feet:
[tex]\begin{gathered} 4in=4in\times\frac{1ft}{12in}=0.333...ft\approx0.33ft \\ \\ 30ft+4in=30ft+0.33ft=30.33ft \end{gathered}[/tex]Then, the wudth of the swimming pool in feet (to the nearest hundredth) is 30.33.
c)
Draw a diagram of the pool with the sidewalk to visualize the situation:
The width and the length of the larger rectangle that includes the sidewalk are 4ft larger than the dimensions of the swimming pool because the sidewalk is 2ft wide.
To find the area of the sidewalk, subtract the area of the smaller rectangle from the area of the larger rectangle:
[tex]\begin{gathered} A_{large}=(34.33ft)\times(16.50ft)=566.445ft^2 \\ A_{small}=(30.33ft)\times(12.50)=379.125ft^2 \\ \\ A_{sidewalk}=A_{large}-A_{small} \\ =566.445ft^2-379.125ft^2 \\ =187.32ft^2 \end{gathered}[/tex]Then, the area of the sidewalk is 187.32 square feet.
Can you pls help me with this question thank you
Answers
The given expression is:
[tex]16+\frac{r^2}{s^2}[/tex]When r=6 and s=2, we need to replace those values into the expression and solve:
[tex]16+\frac{6^2}{2^2}=16+\frac{36}{4}=16+9=25[/tex]The answer is a. 25
in circle C, BC = 6 and angle ACB = 120 degrees. what is the area of the shaded sector?
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Area of sector = θ/360 × πr²
θ = m∠ACB = 120°
radius = r = 6
[tex]\text{Area of sector = }\frac{120}{360}\times\pi\times6^2[/tex][tex]\begin{gathered} \text{Area = }\frac{1}{3}\times\pi\times36 \\ Area\text{ = }\pi\times\frac{36}{3}\text{ } \end{gathered}[/tex][tex]\begin{gathered} Area\text{= }\pi\times12 \\ \text{Area of sector = 12}\pi\text{ (option C)} \end{gathered}[/tex]In which direction does the left side of the graph of this function point?f(x) = 3x3 - x2 + 4x - 2Answer hereSUBMIT
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The given function is expressed as
f(x) = 3x^3 - x^2 + 4x - 2
This is a cubic function which also means that it is an odd function.
The graph of the function is shown below
Harrison saved $38.97 from his first paycheck and $65.04 from his second paycheck. How much did he save from the paychecks?A.$26.07B.$94.01C.$94.91D.$104.01
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Answer:
D.$104.01
Explanation:
Here is what we are told
Harrison got the following amount from his paychecks.
Paycheck 1: $38. 97
Paycheck 2: $65.04
Therefore total amount Harrison saved on his paychecks was
$38. 97 + $65.04 = $104.01
Now looking at the answer choices we see that choice D gives the correct answer.
Therefore choice D is the correct answer.
Answer:
it’s D sir =)
Step-by-step explanation:
Is 4/5 closer to 0, 1/2 or 1 and how?
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It is important to know that 4/5 = 0.8 and 1/2 = 0.5. So, 0.5 is closer to zero because it less than 1/2 or 1.
Hence, the answer is 1/2.The two rectangles below have the same height-to-width ratio. What is thevalue of w? (If necessary, round your answer to two decimal places.)21W=OA. 11.33OB. 17OC. 20OD. 4.3234
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Solution
- Since the height-to-width ratio of both rectangles are the same, we can solve the question as follows:
[tex]\begin{gathered} \frac{H}{W}=\frac{h}{w} \\ where, \\ H=height\text{ of big rectangle} \\ W=width\text{ of big rectangle} \\ h=height\text{ of small rectangle} \\ w=width\text{ of small rectangle} \\ \\ \text{ Thus, we have:} \\ \frac{21}{34}=\frac{7}{w} \\ \\ \text{ Make w the subject of the formula} \\ w=\frac{34\times7}{21} \\ \\ w=11.333\bar{3} \end{gathered}[/tex]Final Answer
OPTION A
List the values in desending order square root 170, 13.5, 64/5, 13 7/8
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Consider the given set of values,
[tex]\sqrt[]{170},13.5,\frac{64}{5},13\frac{7}{8}[/tex]First of all, we need to convert each value into decimal form,
[tex]\begin{gathered} \sqrt[]{170}\approx13.0384 \\ 13.5 \\ \frac{64}{5}\approx12.8 \\ 13\frac{7}{8}=\frac{(13\cdot8)+7}{8}=\frac{111}{8}\approx13.875 \end{gathered}[/tex]So the set of values in decimal form are,
[tex]13.0384,13.5,12.8,13.875[/tex]It is known that,
[tex]13.875>13.5>13.0384>12.8[/tex]The total cost of 3 1/2 pounds of meat at $1.69 per pound and 20 lemons at 60 cents per dozen will bea.$6.00b.$6.40c.$6.52d.$6.82e.$6.92
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Answer:
e. $6.92
Explanation:
The cost of 1 pound of meat = $1.69
[tex]\text{ The cost of }3\frac{1}{2}\text{ pounds of meat }=3.5\times1.69=\$5.92[/tex]Similarly, the cost per dozen of lemons = 60 cents
[tex]\begin{gathered} \text{ The cost of 1 lemon}=\frac{60}{12}=5\text{ cents} \\ \implies\text{ The cost of 20 lemons}=5\times20=100\text{ cents}=\$1 \end{gathered}[/tex]Add the two to find the total cost:
[tex]\begin{gathered} \text{ Total Cost}=5.92+1 \\ =\$6.92 \end{gathered}[/tex]The total cost of 3 1/2 pounds of meat at $1.69 per pound and 20 lemons at 60 cents per dozen will be $6.92.
Option E is correct.
six boys earned $58 selling magazines on Saturday they paid $16 for the magazines if they share the profit equally how much will each boy receive
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Answer:
The amount each boy receive is $7.
[tex]A=\text{ \$7}[/tex]Explanation:
Given;
six boys earned $58 selling magazines on Saturday they paid $16 for the magazines;
[tex]\begin{gathered} \text{number of boys = 6} \\ \text{ selling price of the magazines = \$58} \\ \text{ Cost price of the magazines = \$16} \end{gathered}[/tex]We know that;
[tex]\begin{gathered} \text{Profit = Selling price - Cost price} \\ \text{Profit = \$58 - \$16} \\ \text{Profit = \$42} \end{gathered}[/tex]Since the profit was shared equally among 6 boys the amount each boy will receive is;
[tex]\begin{gathered} A=\frac{\text{Profit}}{\text{6}}=\frac{\text{ \$42}}{6} \\ A=\text{ \$7} \end{gathered}[/tex]Therefore, the amount each boy receive is $7.
[tex]A=\text{ \$7}[/tex]Directions: for questions 1 through 4, find the diagonal length of each solid figure.
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Here, we want to find the diagonal of the given solid
To do this, we need the appropriate triangle
Firstly, we need the diagonal of the base
To get this, we use Pythagoras' theorem for the base
The other measures are 6 mm and 8 mm
According ro Pythagoras' ; the square of the hypotenuse equals the sum of the squares of the two other sides
Let us have the diagonal as l
Mathematically;
[tex]\begin{gathered} l^2=6^2+8^2 \\ l^2\text{ = 36 + 64} \\ l^2\text{ =100} \\ l\text{ = }\sqrt[]{100} \\ l\text{ = 10 mm} \end{gathered}[/tex]Now, to get the diagonal, we use the triangle with height 5 mm and the base being the hypotenuse we calculated above
Thus, we calculate this using the Pytthagoras' theorem as follows;
[tex]\begin{gathered} d^2=5^2+10^2 \\ d^2\text{ = 25 + 100} \\ d^2\text{ = 125} \\ d\text{ = }\sqrt[]{125} \\ d\text{ = }11.2\text{ mm} \end{gathered}[/tex]The two-way frequency table below shows how many students passed their driving test aswell as the number who took a preparatory class. Find the probability that a student passedthe test, given that he or she took the class.Passed82Failed33ClassNo class232133115o8211582159oNone of the other answers are correct
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Answer
Option B is correct.
The probability that a student passes the test given that he or she took the class = P(A | B) = (82/115)
Explanation
The probability that a student passes the test given that he or she took the class is called conditional probability.
In conditional probability, the probability that event A occurs given that event B has already occured is presented as P(A | B) and it is given mathematically as
P(A | B) = P(A n B) ÷ P(B)
For this question, we are asked to find the probability that a student passes the test given that he or she took the class
The probability that a student passes the test
= P(A) = (82 + 23) ÷ (82 + 33 + 23 + 21) = (105) ÷ (159) = (105/159) = (35/53) = 0.6604
The probability that a student took the class
= P(B) = (82 + 33) ÷ (82 + 33 + 23 + 21) = (115) ÷ (159) = (115/159) = 0.7233
The probability that a student passes the test and took the class
= P(A n B) = (82) ÷ (82 + 33 + 23 + 21) = 82 ÷ 159 = (82/159) = 0.5157
The probability that a student passes the test given that he or she took the class = P(A | B) = P(A n B) ÷ P(B) = (82/159) ÷ (115/159)
[tex]\begin{gathered} P(A|B)=\frac{82}{159}\div\frac{115}{159} \\ =\frac{82}{159}\times\frac{159}{115} \\ =\frac{82}{115} \end{gathered}[/tex]P(A | B) = (82/115)
Hope this Helps!!!
A cone has a volume of 367 cm and height h. Complete this table for volume of cylinders with the same radius but different heights. height (cm) h volume (cm) 367 2h 5h h 2 Write your answer in terms of 77, just like the first example 367. When typing your answer, use the word "pi" to represent the symbol 7. For example, 36 would be typed as 36pi
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20. Solve the equation, 3/5 * a = 1/4
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We have to solve the equation:
[tex]\begin{gathered} \frac{3}{5}a=\frac{1}{4} \\ a=\frac{1}{4}\cdot\frac{5}{3} \\ a=\frac{5}{12} \end{gathered}[/tex]We just divide both sides by 5/3 and multiply. One side is clear a, and the other side results in 5/12.
Answer: a=5/12 (Option C)
A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi samples at different stores in a major city. Construct a 98% confidence interval estimate of the mean amount of mercury in the population. 0.52 0.79 0.10 0.87 1.32 0.50 0.90 What is the confidence interval estimate of the population mean u?
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From the question,
n=7
c=98% =0.98
First we calculate the mean
[tex]\operatorname{mean}(\bar{x})=\frac{0.52+0.79+0.10+0.87+1.32+0.50+0.90}{7}[/tex][tex]=\frac{5}{7}[/tex][tex]\approx0.7143[/tex]if the height of a cine is 26 and the radius of the base is 10, find the volume of the cone to the nesrest tenth.
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Use the table of values to evaluate the expressions below.
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SOLUTION
(a)
f(g(0)) From the table
[tex]\begin{gathered} g(0),\text{ that is the value of g\lparen x\rparen when x = 0 is 6} \\ put\text{ into f\lparen x\rparen, f\lparen6\rparen, that is value of f\lparen x\rparen when x = 6 is 2} \end{gathered}[/tex]Hence f(g(0)) = 2
(b)
g(f(9)) from the table
[tex]\begin{gathered} f(9)=8 \\ g(8)=1,\text{ } \\ \end{gathered}[/tex]Hence g(f(9)) = 1
(c)
f(f(1))
[tex]\begin{gathered} f(1)=7 \\ f(7)=5 \end{gathered}[/tex]hence f(f(1)) = 5
(d)
g(g(2))
[tex]\begin{gathered} g(2)=4 \\ g(4)=2 \end{gathered}[/tex]Hence g(g(2)) = 2