Given the following data
Base area of the cone = 6cm
Height of the cone = 9 cm
The volume of a cone is given as
[tex]\begin{gathered} V\text{ = }\frac{1}{3}\cdot\text{ }\pi\cdot r^2\cdot\text{ h} \\ \text{where }\pi\text{ = 3.14},\text{ r = 6cm , and h = 9cm} \\ V\text{ = }\frac{1}{3}\cdot\text{ 3.14 }\cdot6^2\cdot\text{ 9} \\ V\text{ = }\frac{1}{3}\text{ x 3.14 x 36 x 9} \\ V\text{ = }\frac{3.14\text{ x 36 x 9}}{3} \\ V\text{ = }\frac{1017.36}{3} \\ V=339.1cm^3 \end{gathered}[/tex]If R is between G and Z, GZ = 12in., and RG =3in., then RZ =
Given R is between G and Z.
GZ=12 inches
RG=3 inches.
Since, R is between G and Z,
[tex]GZ=GR+RZ[/tex]It follows
[tex]\begin{gathered} RZ=GZ-GR \\ =12-3 \\ =9 \end{gathered}[/tex]So, RZ is 9 inches.
14. Given: JM bisects JL JM perpendicular to KLProve: TRIANGLE JMK congruent to TRIANGLE JML
1) is already written, so we start with the second line.
2)
JM is parallel to KL ----> Given
3) ∠KML = ∠JML ----> They are angles on two perpendicular lines, and Since JM bisects LK, they are equal.
4) ∠KJL=∠MKL ---> Since JM bisects ∠J, the angles KJL and MKL are equal
5) ∠JKM=∠JLM ----> Since 3) and 4), the angles JKM and JLM must also be equal so that the sum of internal angles of each triangle will be 180°
Thus: Triangle JMK is congruent to triangle JML
Be specific with your answer thank you thank you thank you bye-bye
The y-axis on the graph, that shows us the cost, goes from 2 to 2 units.
To find the cost at option one, the red line, we look in the graph where the line is when x = 80.
For x= 80, y= 58
Now, the same for option 2:
For x = 80, y= 44.
58-44 = 14
Answer: The difference is 14.
An insurance company offers flood insurance to customers in a certain area. Suppose they charge $500 fora given plan. Based on historical data, there is a 1% probability that a customer with this plan suffers aflood, and in those cases, the average payout from the insurance company to the customer was $10,000.Here is a table that summarizes the possible outcomes from the company's perspective:EventFloodPayout Net gain (X)$10,000 -$9,500$0$500No floodLet X represent the company's net gain from one of these plans.Calculate the expected net gain E(X).E(X) =dollars
The given is a discrete random variable.
For a discrete random variable, the expected value is calculated by summing the product of the value of the random variable and its associated probability, taken over all of the values of the random variable.
It is given that the probability of a flood is 1%=0.01.
It follows that the probability of no flood is (100-1)%=99%.
Hence, the expected net gain is:
[tex]E(X)=0.01(-9500)+0.99(500)=-95+495=400[/tex]Hence, the expected net gain is $400.
The expected net gain is E(X) = $400.
Given the functions, f(x) = 6x+ 2 and g(x)=x-7, perform the indicated operation. When applicable, state the domain
restriction.
The domain restriction for (f/g)(x) is x=7
What are the functions in mathematics?a mathematical phrase, rule, or law that establishes the link between an independent variable and a dependent variable.
What does a domain math example mean?The collection of all potential inputs for a function is its domain. For instance, the domain of f(x)=x2 and g(x)=1/x are all real integers with the exception of x=0.
Given,
f(x) = 6x+2
g(x) = x-7
So,
(f/g)(x) = 6x+2/x-7
Remember that the denominator can not be equal to zero
Find the domain restriction
x-7=0
x=7
Therefore, the domain is all real numbers except the number 7
(-∞,7)∪(7,∞)
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Which of the following is the result of using the remainder theorem to find F(-2) for the polynomial function F(x) = -2x³ + x² + 4x-3?
Solution
We have the polynomial
[tex]f(x)=-2x^3+x^2+4x-3[/tex]Usin the remainder theorem, we find f(-2) by substituting x = -2
So we have
[tex]\begin{gathered} f(x)=-2x^{3}+x^{2}+4x-3 \\ \\ f(-2)=-2(-2)^3+(-2)^2+4(-2)-3 \\ \\ f(-2)=-2(-8)+4-8-3 \\ \\ f(-2)=16+4-8-3 \\ \\ f(-2)=20-11 \\ \\ f(-2)=9 \end{gathered}[/tex]Therefore, the remainder is
[tex]9[/tex]I need help creating a tree diagram for this probability scenario
We need to draw a tree diagram for the information given
The total is 400
120 in finance course
220 in a speech course
55 in both courses
Then we start for a tree for the given number
Then to make the tree for probability we will divide each number by a total 400
Then the probability of finance only is 65/400
The probability of speech only is 165/400
The probability of both is 55/400
The probability of neither is 5/400
The probability of finance or speech is 285/400
How much of the wall does the mirror cover? Use the π button in your calculations and round your answer to the nearest hundredths. Include units.
Since the diameter of the mirror is given, calculate the area of the mirror using the formula
[tex]A=\frac{1}{4}\pi\cdot(D)^2[/tex]replace with the information given
[tex]\begin{gathered} A=\frac{1}{4}\pi\cdot24^2 \\ A=144\pi\approx452.39in^2 \end{gathered}[/tex]The mirror covers 452.39 square inches.
4 5 3 7 89 65Each time, you pick one card randomly and then put it back.What is the probability that the number on the card you pickfirst time is odd and the number on the second card you take isa multiple of 2? Keep your answers in simplified improperfraction form.Enter the answer
We have a total of 8 cards, where 3 of them are a multiple of 2, and 5 is an odd number. Consider that event A represents the probability of picking an odd number and event B is picking a multiple of 2. We know that the events are independent (because we put the cards back), therefore the probability of A and B can be expressed as
[tex]P(A\text{ and }B)=P(A)\cdot P(B)[/tex]Where
[tex]\begin{gathered} P(A)=\frac{5}{8} \\ \\ P(B)=\frac{3}{8} \end{gathered}[/tex]Therefore
[tex]P(A\text{ and }B)=\frac{5}{8}\cdot\frac{3}{8}=\frac{15}{64}[/tex]The final answer is
[tex]P(A\text{ and }B)=\frac{15}{64}[/tex]A rectangular parking lot has length that is 3 yards less than twice its width. If the area of the land is 299 square yards, what are the dimensions of the land?The parking lot has a width of square yards.
Answer:
• Width = 13 yards
,• Length = 23 yards
Explanation:
Let the width of the parking lot = w yards.
The length is 3 yards less than twice its width.
[tex]\implies\text{Length}=(2w-3)\text{ yards}[/tex]The area of the land = 299 square yards.
[tex]w(2w-3)=299[/tex]We then solve the equation above for w.
[tex]\begin{gathered} 2w^2-3w=299 \\ \implies2w^2-3w-299=0 \end{gathered}[/tex]Factor the resulting quadratic expression.
[tex]\begin{gathered} 2w^2-26w+23w-299=0 \\ 2w(w-13)+23(w-13)=0 \\ (2w+23)(w-13)=0 \end{gathered}[/tex]Solve for w.
[tex]\begin{gathered} 2w+23=0\text{ or }w-13=0 \\ 2w=-23\text{ or }w=13 \\ w\neq-\frac{23}{2},w=13 \end{gathered}[/tex]Since w cannot be negative, the parking lot has a width of 13 yards.
Finally, find the length of the parking lot.
[tex]\begin{gathered} 13l=299 \\ l=\frac{299}{13}=23\text{ yards} \end{gathered}[/tex]The length of the parking lot is 23 yards.
Translate to an equation and solve W divided by 6 is equal to 36 w=
Answer:
[tex]w\text{ = 216}[/tex]Explanation:
Here, we want to translate it into an equation and solve
W divided by 6 equal to 36:
[tex]\begin{gathered} \frac{w}{6}\text{ = 36} \\ \\ w\text{ = 6}\times36 \\ w\text{ = 216} \end{gathered}[/tex]explain why 4 x 3/5=12x 1/5
Answer:
They equal because when you simplify each side, you will arrive at the same answer.
[tex]\begin{gathered} 4\times\frac{3}{5}=\frac{4\times3}{5} \\ =\frac{12}{5} \end{gathered}[/tex]also;
[tex]\begin{gathered} 12\times\frac{1}{5}=\frac{12\times1}{5} \\ =\frac{12}{5} \end{gathered}[/tex]Explanation:
We want to explain why;
[tex]4\times\frac{3}{5}=12\times\frac{1}{5}[/tex]They equal because when you simplify each side, you will arrive at the same answer.
[tex]\begin{gathered} 4\times\frac{3}{5}=\frac{4\times3}{5} \\ =\frac{12}{5} \end{gathered}[/tex]also;
[tex]\begin{gathered} 12\times\frac{1}{5}=\frac{12\times1}{5} \\ =\frac{12}{5} \end{gathered}[/tex]So, they give the same answer when simplified.
Also you can derive one from the other;
[tex]\begin{gathered} 4\times\frac{3}{5}=12\times\frac{1}{5} \\ 4\times3\times\frac{1}{5}=12\times\frac{1}{5} \\ 12\times\frac{1}{5}=12\times\frac{1}{5} \\ \frac{12}{5}=\frac{12}{5} \end{gathered}[/tex]Therefore, both sides are equal.
Consider the graph below.(3,1) (4,2) (6,3) (4,4) (8,5) Which correlation coefficient and interpretation best represent the given points?1.) 0.625, no correlation 2.) 0.791. no correlation 3.) 0.625, positive correlation4.) 0.791. positive correlation
Given the information on the problem,we have that the correlation coefficient of the data given is:
[tex]r=\frac{\sum^{}_{}(x-\bar{y})(y-\bar{x})}{\sqrt[]{SS_x\cdot SSy}}=\frac{10}{\sqrt[]{16\cdot10}}=0.79[/tex]therefore, the value of the correlation coeficient is 0.79, which shows a strong positive correlation
hello I'm stuck on this question and need help thank you
Explanation
[tex]\begin{gathered} -2x+3y\ge9 \\ x\ge-5 \\ y<6 \end{gathered}[/tex]Step 1
graph the inequality (1)
a) isolate y
[tex]\begin{gathered} -2x+3y\geqslant9 \\ add\text{ 2x in both sides} \\ -2x+3y+2x\geqslant9+2x \\ 3y\ge9+2x \\ divide\text{ both sides by 3} \\ \frac{3y}{3}\geqslant\frac{9}{3}+\frac{2x}{3} \\ y\ge\frac{2}{3}x+3 \end{gathered}[/tex]b) now, change the symbol to make an equality and find 2 points from the line
[tex]\begin{gathered} y=\frac{2}{3}x+3 \\ i)\text{ for x=0} \\ y=\frac{2}{3}(0)+3 \\ \text{sp P1\lparen0,3\rparen} \\ \text{ii\rparen for x=3} \\ y=\frac{2}{3}(3)+3=5 \\ so\text{ P2\lparen3,5\rparen} \end{gathered}[/tex]now, draw a solid line that passes troguth those point
(0,3) and (3,5)
[tex]y\geqslant\frac{2}{3}x+3\Rightarrow y=\frac{2}{3}x+3\text{\lparen solid line\rparen}[/tex]as we need the values greater or equatl thatn the function, we need to shade the area over the line
Step 2
graph the inequality (2)
[tex]x\ge-5[/tex]this inequality represents the numbers greater or equal than -5 ( for x), so to graph the inequality:
a) draw an vertical line at x=-5, and due to we are looking for the values greater or equal than -5 we need to use a solid line and shade the area to the rigth of the line
Step 3
finally, the inequality 3
[tex]y<6[/tex]this inequality represents all the y values smaller than 6, so we need to draw a horizontal line at y=6 and shade the area below the line
Step 4
finally, the solution is the intersection of the areas
I hope this helps you
can you help me figure out the equation in the drop down menus
To find:
The piecewise function for the graph.
Solution:
From the graph, it is clear that when x is less than -1, the graph passes through (-1, -3) and (-2, -5).
It is known that the equation of a line passes through two points is given by:
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]So, the equation of line passing through (-1, -3) and (-2, -5) is:
[tex]\begin{gathered} y-(-3)=\frac{-5-(-3)}{-2-(-1)}(x-(-1)) \\ y+3=\frac{-2}{-1}(x+1) \\ y+3=2x+2 \\ y=2x-1 \end{gathered}[/tex]So, the first drop down is "2x - 1", and second drop down is "x is less than or equal to -1".
Now, the graph passes through (1, 5) and (2, 6). So, the equation of the line is:
[tex]\begin{gathered} y-5=\frac{6-5}{2-1}(x-1) \\ y-5=x-1 \\ y=x+4 \end{gathered}[/tex]So, the third drop down menu is "x + 4" and the fourth drop down menu is "x is greater than or equal to 1".
The next algebra test is worth 100 points and contains 35 problems. Multiple-Choice questions are worth 2 points each and word problems are 7 points each. How many of each type equation are there?
Let
x ----->number of multiple-choice questions
y ----> number of word problems
so
we have
x+y=35 --------> equation 1
2x+7y=100 -----> equation 2
solve the system of equations
Solve by graphing
using a graphing tool
see the attached figure
therefore
x=29
y=6
number of multiple-choice questions is 29
number of word problems is 6
A coin is tossed an eight sided die numbered 1 through 8 is rolled find the probability of tossing a head and then rolling a number greater than 6. Round to three decimal places if needed
We are given that a coin is tossed and a die numbered from 1 through 8 is rolled. To determine the probability of tossing head and then rolling a number greater than 6 is given by the following formula:
[tex]P(\text{head and n>6)=p(head)}\cdot p(n>6)[/tex]This is because we are trying to determine the probability of two independent events. The probability of getting heads is given by:
[tex]P(\text{heads})=\frac{1}{2}[/tex]This is because there are two possible outcomes, heads or tails and we are interested in one of the outcomes.
Now we determine the probability of getting a number greater than 6 when rolling the dice. For this, there are 8 possible outcomes and we are interested in two of them, these are the numbers greater than 6 on the die (7, 8). Therefore, the probability is:
[tex]P(n>6)=\frac{2}{8}=\frac{1}{4}[/tex]Now we determine the product of both probabilities:
[tex]P(\text{head and n>6)=}\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}[/tex]Now we rewrite the answer as a decimal:
[tex]P(\text{head and n>6)=}0.125[/tex]Therefore, the probability is 0.125.
Kara categorized her spending for this month into four categories: Rent, Food, Fun, and Other. Theamounts she spent in each category are pictured here.Food$333Rent$417Other$500Fun$250What percent of her total spending did she spend on Fun? Answer to the nearest whole percent.
In this problem we have to calculate the total spences so we add all the costs so:
[tex]\begin{gathered} T=333+417+500+250 \\ T=1500 \end{gathered}[/tex]So 1500 is the 100% so now we can calculate which percentage correspount to 250 so:
[tex]\begin{gathered} 1500\to100 \\ 250\to x \end{gathered}[/tex]so the equation is:
[tex]\begin{gathered} x=\frac{250\cdot100}{1500} \\ x=16.66 \end{gathered}[/tex]So she spend 16.66% in fun
using the converse of the same-side interior angles postulate what equation shows that g∥h
Answer: [tex]\angle 2+\angle 4=180^{\circ}[/tex] or [tex]\angle 1+\angle 3=180^{\circ}[/tex]
0.75 greater than 1/2
True
0.75 is greater than 0.5
Explanation
Step 1
remember
[tex]\frac{a}{b}=\text{ a divided by b}[/tex]then
[tex]\frac{1}{2}=\text{ 1 divided by 2 = 0.5}[/tex]Step 2
compare
0.75 and 0.5
[tex]0.75\text{ is greater than 0.5}[/tex]I hope this helps you
FOR GREATER THAN WE ADD THE TERMS.
MATHEMATICALLY THIS MEANS
[tex] = 0.75 + \frac{1}{2} \\ = 0.75 + 0.5 \\ = 1.25[/tex]
1.25 is the answer.
Anna weighs 132 lb. Determine her mass in kilograms using the conversion 1 kg equal 2.2 lb. Use this mass to answer this question. calculate Anna's weight on Jupiter. (G= 25.9 m/ S2) must include a unit with your answer
Input data
132 lb
132 lb * 1kg / 2.2lb = 60 kg
Anna's weight on Jupiter
w = 60 kg * 25.9 m/S2
w = 1554 N
HELP ASAP!!!
Find the square of 1-4i.
ANSAWER:
−15+8i
Explanation:
First, you can expand the square of the bynomial:
"Solve for x. Enter as a decimal not as a fraction. Round to the nearest hundredth if necessary."
Answer:
x =
5
Explanation
From the given diagram, it can be infered that WY = 2QR
From the diagram
WY = x+9
QR = 2x-3
substitute into the expression
x+9 = 2(2x-3)
x+9 = 4x - 6
Collect the like terms
x-4x = -6-9
-3x = -15
x = -15/-3
x = 5
Hence the value of x is 5
Sparkles the Clown makes balloon animals for children at birthday parties. At Bridget's party, she made 5 balloon poodles and 1 balloon giraffe, which used a total of 15 balloons. For Eduardo's party, she used 7 balloons to make 1 balloon poodle and 1 balloon giraffe. How many balloons does each animal require?
Let p be the number of balloons required to make one balloon poodle and g the number of balloons required to make one balloon giraffe.
Then we have:
I) 5p + g = 15
II) p + g = 7
Subtracting equation II from equation I, we have:
5p - p + g - g = 15 - 7
4p = 8
p = 8/4
p = 2
Replacing p with 2 in equation II we have:
2 + g = 7
g = 7 - 2
g = 5
Answer: Each poodle requires 2 balloons and each giraffe requires 5 balloons.
The length of the hypotenuse in a 30°-60°-90° triangle is 6√10yd. What is thelength of the long leg?
In order to calculate the length of the long leg, we can use the sine relation of the 60° angle.
The sine relation is the length of the opposite side to the angle over the length of the hypotenuse.
So we have:
[tex]\begin{gathered} \sin (60\degree)=\frac{x}{6\sqrt[]{10}} \\ \frac{\sqrt[]{3}}{2}=\frac{x}{6\sqrt[]{10}} \\ 2x=6\sqrt[]{30} \\ x=3\sqrt[]{30} \end{gathered}[/tex]So the length of the long leg is 3√30 yd.
Determine which is the better investment 3.99% compounded semi annually Lee 3.8% compounded quarterly round your answer 2 decimal places
Remember that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]In the 3.99% compounded semiannually
we have
r=3.99%=0.0399
n=2
substitute
[tex]\begin{gathered} A=P(1+\frac{0.0399}{2})^{2t} \\ \\ A=P(1.01995)^{2t} \end{gathered}[/tex]and
[tex]\begin{gathered} A=P[(1.01995)^2]^t \\ A=P(1.0403)^t \end{gathered}[/tex]the rate is r=1.0403-1=0.0403=4.03%
In the 3.8% compounded quarterly
we have
r=3.8%=0.038
n=4
substitute
[tex]\begin{gathered} A=P(1+\frac{0.038}{4})^{2t} \\ A=P(1.0095)^{2t} \\ A=P[(1.0095)^2]^t \\ A=P(1.0191)^t \end{gathered}[/tex]the rate is r=1.0191-1=0.0191=1.91%
therefore
the 3.99% compounded semiannually is a better investmentImagine you asked students to draw an area model for the expression 5+4x2.
Walking around the room, you see the following three area models.
First, briefly explain the student thinking process you think might be behind each answer.
Answer Describe the thinking process
Which order would you call students A, B and C to present their work to the class and how would you guide the discussion?
Answer:
area 1
Step-by-step explanation:
Which of the following could be the points that Jamur plots?
To solve this problem, we need to calculate the midpoint for the two points in each option and check if it corresponds to the given midpoint (-3,4).
Calculating the midpoint for the two points of option A.
We have the points:
[tex](-1,7)and(2,3)[/tex]We label the coordinates as follows:
[tex]\begin{gathered} x_1=-1 \\ y_1=7 \\ x_2=2 \\ y_2=3 \end{gathered}[/tex]And use the midpoint formula:
[tex](\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]Substituting our values:
[tex](\frac{-1_{}+2_{}}{2},\frac{7_{}+3_{}}{2})[/tex]Solving the operations:
[tex](\frac{1_{}}{2},\frac{10_{}}{2})=(\frac{1_{}}{2},5)[/tex]Since the midpoint is not the one given by the problem, this option is not correct.
Calculating the midpoint for the two points of option B.
We have the points:
[tex](-2,6)and(-4,2)[/tex]We follow the same procedure, label the coordinates:
[tex]\begin{gathered} x_1=-2 \\ y_1=6 \\ x_2=-4 \\ y_2=2 \end{gathered}[/tex]And use the midpoint formula:
[tex]\begin{gathered} (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ \text{Substituting our values} \\ (\frac{-2-4_{}}{2},\frac{6+2_{}}{2}) \\ \text{Solving the operations:} \\ (\frac{-6}{2},\frac{8}{2}) \\ (-3,4) \end{gathered}[/tex]The midpoint for the two points in option B is (-3,4) which is the midpoint given by the problem.
Answer: B (-2,6) and (-4,2)
find a slope of the line that passes through (8,8) and (1,9)
The slope formula is
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]we can use this formula by introducing the values of the given points. In our case
[tex]\begin{gathered} (x_1,y_1)=(8,8) \\ (x_2,y_2)=(1,9) \end{gathered}[/tex]Hence, we have
[tex]m=\frac{9-8}{1-8}[/tex]It yields,
[tex]m=\frac{1}{-7}[/tex]hence, the answer is
[tex]m=-\frac{1}{7}[/tex]quadrilateral WXYZ is reflected across the line y=x to create quadrilateral W’X’Y’Z'. What are the coordinates of quadrilateral W’X’Y’Z'.
Explanation
We are required to determine the coordinates of W’X’Y’Z' when WXYZ is reflected across the line y = x.
This is achieved thus:
From the image, we can deduce the following:
[tex]\begin{gathered} W(-7,3) \\ X(-5,6) \\ Y(-3,7) \\ Z(-2,3) \end{gathered}[/tex]We know that the following reflection rules exist:
Therefore, we have:
[tex]\begin{gathered} (x,y)\to(y,x) \\ W(-7,3)\to W^{\prime}(3,-7) \\ X(-5,6)\to X^{\prime}(6,-5) \\ Y(-3,7)\to Y^{\prime}(7,-3) \\ Z(-2,3)\to Z^{\prime}(3,-2) \end{gathered}[/tex]Hence, the answers are:
[tex]\begin{gathered} \begin{equation*} W^{\prime}(3,-7) \end{equation*} \\ \begin{equation*} X^{\prime}(6,-5) \end{equation*} \\ \begin{equation*} Y^{\prime}(7,-3) \end{equation*} \\ \begin{equation*} Z^{\prime}(3,-2) \end{equation*} \end{gathered}[/tex]This is shown in the graph bwlow for further undertanding: