Given,
The numbers of times 14 people ate out last month is,
7,5, 3, 6, 3, 3, 6, 5, 6, 3, 6, 5, 5,4
Arranging the data is ascending order,
3, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7
The frequency of 3, 5 and 6 is same.
The number with maximum frequency is called the mode of the data.
The mode of the data set is 3, 5, 6.
What are the domain and range of the function f of x is equal to the quantity x squared plus 6x plus 8 end quantity divided by the quantity x plus 4 end quantity?
A. D: {x ∊ ℝ | x ≠ 4}; R: {y ∊ ℝ | y ≠ 0}
B. D: {x ∊ ℝ | x ≠ −4}; R: {y ∊ ℝ | y ≠ −2}
C. D: {x ∊ ℝ | x ≠ 4}; R: {y ∊ ℝ | y ≠ 2}
D. D: {x ∊ ℝ | x ≠ −2}; R: {y ∊ ℝ | y ≠ 0}
The domain and range of [tex]f(x) = \frac{x^{2}+ 6x+8}{x+4}[/tex] is D: {x ∊ ℝ | x ≠ −4}; R: {y ∊ ℝ | y ≠ −2} .
The domain of a function f(x) is set of the value of x for which it is defied and Range of function is set of values f takes .
The rational number [tex]f(x) = \frac{p(x)}{q(x)}[/tex] where p(x) and q(x) are polynomial in terms of x and q(x) ≠ 0 . The domain of rational number is set of values that do not cause denominator equal to zero .
The given function is :
[tex]f(x) = \frac{x^{2}+ 6x+8}{x+4}[/tex]
we need to find domain and range of function,
For domain put denominator equals to zero
x+4 = 0
x = -4
so, domain is every real number except number making it zero
domain = (-∞,-4)∪(-4,∞) and x ≠ -4
For range ,factoring numerator
x²+6x+8 = x²+2x+4x+8
x(x+2)+4(x+2) = (x+4)(x+2)
Putting numerator back,
[tex]f(x) = \frac{(x+4)(x+2)}{x+4}[/tex]
Cancelling factor (x+4) from numerator and denominator
[tex]f(x) = x+2[/tex]
Putting x+2 = 0
[tex]x +2 = 0\\x = -2[/tex]
range = (-∞,-2)∪(-2,∞) and f(x) ≠ -2
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D: {x ∊ ℝ | x ≠ −4}; R: {y ∊ ℝ | y ≠ −2}
I took the test
The total profit (in dollars) from the sale of a espresso machines is P(x) = 160x - 0.85x^2 + 25.Evaluate the marginal profit at the following values to 4 decimal place accuracy if necessary:(A) P'(60) =(B) P'(65) =
We will have the following:
[tex]\begin{gathered} P(x)=160x-0.85x^2+25 \\ \\ \Rightarrow P^{\prime}(x)=160-1.7x \end{gathered}[/tex]So:
A) P'(60):
[tex]P^{\prime}(60)=160-1.7(60)\Rightarrow P^{\prime}(60)=58[/tex]B) P'(65):
[tex]P^{\prime}(65)=160-1.7(65)\Rightarrow P^{\prime}(65)=49.5[/tex]I need to know number 6SP please I need to find the mean
The mean absolute deviation is given by the next formula:
[tex]\text{MAD}=\frac{1}{n}\sum ^{\square}_i|x_i-\bar{x}|[/tex]Where n is the number of points in the data set and x with a bar on top is the mean.
In our case,
[tex]\bar{x}=\frac{1}{25}(5\cdot1+5\cdot2+4\cdot3+4\cdot4+6\cdot5+6\cdot1)=\frac{79}{25}[/tex]and n=25.
Then,
[tex]\text{MAD}=\frac{1}{25}(5|1-\frac{79}{25}|+5|2-\frac{79}{25}|+4|3-\frac{79}{25}|+4|4-\frac{79}{25}|+6|5-\frac{79}{25}|+|6-\frac{79}{25}|)[/tex]Finally,
[tex]\text{MAD}=\frac{862}{625}\approx1.4[/tex]The answer is 1.4 once rounded.
Is (5, 1) a solution to the equation y = 1? yes no
Since the coordinates are written as (x,y), for the point (5,1) we have that:
[tex]\begin{gathered} x=5 \\ y=1 \end{gathered}[/tex]Therefore, this is a solution to the equation y=1.
Then, the answer is yes.
Compare the functions f(x) = 20x² and g(x) = 4* by completing parts (a) and (b).(a) Fill in the table below. Note that the table is already filled in for x = 2.(The ALEKS calculator can be used to make computations easier.)f(x)=20x²g(x) = 4*(b) For x ≥ 5, the table suggests that f(x) is (Choose one) ✓ greater than g(x).I need help with this math problem.
ANSWERS
(a) Table:
(b) For x ≥ 5, the table suggests that f(x
EXPLANATION
(a) First, we have to fill in this table. To do so, we have to substitute x with each value from the first column for each function,
[tex]f(3)=20\cdot3^2=20\cdot9=180[/tex][tex]g\mleft(3\mright)=4^3=64[/tex]Repeat for all the other x-values,
(b) As we can see in the table, for x = 2, x = 3, and x = 4, f(x) is greater than g(x). But, for x = 5 and x = 6, f(x) is less than g(x).
Hence, for x ≥ 5, f(x) is never greater than g(x).
Rewrite the equation by completing the square.X^2 + 11x + 24 = 0
Solution
For this case we can complete the square on this way:
[tex]x^2+11x+(\frac{11}{2})^2+24-(\frac{11}{2})^2[/tex]and if we simplify we got:
[tex](x+\frac{11}{2})^2+24-\frac{121}{4}=(x+\frac{11}{2})^2-\frac{25}{4}[/tex]We can use this property:
[tex](a^2+2ab+b^2)=(a+b)^2[/tex]For this case:
[tex]a=x,b=\frac{11}{2}[/tex]Answer:
(x+11/2)^2 = 25/4
did it on khan, hope this helps<3
Determine the graph of line with the given y-intercept and slope
The form of the equation of the line is
y=mx+b
where m is the slope and
b is the y-intercept
In our case
m=3/2
and
b=-5
the equation of the line is
y=3/2 x -5
In order to graph we need at least another point and connect it with the y-intercept
if x=2
y=3/2 (2)-5=3-5=-2
y=-2
The point is
(2,-2)
Then we can graph the line
What do you notice about these two expressions? 4 50 8 + 15 50 8 + 7 15 1 They both show subtracting 2 4 They both show adding 3 50 8 They both contain the sum 7 15
The 3rd option is right as both contain the sum
50/7 + 8/15
What type of wording in a problem statement or description of a situation tells you that you have a rate of change?
There different types of wording that tells us that we have a rate of change.
For example, when talking about speed, or in general, when talking about any measure per unit of time (meters per second, gallons per minute, miles per hour) all these are rates of change.
Other example, not as common as the first one, is when the statement refers to a slope, which is in other words, a rate of change.
Similar to the first one, when the statement talks about amounts and/or growth over time, for example: On the first day the plant was 10 cm high, on the second day it was 15 cm high, on the fifth day...
During a special one-day sale, 600 customers bought the on-sale sandwich. Ofthese customers, 20% used coupons. The manager will run the sale again the nextday if more than 100 coupons were used. How many coupons were used andshould she run the sale again?80 coupons were used; no, she should not run the sale again140 coupons were used; yes, she should run the sale againOOO115 coupons were used; yes, she should run the sale again120 coupons were used; yes, she should run the sale again
Answer:
120 coupons were used; yes, she should run the sale again
Explanation:
The total number of customers = 600
The percentage that used coupon = 20%
[tex]\begin{gathered} 20\%\text{ of 600} \\ =\frac{20}{100}\times600 \\ =120 \end{gathered}[/tex]Thus, 120 coupons were used.
Since more than 100 coupons were used, the manager should run the sale again.
Need help solving this problem This is from my ACT prep guide
Answer:
[tex]h=101.7\text{ ft}[/tex]Step-by-step explanation:
To approach this situation, let's make a diagram of it;
To solve this problem we can use trigonometric relationships since there are right triangles involved; trigonometric relationships are represented as
[tex]\begin{gathered} \sin (\text{angle)}=\frac{\text{opposite}}{\text{ hypotenuse }} \\ \cos (\text{angle)}=\frac{\text{adjacent}}{\text{hypotenuse}} \\ \tan (\text{angle)}=\frac{\text{opposite}}{\text{adjacent}} \end{gathered}[/tex]Then, since we can divide the neighbor building into two parts compounded by two right triangles, use the tan relationship for the 56 degrees triangle, and sin relationship for the 32 degrees triangle:
[tex]\begin{gathered} \tan (56)=\frac{h_1}{150_{}} \\ h_1=150\cdot\tan (56) \\ h_1=22.2\text{ ft} \\ \\ \sin (32)=\frac{h_2}{150} \\ h_2=150\cdot\sin (32) \\ h_2=79.5\text{ ft} \end{gathered}[/tex]Hence, for the total height of the building:
[tex]\begin{gathered} h=h_1+h_2 \\ h=22.2+79.5 \\ h=101.7\text{ ft} \end{gathered}[/tex]Suppose a charity received a donation of $17.3 million. If this represents 41% of the charity's donated funds, what is the total amount of its donated funds? Round your answer to the nearest million dollars.
Using the concept of percentage, the total amount that was donated to charity is $42.195 million.
What is meant by percentage?A figure or ratio that is stated as a fraction of 100 is referred to as a percentage in mathematics. The abbreviation used to represent percentages is the symbol "%". A% has neither a recognized unit of measurement nor any dimensions. Take this as an example: There will be 50 men in the class if there are 100 students total and there are 50 men in the class. There are 250 male students overall, or 250 out of 500.
According to the data, the charity got a donation of $17.3 million, or 41% of its total donations.
Let x serve as a symbol for the total sum.
This will be shown as follows:
41% × x = $17.3million
0.41x = $17.3 million
Divide by 0.41
x = $17.3million / 0.41
x = $42.195 million
The total amount that is donated to charity is $42.195 million
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Determine whether each equation has a solution. Justify your answer.a. (a + 4) / (5 + a) = 1b. (1 + b)/ (1 - b) = 1c. (c - 5 )/ (5 - c) = 1
Leave the variable in each equation in one side of the equation to find if it has a solution:
a.
[tex]\begin{gathered} \frac{a+4}{5+a}=1 \\ \\ Multiply\text{ both sides by \lparen5+a\rparen:} \\ \frac{a+4}{5+a}(5+a)=1(5+a) \\ \\ a+4=5+a \\ \\ Subtract\text{ a in both sides of the equation:} \\ a-a+4=5+a-a \\ 4=5 \end{gathered}[/tex]As 4 is not equal to 5, the equation has no solution.b.
[tex]\begin{gathered} \frac{1+b}{1-b}=1 \\ \\ Multiply\text{ both sides by \lparen1-b\rparen:} \\ \frac{1+b}{1-b}(1-b)=1(1-b) \\ \\ 1+b=1-b \\ \\ Add\text{ b in both sides of the equation:} \\ 1+b+b=1-b+b \\ 1+2b=1 \\ \\ Subtract\text{ 1 in both sides of the equation:} \\ 1-1+2b=1-1 \\ 2b=0 \\ \\ Divide\text{ both sides of the equation by 2:} \\ \frac{2b}{2}=\frac{0}{2} \\ \\ b=0 \\ \\ Prove\text{ if b=0 is the solution:} \\ \frac{1+0}{1-0}=1 \\ \\ \frac{1}{1}=1 \\ \\ 1=1 \end{gathered}[/tex]The solution for the equation is b=0c.
[tex]\begin{gathered} \frac{c-5}{5-c}=1 \\ \\ Multiply\text{ both sides by \lparen5-c\rparen:} \\ \frac{c-5}{5-c}(5-c)=1(5-c) \\ \\ c-5=5-c \\ \\ Add\text{ c in both sides of the equation:} \\ c+c-5=5-c+c \\ 2c-5=5 \\ \\ Add\text{ 5 in both sides of the equation:} \\ 2c-5+5=5+5 \\ 2c=10 \\ \\ Divide\text{ both sides by 2:} \\ \frac{2c}{2}=\frac{10}{2} \\ \\ c=5 \\ \\ Prove\text{ if c=5 is a solution:} \\ \frac{5-5}{5-5}=1 \\ \\ \frac{0}{0}=1 \\ \\ \frac{0}{0}\text{ is undefined} \end{gathered}[/tex]As the possible solution (c=5) makes the expression has a undefined part (0/0) it is not a solution.
The equation has no solutionThe school store is running a promotion on school supplies. Different supplies are placed on two shelves.• You can purchase 3 items from shelf A and 2 from shelf B for $26, or• You can purchase 2 items from shelf A and 5 from shelf 8 for $32.Let z represent the cost of an item from shelf A and let y represent the cost of an item from shelf B. Write and solve asystem of equations to find the cost of items from shelf A and shelf B. Show your work and thinking!
We define the following notation:
• z = cost of an item from shelf A,
,• y = cost of an item from shelf B.
From the statement, we know that:
• 3 items from shelf A and 2 from shelf B cost $26, so we have:
[tex]3z+2y=26,[/tex]• 2 items from shelf A and 5 from shelf B for $32, so we have:
[tex]2z+5y=32.[/tex]We have the following system of equations:
[tex]\begin{gathered} 3z+2y=26, \\ 2z+5y=32. \end{gathered}[/tex]1) To solve this system, we multiply the first equation by 2 and the second equation by 3:
[tex]\begin{gathered} 2\cdot(3z+2y)=2\cdot26\rightarrow6z+4y=52, \\ 3\cdot(2z+5y)=3\cdot32\rightarrow6z+15y=96. \end{gathered}[/tex]2) We subtract equation 1 to equation 2, and then we solve for y:
[tex]\begin{gathered} (6z+15y)-(6z+4y)=96-52, \\ 11y=44, \\ y=\frac{44}{11}=4. \end{gathered}[/tex]We found that y = 4.
3) We replace the value y = 4 in the first equation, and then we solve for z:
[tex]\begin{gathered} 3z+2\cdot4=26, \\ 3z+8=26, \\ 3z=26-8, \\ 3z=18, \\ z=\frac{18}{3}=6. \end{gathered}[/tex]Answer
The cost of the items are:
• z = 6, for items from shelf A,
,• y = 4, for items from shelf B.
The function g is graphed below. At what numbers in the interval (-4,4) is g discontinuous?
The graph has discontinuity if the curve or the line is not continuous.
From the graph shown, The graph stopped at (-1, 0) and then starts again at (-1, 4), stopped again at (3, 2)
then gone with (3, 3) and continues up to (4, -2)
So graph g discontinuous at -1 and 3
You have $ 24 and you go to the grocery store to buy a Kit Kats snack and a drink You spent $ 1.75 on a drink If each bar of Kit Kat costs $1.49write an inequality that represents how many bars of Kit Kat you can buy
The inequality that represents the situation is:
[tex]1.49x+1.75\le24[/tex]Where x is the number of Kit Kat we could buy.
(0,2)-5The function g(x) = -3x - 6. Compare the slopes.A. The slope of f(x) is the same as the slope of g(x).B. The slope of f(x) is undefined, and the slope of g(x) is negative.Ο ΟC. The slope of f(x) is greater than the slope of g(x).D. The slope of f(x) is less than the slope of g(x).
To solve the exercise, first we are going to find the slope of the function f(x). Since we have a graph of the function, we can see two points through which the line passes:
[tex]\begin{gathered} (x_1,y_1)=(0,2) \\ (x_2,y_2)=(1,-1) \end{gathered}[/tex]Now we can use this formula to find the slope:
[tex]\begin{gathered} m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \text{ Where m is the slope and} \\ (x_1,y_1)\text{ and }(x_2,y_2)\text{ are two points through which the line passes} \end{gathered}[/tex][tex]\begin{gathered} m_{f(x)}=\frac{-1-2}{1-0} \\ m_{f(x)}=\frac{-3}{1} \\ m_{f(x)}=-3 \end{gathered}[/tex]Then, the slope of the function f(x) is -3.
On the other hand, the function g(x) also describes a line and is written in slope-intercept form, that is:
[tex]\begin{gathered} y=mx+b\Rightarrow\text{ slope-intercept form} \\ \text{ Where m is the slope and} \\ b\text{ is the y-intercept of the line} \end{gathered}[/tex]Then, you can see that the slope of the function g(x) is -3, because
[tex]\begin{gathered} g(x)=-3x-6 \\ m_{g(x)}=-3 \\ \text{and} \\ b=-6 \end{gathered}[/tex]Therefore, the slope of f(x) is the same as the slope of g(x) and the correct answer is option A.
I need to find the solution to 5x-2 divided by 6
Given:
[tex](5x-2)\div6[/tex]To find:
Solve the given expression.
Explanation:
Rewrite the given expression,
[tex]\frac{5x-2}{6}[/tex]Expand the fraction into 2 simpler fractions with common denominator 6,
[tex]\frac{5x}{6}+\frac{-2}{6}[/tex]Simplify the expression, we will get:
[tex]\frac{5}{6}x-\frac{1}{3}[/tex]Final answer:
Hence, the required solution is:
[tex]\frac{5}{6}x-\frac{1}{3}[/tex]
Find all solutions in the set of real numbers. Show all your work. cos2θ=−sin2θ
Equate the given to zero
[tex]\begin{gathered} \cos 2\theta=-\sin ^2\theta\Longrightarrow\cos 2\theta+\sin ^2\theta=0 \\ \\ \text{Use the identity }\cos 2\theta=1-2\sin ^2\theta,\text{ THEN the equation becomes} \\ \cos 2\theta+\sin ^2\theta=0 \\ (1-2\sin ^2\theta)+\sin ^2\theta=0 \\ \\ \text{simplify} \\ (1-2\sin ^2\theta)+\sin ^2\theta=0 \\ 1-2\sin ^2\theta+\sin ^2\theta=0 \\ 1-\sin ^2\theta=0 \\ \\ \text{Add }\sin ^2\theta\text{ to both sides} \\ 1-\sin ^2\theta+\sin ^2\theta=0+\sin ^2\theta \\ 1\cancel{-\sin ^2\theta+\sin ^2\theta}=\sin ^2\theta \\ 1=\sin ^2\theta \\ \text{OR} \\ \sin ^2\theta=1 \\ \\ \text{get the square root of both sides} \\ \sqrt[]{\sin ^2\theta}=\sqrt[]{1} \\ \sin \theta=\pm1 \\ \\ \text{The values for which }\sin \theta\text{ is equal to }+1\text{ or }-1\text{ is} \\ \theta=\frac{\pi}{2}+2\pi n,\theta=\frac{3\pi}{2}+2\pi n \end{gathered}[/tex]Help me with my Algebra 2 work :) -- (Multiplying and Dividing Rational Expressions)(Drag and Drop question)
a. For the expression P(x) • Q(x), we multiply the two functions.
[tex]\frac{2}{3x-1}\times\frac{6}{-3x+2}[/tex]When multiplying rational expressions, simply multiply the numerator to the other numerator and the denominator to the other denominator.
Hence, P(x) • Q(x) is equal to:
[tex]\frac{12}{(3x-1)(-3x+2)}[/tex]
b. For the expression, P(x) ÷ Q(x), we multiply P(x) to the reciprocal of Q(x).
[tex]\frac{2}{3x-1}\times\frac{-3x+2}{6}[/tex]As mentioned above, when multiplying rational expressions, simply multiply the numerator to the other numerator and the denominator to the other denominator. The expression becomes:
[tex]\frac{2(-3x+2)}{6(3x-1)}[/tex]Then, simplify 2/6 to 1/3.
Hence, P(x) ÷ Q(x) is equal to:
[tex]\frac{-3x+2}{3(3x-1)}[/tex]Find the measureOf an act intercept by and inscribed whose measure is 75°Hent, draw a picture and label it,
From the arc-angle relationships, we know that the inscribed angle is half of the intercepted arc, that is,
[tex]75=\frac{arcAB}{2}[/tex]Then, by moving the denominator to the left hand side, we get
[tex]\begin{gathered} 2\times75=arcAB \\ \text{arcAB}=150 \end{gathered}[/tex]then, the arcAB measure 150 degrees.
Find the y - intercept of the equation -7x - 5y = -140
Given -
-7x - 5y = -140
To Find -
The y-intercept of the equation =?
Step-by-Step Explanation -
The Y-intercept is a point where the graph of the equation cuts the y-axis.
At y-axis x = 0.
So, we know that at y-intercept x = 0
Put x = 0 in -7x - 5y = -140
= -7(0) - 5y = -140
= -5y = -140
= y = 140/4
y = 28
Final Answer -
The y-intercept of the equation = (0, 28)
Answer:
4
Step-by-step explanation:
-7x-5y=-140
Y=4
Because 7 times 5 =35
140 divided by 35=4
5(4)=20
20 times 7=140. So y=4
and the missing numbers.
11. 360 ÷ 60 = 36 tens ÷ 6 tens =
Answer:
360÷60=36÷6=6÷0.6=10 not sure if this is right
the function h(t)=-4.9t^2+19t+1.5 describes the height in meters of a basketball t seconds after it has been thrown vertically into the air.what is the maximum height of the basketball? round your answer to the nearest tenth
we have
h(t)=-4.9t^2+19t+1.5
This function represent a vertical parabola open downward
The vertex represent a maximum
using a graphing tool
see the attached figure
please wait a minute
The vertex is the point (1.94, 19.92)
therefore
the maximum height of the basketball is 19.9 meters
can someone please help me find the Answer to the following
First, to find the slant height, we use Pythagorean's Theorem to find the slant height, as the image below shows.
Then,
[tex]\begin{gathered} h^2=10^2+7.5^2 \\ h^2=100+56.25 \\ h=\sqrt[]{156.25} \\ h=12.5ft \end{gathered}[/tex]The slant height is 12.5 feet.Now, the surface area formula is
[tex]SA=2bs+b^2[/tex]Where s = 12.5 and b = 15.
[tex]\begin{gathered} SA=2\cdot15\cdot12.5+15^2 \\ SA=375+225 \\ SA=600ft^2 \end{gathered}[/tex]Hence, the surface area of the pyramid is 600 square feet.Find the solution to the following equation.4(x - 1) + 11.1 = 7(x - 4)
Answer: x = -3
Explanation:
The expression we have is:
[tex]4(x-1)+11.1=0.7(x-4)[/tex]Step 1: Use the distributive property, multiply the number outside the parenthesis by the numbers or terms inside the parenthesis
[tex]4x-4+11.1=0.7x-2.8[/tex]Step 2: Move all of the terms with x to the left, and all of the independent terms to the right side of the equation
[tex]4x-0.7x=-2.8+4-11.1[/tex]Note that we we move a term to the other side of the equation they change of sig.
step 3: combine like terms
[tex]3.3x=-9.9[/tex]Step 4: Divide both sides of the equation by 3.3 to solve for x
[tex]\begin{gathered} \frac{3.3x}{3.3}=\frac{-9.9}{3.3} \\ x=\frac{-9.9}{3.3} \\ x=-3 \end{gathered}[/tex]were would he be at in 6 seconds
6 seconds later we can say that dwyane was running to the left of zero, because if he was running at 4 meters per second to the right when he passed the zero point, hence 6 seconds later he was on the left of the zero at the point -2.
Samantha have six times as many fish in her aquarium as Lily. Lily and Samantha have 70 fishes together. how many fish does each girl have?
The problem indicates that Samantha has 6 times as many fish as Lily.
If we call the number of fishes that Lily has "x", then the number of fishes that Samantha has will be "6x".
Lily: x
Samantha: 6x
Now, since together they have 70 fishes, we need to add the two quantities (x and 6x) and equal them to 70:
[tex]x+6x=70[/tex]To solve this problem we need to solve this equation for x.
The first step is to add the like terms on the left of the equation:
[tex]7x=70[/tex]And then, divide both sides of the equation by 7:
[tex]\begin{gathered} \frac{7x}{7}=\frac{70}{7} \\ \end{gathered}[/tex]As you can see 7 in the numerator and denominator on the left side cancel each other:
[tex]x=\frac{70}{7}[/tex]And on the right side, 70/7 is equal to 10:
[tex]x=10[/tex]Let's remember our initial definitions:
Lily: x
Samantha: 6x
Since we now know that x=10, we conclude that Lily has 10 fishes.
To calculate the amount that Samantha has multiply 6 by 10:
[tex]6x=6(10)=60[/tex]Samantha has 60 fishes.
Answer:
Lily: 10
Samantha: 60
Number of balls madeHow many games had 11 or fewer balls made?
The number of games that had 11 or fewer balls made = 6
Explanations:The number of games that have 11 or fewer balls made will be the number of games that have between 0 to 11 balls made.
Number of games that had 0-3 balls made = 0
Number of games that had 4 - 7 balls made = 1
Number of games that had 8-11 balls made = 5
The number of games that had 11 or fewer balls made = (Number of games that had 0-3 balls made) + (Number of games that had 4-7 balls made) + (Number of games that had 8-11 balls made)
The number of games that had 11 or fewer balls made = 0 + 1 + 5
The number of games that had 11 or fewer balls made = 6
Answer:
6
Step-by-step explanation:
use the information to answer the following questionsmr. ramirez purchased 20 concert tickets for a total of $225.the concert tickets cost $15 for adults and $10 for children under the age of 12 part A: write a system of equations to represent the given scenario. use the variables "a" for adults and "c" for children.
Answer:
a + c = 20
15a + 10c = 225
Explanation:
If 'a' is the number of tickets for adults and 'c' is the number of tickets for children, we can write the following equation:
a + c = 20
Because Mr. Ramirez purchased a total of 20 concert tickets.
In the same way, if the cost of the tickets is $15 for adults and $10 for children, we can write the following equation:
15a + 10c = 225
Because 15a is the total cost for adults, 10c is the total cost for children and 225 is the total cost in general.
Therefore, the system of equations that represents the given scenario is:
a + c = 20
15a + 10c = 225