To find the probability of choosing a senior student, we have to divide the total number of seniors by the total number of students.
According to the problem, there are 102 seniors. Let's sum to find the total number of students.
[tex]165+145+114+102=526[/tex]Then, we divide these numbers.
[tex]P_{\text{senior}}=\frac{102}{526}=\frac{51}{263}\approx0.19[/tex]The probability of choosing a senior is 51/263 or 19%, approximately.Jimmy's school is selling tickets to a play. On the first day of ticket sales the school sold 12 senior citizentickets and 5 child tickets for a total of $173. The school took in $74 on the second day by selling 1 seniorcitizen ticket and 5 child tickets. Find the price of a senior citizen ticket and the price of a child ticket.
SOLUTION
Let s represent senior citizen tickets
and let c represent child ticket
On the first day the school sold 12 senior tickets and 5 child's tickets for $173.
This will be represented algebraically by 12s + 5c = 173 ..... equation 1
The following day they sold 1 senior ticket and 5 child's tickets for $74.
This will be represented by s + 5c = 74 ....... equation 2
Now we will solve equations 1 and 2 simultaneously we have
[tex]undefined[/tex]Find the volume of the sphere. Round your answer to the nearest tenth.A) 2,289.1 m^3B) 3,052.1 m^3C) 24,416.6 m^3D) 12,437.4 m^3
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
sphere:
diameter = 18 m
Step 02:
geometry:
volume of the sphere:
v = 4/3 π r³
π = 3.14
r = d / 2 = 18 m / 2 = 9 m
v = 4/3 π (9 m)³ = 3052.1 m³
The answer is:
3052.1 m³
Can someone help me with this geometry question I don’t know if I am right.
Let us find out if the given two triangles ABC and DEF are similar triangles or not.
Triangle ABC is a right-angled triangle so we can apply the Pythagorean theorem to find the missing side.
[tex]a^2+b^2=c^2[/tex]Where a and b are the shorter sides and c is the longest side (hypotenuse)
[tex]\begin{gathered} 20^2+21^2=c^2 \\ 400+441=c^2 \\ 841=c^2 \\ \sqrt[]{841}=c \\ 29=c \\ c=29 \end{gathered}[/tex]Similarly, we can apply the Pythagorean theorem to triangle DEF to find the missing side.
[tex]\begin{gathered} d^2+e^2=f^2 \\ 40^2+e^2=58^2 \\ e^2=58^2-40^2 \\ e^2=3364-1600 \\ e^2=1764 \\ e=\sqrt[]{1764} \\ e=42 \end{gathered}[/tex]Now, recall that two triangles are similar if the ratio of the corresponding sides is equal.
The corresponding sides are
AB = DE
BC = EF
AC = DF
[tex]\begin{gathered} \frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC} \\ \frac{40}{20}=\frac{42}{21}=\frac{58}{29} \\ \frac{2}{1}=\frac{2}{1}=\frac{2}{1} \end{gathered}[/tex]As you can see, the ratio of the corresponding sides of the two triangles is equal.
Hence, the triangles ABC and DEF are similar.
Please help if you can. I will only accept answers with work shown. Will give Brainliest.
Initial subscribers: 285
Increase rate : 75 % = 75/100 = 0.75 (decimal form)
years passed = 1994-1985 = 9 years
Apply the formula:
A = P (1 +r ) ^ t
Where:
A = number of cell phone subscribers after t years
P = initial suscribers
r= increase rate in decimal form
t= years
Replacing:
A = 285 (1 +0.75)^9 = 43,872
Please solve equation for maximum and minimum
Answer:
17^(1/11), which occurs at x=9
Step-by-step explanation:
To find the Absolute Extrema in a set of points, you need to evaluate (plug in) the endpoints, and maxima/minima of the equations and figure out the greatest and lowest ones.
1.) By using this method, the first step is to find the Relative Maximums/Minimums of these areas. We can do this by finding the derivative of the equation, and setting that equal to 0 and solving. [tex]\frac{d}{dx} (x^2-64)x^{\frac{1}{11}} = (x^2-64)^{-10/11} * 2x[/tex]. If we set this equal to 0, we will find that x = 0. Therefore, x=0 is a minimum. Since this point belongs to the interval of [-8, 9], we can use it.
2.) Plug the endpoints of the interval and the result from our calculations. If we do this, we get f(-8)=0, f(9) = 17^(1/11), f(0)=0
3.) Since we are finding the Maxima, we look for the greatest value, which is 17^(1/11), which occurs at x=9
Solve in R Sin (x/4) = √2/2
Trigonometric Equations
Solve for x in R:
[tex]sin\text{ }\frac{x}{4}=\frac{\sqrt{2}}{2}[/tex]There are two angles whose sine is the given value: 45° and 135°. We need to express them in radians:
[tex]45^o=45\text{ }\frac{\pi}{180}=\frac{\pi}{4}[/tex][tex]135^o=135\frac{\pi}{180}=\frac{3\pi}{4}[/tex]Thus, we have two solutions:
[tex]\begin{gathered} \frac{x}{4}=\frac{\pi}{4} \\ x=\pi \\ And: \\ \frac{x}{4}=\frac{3\pi}{4} \\ x=3\pi \end{gathered}[/tex]Both solutions point to the same terminal angle, so we only have one solution in the first rotation of the angle:
x = π
Since it's required to find the solution for all the real numbers, we must account for all the possible angles in any number of rotations clockwise or counterclockwise as follows:
x = π + 2kπ
Where k is an integer number. For example, for k = 1, we have the already-found solution above x = 3π
A training field is formed by Joining a rectangle and two semicircles, as shown below. The rectangle is 96 m long and 74 m wide.Find the area of the training fleld. Use the value 3.14 for it, and do not round your answer. Be sure to include the correct unit in your answer.
To calculate the area of the training field, the first step is to calculate the area of the rectangular portion. The formula for calculating the area of a rectangle is expressed as
Area = length x width
From the information given,
length = 96
width = 74
Area of rectangular portion = 96 x 74 = 7104
The two semicircles would add up to form a complete circle because a semicircle is half of a circle. The diameter of the circle would be the width of the rectangular portion. The formula for calculating the area of a circle is expressed as
Area = pi x radius^2
From the information given,
pi = 3.14
diameter = 74
radius = diameter/2 = 74/2 = 37
By substituting the values into the formula, we have
Area of the two semicircular portions = 3.14 x 37^2 = 4298.66
Area of the training field = area of rectangular portion + area of the two semicircular portions = 7104 + 4298.66
Area of the training field = 11402.66 m^2
4.
The value of a truck decreases exponentially since its purchase. The two points on the
graph shows the truck's initial
value and its value a decade afterward.
[6040,000)
a) Express the car's value, in dollars, as a function of time
d, in decades, since purchase.
(1 24,000)
b) Write an expression to represent the car's value 4 years
after purchase.
c) By what factor is the value of the car changing each year? Show your reasoning.
Answer:
a. v = 40 000 (3/ 5)^d
b. v = 40 000 (3/5)^(4/10)
c. 0.95
Explanation:
The exponential growth is modelled by
[tex]v=A(b)^d[/tex]We know that points (0, 40 000) and (1, 24 000) lie on the curve. This means, the above equation must be satsifed for v = 40 000 and d = 0. Putting v = 40 000 and d = 0 into the above equation gives
[tex]40\; 000=Ab^0[/tex][tex]40\; 000=A[/tex]Therefore, we have
[tex]v=40\; 000b^d[/tex]Similarly, from the second point (1, 24 000) we put v = 24 000 and d = 1 to get
[tex]24\; 000=40\; 000b^1[/tex][tex]24\; 000=40\; 000b^{}[/tex]dividing both sides by 40 000 gives
[tex]b=\frac{24\; 000}{40\; 000}[/tex][tex]b=\frac{3}{5}[/tex]Hence, our equation that models the situation is
[tex]\boxed{v=40\; 000(\frac{3}{5})^d\text{.}}[/tex]Part B.
Remember that the d in the equation we found in part A is decades. Since there are 10 years in a decade, we can write
t = 10d
or
d = t/10
Where t = number of years
Making the above substitution into our equation gives
[tex]v=40\; 000(\frac{3}{5})^{\frac{t}{10}}[/tex]Therefore, the car's value at t = 4 is
[tex]\boxed{v=40\; 000(\frac{3}{5})^{\frac{4}{10}}}[/tex]Part C:
The equation that gives the car's value after t years is
[tex]v=40\; 000(\frac{3}{5})^{\frac{t}{10}}[/tex]which using the exponent property that x^ab = (x^a)^b we can rewrite as
[tex]v=40\; 000\lbrack(\frac{3}{5})^{\frac{1}{10}}\rbrack^t[/tex]Since
[tex](\frac{3}{5})^{\frac{1}{10}}=0.95[/tex]Therefore, our equation becomes
[tex]v=40\; 000\lbrack0.95\rbrack^t[/tex]This tells us that the car's value is changing by a factor of 0.95 each year.
which operation is applied to 3 and ×+5 in the expression 3(x+5) over 0.2
In the expression " 3(x+5) over 0.2" the word "over" indicates that 0.2 is dividing the first term 3(x+5), you can write the calculation as follows:
[tex]\begin{gathered} \frac{3(x+5)}{0.2} \\ \cdot-\cdot or\cdot-\cdot \\ 3(x+5)\div0.2 \end{gathered}[/tex]The operation is a division.
find the perimeter of the given triangle. round to the nearest tenth
The perimeter of the triangle to the nearest tenth is 13.8 units.
How to find the perimeter of a triangle?The perimeter of a triangle is the sum of the whole sides of the triangle.
Therefore, let's find the perimeter of the triangle.
let's find the two missing sides of the triangle as follows;
tan 18 = opposite / adjacent
tan 18 = y / 5.8
y = 5.8 tan 18
y = 1.88453423815
y = 1.88
cos 18 = adjacent / hypotenuse
cos 18 = 5.8 / x
x = 5.8 / cos 18
x = 6.09848421123
x = 6.09
Therefore,
perimeter of the triangle = 5.8 + 1.9 + 6.1
perimeter of the triangle = 13.8 units
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The equation y = 40 + 3x represents the amount a company will pay to have stickers made, where x represents the item published and y represents the total cost in dollars in dollars. the equation y = 5x represents the company's income from selling the sticker, where y represents the money earned in dollars and x represents the number of items sold.5. At at one point the lines intersect?6. when will the company make a profit?
we have the equations
y=40+3x -----> blue line
and
y=5x -----> red line
Part 5
intersection point
Equate both equations
5x=40+3x
5x-3x=40
2x=40
x=20
Find the value of y
y=5(20)=100
the intersection point is (20,100)
Part 6
when will the company make a profit?
the company make a profit when 5x > 40+3x
Remember that
For x=20------> the profit is zero
so
the company make a profit when x>20
Verify
solve the inequality
5x > 40+3x
5x-3x > 40
2x > 40
x > 20 ----> is ok
the volume of a recrangular prism is 72 centimeters.the prism is 2 centimeters wide and 4 centimeters high. what is the length of the prism
The volume of a rectangular prism is given by:
[tex]V=w\cdot l\cdot h[/tex]We know that the volume is 72 cub cm, the width is 2 cm and the height is 4 cm. Plugging this value in the equation we have:
[tex]72=2\cdot l\cdot4[/tex]Solving the equation for l we have:
[tex]\begin{gathered} 72=2\cdot l\cdot4 \\ 72=8l \\ l=\frac{72}{8} \\ l=9 \end{gathered}[/tex]Therefore the lenght of the prism is 9 cm.
Translate this sentence into an equation.The sum of 21 and Mabel's score is 66.
The sum of 21 and Mabel's score is 66:
This means 21 was added to Mabel's score to give 66
let Mabel's score = m
21 + Mabel's score = 66
In the from equation:
[tex]\text{21 + m = 66}[/tex]Find the value of tan X rounded to the nearest hundredth, if necessary.
5
W
1
√26
X
The value of tan x is 5
We need to find the value of tan x
Tan is one of the trignometric functions and the range of a tan function varies from 0 ≤ tan x ≤ 2π
In the triangle vwx, the perpendicular of x is VW and the base is WX
tan x = perpendicular / base
Here, the perependicular is 5 and base is 1
tan x = 5/1
tan x = 5
Therefore, the value of tan x is 5
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A certain strain of bacteria is growing at a rate of 44% per hour, and with 2,000 bacteria initially, this event can be modeled by the equation B(t) = 2,000(1.44)t. With this fast growth rate, scientists want to know what the equivalent growth rate is per minute. Using rational exponents, what is an equivalent expression for this bacterial growth, expressed as a growth rate per minute?
The given equation for the growth rate per hour is:
[tex]B(t)=2,000(1.44)^t[/tex]Where t is the time in hours.
The equivalent growth rate per minute would be the equivalent in minutes for hours, then:
[tex]1\min \cdot\frac{t\text{ hours}}{60\min }=\frac{t}{60}[/tex]Where t is the time in minutes, then the answer is:
[tex]B(t)=2,000(1.44)^{\frac{t}{60}}[/tex]Identify how many solutions there are to the system of equations represented on the following graph. Treat the red and black graphs as one circle. H H
The solution of two or more equations is the point where the equation intersects. The more that the graph of the equation intersects, the more its solutions.
From the given graph of circle and parabola, they intersect three times. Therefore, there are 3 solutions in this given system of equations.
Ayana drew a scale drawing of a house and its lot. The backyard, which is 70 feet long in real life, is 203 inches long in the drawing. What scale did Ayana use for the drawing?29 inches : [ ] feet
Let m be scale used by individual for drawing.
Then the product of scale factor and original length is equal to the length in drawing. So,
[tex]\begin{gathered} 70\times m=203 \\ m=\frac{203}{70} \\ =\frac{29}{10} \end{gathered}[/tex]So, 29 inches of drawing is corresponding to 10 feet of house.
The length of a rectangle is given by the function l(x)=12x2+2x+4. The width of the rectangle is given by the function w(x)=3x−1.Which function represents the area of the rectangle?
The area of a rectangle is given by the next equation:
Area = Length * Width
Where
Length =1/2x²+2x+4.
and
Width = 3x−1
Replacing:
Area = (1/2x²+2x+4)* (3x-1)
Solve the operation:
(1/2x²* 3x) + (1/2x²*-1)+(2x*3x)+(2x*-1)+(4*3x)+(4*-1)
Simplify:
3/2x³-1/2x²+6x²-2x+12x-4
= 3/2x³+11/2x²+10x-4
Hence, the correct answer is a(x)=3/2x³+11/2x²+10x-4
Answer:a(x)=3/2x³+11/2x²+10x-4
Step-by-step explanation:
Need help with the question I need to understand so I do good on my test
Given:-
There are two types of log to measure the density.
Pine log of radius 5-inch and 30-inch long.
Oak board that is 5.5 inch long and 1.5 inch thickness and 3 feet long.
To find the shapes which can be used.
The pine long is round in shape so the shape formed in pine log is CYLINDRICAL SHAPE.
Oak board is 5.5 inch long and 1.5 inch thickness and 3 feet long so the shape formed will be CUBOID.
So the required solutions are CYLINDER AND CUBOID.
Find the solution for the given the system of equations:Y= (1/2)x - 1/2 and y=2^(x+3)
Answer:
This system has no solution.
Step-by-step explanation:
The solution of this system is the ordered pair that is the solution to both equations, we can solve this using the graphical method, which consists of graphing both equations in the same coordinate system.
The solution to the system will be at the point where the two functions intersect.
Since the functions do not intersect, this system has no solution.
Solve Step 3 onlyTherefore, the solutions of the original equation are the following. (Enter your answers as a comma-separated list. Use n as an integer constant. Enter your response in radians.)
ANSWER:
[tex]x=\pi n, \frac{3\pi}{2}+2\pi n[/tex]EXPLANATION"
Given:
[tex]\sin x(\sin x+1)=0[/tex]Having solved Step 1 and Step 2 as seen above, we can go ahead and write the solutions of the equation as seen below;
[tex]\begin{gathered} If\text{ }\sin x=0 \\ \therefore x=\pi \\ \\ If\text{ }\sin x=-1 \\ then\text{ }x=\frac{3\pi}{2} \\ \\ So\text{ }the\text{ }solution\text{ }will\text{ }be; \\ x=\pi n,\frac{3\pi}{2}+2\pi n \end{gathered}[/tex]The base of a triangle is given by a number, x (metres). The height of the triangle is ten metres less than the product of two and the number. The area of the triangle is equal to the product of seven and the base length.
According to the question the base of the triangle is x, the height is ten less than the product of two and x, this is 2x-10. The area of the triangle is the product of seven and the base, this is 7x.
The area of a triangle is given by:
[tex]A=\frac{b\cdot h}{2}[/tex]Replace each variable for the given expressions:
[tex]\begin{gathered} 7x=\frac{x\cdot(2x-10)}{2} \\ 7x=\frac{2x^2-10x}{2} \\ 7x=x^2-5x \\ 7=x-5 \\ x=7+5 \\ x=12 \end{gathered}[/tex]x has a value of 12.
(b)Dan leaves his house on his bike. He rides at a constant speed until he reaches a lemonade stand, where he parks his bike and takes a rest. Then he turns around and bikes home as fast as he can.
Answer:
The correct answer is the second graph.
Explanation:
Dan leaves his house at a constant speed. Then, the time starts counting, and his graph is a line that starts in (0, 0)
Then, he stops, we can see this part in the graph constant. Because he is not moving, the distance from his house remains the same.
Finally, goes back and we can this as a line. Since he is going at a faster speed, the line is more steep.
This is what we can see in the second graph
graph g(x) where f(x) = 2x-5 and g(x) = f(x+1)
The graph therefore is shown below;
Answer this question based on the knowledge of angle in a Circle.
From the given diagram, PQ is a diameter and since the triangle inside a semicircle is right angled, hence triangle PQR is a right triangle.
Also since the side QR is parallel to OS, hence the line PR is perpendicular to the lines QR and OS
Hence;
PR ⊥ QR and OS ⊥ PR
Proof that From △PRQ,
Since OS ⊥ PR, hence △PSR is divided into two similar triangles by the line OS showing that the base angles (
Recall that for an isosceles triangle, the base angles and two opposite sides are equal. Based on the proof above, we can conclude that;△PSR is isosceles triangle showing that SP = SRThe table gives the temperature( in Fahrenheit) in five cities at 6 am on the same day please zoom in pic so its not blurry
(a)
Temp. in fairbanks is -29 if the temp. risen by 17 then temp. is:
[tex]\begin{gathered} \text{Present temp. =initial temp. }+\text{ change in temp.} \\ =-29+17 \\ =-12 \end{gathered}[/tex]In Noon the temp in fairbanks is -12 degree fehrebheit.
(b)
6 A.M temp in Santa =74
6 A.M. temp in toronto =-19
[tex]\begin{gathered} \text{change in temp.= high temp. - low temp.} \\ =74-(-19) \\ =74+19 \\ =93 \end{gathered}[/tex]In 6 A.M. temp 93 fehre
Phil wants to play full-back for his football team. The decision depends on who serves as head coach for a given game. Coach Sal is head coach about 75% of the time, and Coach Benny is head coach other 25% of the time. Coach Sal has faith in Phil, so he starts him at full-back in 70% of the games he coaches. Coach Benny is not so sure, so he starts Phil at full-back 30% of the time. What is the probability of Phil starting as full-back for the next game?0.3980.60.40.24
Given that
Coach Sal is the head coach about 75% of the time and coach Benny is the coach for the remaining 25% of the time.
Sal has faith in Phil, so he full-back in 70% of the time and Benny had the faith, so he full-back 30% of the time.
Explanation -
Since the coach, Sal is the coach for 75% of the time and Benny is for 25% of the time.
Also, Sal's faith in full-back is 70% and Benny's faith in full-back is 30%.
Then,
75% of 70% = 75/100 x 70/100 = 0.525
25% of 30% = 25/100 x 30/100 = 0.075
So the final probability of Phil will be 0.525 + 0.075 = 0.6
Final answer -
So the final answer is 0.6.Hence option B is correct.Solve for x. I think u have to do a portion I’m not sure
Answer:
Explanation:
Based on the given figure, we can form similar triangles:
Triangle 1:
Triangle 2:
To solve for the value of x, we use ratio.
[tex]\begin{gathered} \frac{10}{6}=\frac{15}{x} \\ \text{Simplify and rearrange} \\ x=\frac{(15)(6)}{10} \\ x=\frac{90}{10} \\ \text{Calculate} \\ x=9 \end{gathered}[/tex]Therefore, the value of x is 9.
If a student got 10 answers out of 15 what’s the percent?
To calculate the percentage we have to write a simple fraction like this:
percentage = number of answers / total of questions
percentage = 10 / 15
percentage = 2 / 3
percentage = 0.667
Now, we have to multiply the result by 100:
percentage = 0.667 x 100
percentage = 66.7%
Answer: 66.7%
Ms. Friedman and Mrs. Elliot both teachsixth grade math. They share a storagecloset. What is the total area of both roomsand the storage closet?
The two classrooms are identical in length and width. On the other hand, the dimensions of the storage closet are
[tex](40-34)\times(36-30)=6\times6[/tex]The shape of both classrooms and the storage closet is rectangular; therefore, their areas are
[tex]\begin{gathered} A_{\text{rectangle}}=l\cdot w\to length\cdot width_{} \\ \Rightarrow A_{\text{Friedman}}=40\cdot36 \\ _{}A_{\text{Elliot}}=40\cdot36 \\ A_{storage}=6\cdot6 \\ \end{gathered}[/tex]Simplifying,
[tex]\begin{gathered} \Rightarrow A_{\text{storage}}=36ft^2 \\ \Rightarrow A_{\text{Friedman}}=A_{\text{Elliot}}=1440ft^2 \end{gathered}[/tex]Finally, the total area of the compound is
[tex]\begin{gathered} A_{\text{total}}=A_{\text{Friedman}}+A_{\text{Elliot}}-A_{\text{storage}} \\ \Rightarrow A_{\text{total}}=2\cdot1440-36=2844 \end{gathered}[/tex]Thus, the total area of the two classrooms plus the closet is 2844ft^2
Then,