Answer:
x = 25
Step-by-step explanation:
Since the triangles are congruent then corresponding angles are congruent.
That is ∠ D and ∠ B are corresponding angles and congruent, thus
4x - 20 = 2x + 30 ( subtract 2x from both sides )
2x - 20 = 30 ( add 20 to both sides )
2x = 50 ( divide both sides by 2 )
x = 25
HELP ASAP I NEED THIS RIGHTNOW 30 points
Answer:
Pretty sure it is c
Step-by-step explanation:
Answer:
C.
Step-by-step explanation:
She will be painting the outsides of the table, so we need to find the surface area of the table.
There is the flat part of the table, which is a rectangular prism. There are also four legs, which are rectangular prism.
So, she will paint C. the surface area of 6 rectangular prisms.
Hope this helps!
among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 played both hockey and volleyball 15 played cricket and volley ball and 10 played all three. if every student played at least 1 game find the no of students and how many students played only cricket, only hockey and only volley ball
Answer:
Cricket only= 30
Volleyball only = 15
Hockey only = 25
Explanation:
Number of students that play cricket= n(C)
Number of students that play hockey= n(H)
Number of students that play volleyball = n(V)
From the question, we have that;
n(C) = 50, n(H) = 50, n(V) = 40
Number of students that play cricket and hockey= n(C∩H)
Number of students that play hockey and volleyball= n(H∩V)
Number of students that play cricket and volleyball = n(C∩V)
Number of students that play all three games= n(C∩H∩V)
From the question; we have,
n(C∩H) = 15
n(H∩V) = 20
n(C∩V) = 15
n(C∩H∩V) = 10
Therefore, number of students that play at least one game
n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)
= 50 + 50 + 40 – 15 – 20 – 15 + 10
Thus, total number of students n(U)= 100.
Note;n(U)= the universal set
Let a = number of people who played cricket and volleyball only.
Let b = number of people who played cricket and hockey only.
Let c = number of people who played hockey and volleyball only.
Let d = number of people who played all three games.
This implies that,
d = n (CnHnV) = 10
n(CnV) = a + d = 15
n(CnH) = b + d = 15
n(HnV) = c + d = 20
Hence,
a = 15 – 10 = 5
b = 15 – 10 = 5
c = 20 – 10 = 10
Therefore;
For number of students that play cricket only;
n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30
For number of students that play hockey only
n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25
For number of students that play volleyball only
n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15
Answer:
Cricket only= 30
Volleyball only = 15
Hockey only = 25
Explanation of the answer:
Number of students that play cricket= n(C)
Number of students that play hockey= n(H)
Number of students that play volleyball = n(V)
From the question, we have that;
n(C) = 50, n(H) = 50, n(V) = 40
Number of students that play cricket and hockey= n(C∩H)
Number of students that play hockey and volleyball= n(H∩V)
Number of students that play cricket and volleyball = n(C∩V)
Number of students that play all three games= n(C∩H∩V)
From the question; we have,
n(C∩H) = 15
n(H∩V) = 20
n(C∩V) = 15
n(C∩H∩V) = 10
Therefore, number of students that play at least one game
n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)
= 50 + 50 + 40 – 15 – 20 – 15 + 10
Thus, total number of students n(U)= 100.
Note;n(U)= the universal set
Let a = number of people who played cricket and volleyball only.
Let b = number of people who played cricket and hockey only.
Let c = number of people who played hockey and volleyball only.
Let d = number of people who played all three games.
This implies that,
d = n (CnHnV) = 10
n(CnV) = a + d = 15
n(CnH) = b + d = 15
n(HnV) = c + d = 20
Hence,
a = 15 – 10 = 5
b = 15 – 10 = 5
c = 20 – 10 = 10
Therefore;
For number of students that play cricket only;
n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30
For number of students that play hockey only
n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25
For number of students that play volleyball only
n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15
▬▬▬▬▬▬▬▬▬▬▬▬
find the maximum value of c=6x+2y
Answer:
∞
Step-by-step explanation:
c can have any value you like.
There is no maximum. We say it can approach infinity.
__
Additional comment
There may be some maximum imposed by constraints not shown here. Since we don't know what those constraints are, we cannot tell you what the maximum is.
Use spherical coordinates. Find the volume of the solid that lies within the sphere x2 + y2 + z2 = 16, above the xy-plane, and below the cone z = x2 + y2 .
The volume is given by the integral,
[tex]\displaystyle\int_0^{2\pi}\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\int_0^4\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta[/tex]
That is, [tex]\rho[/tex] ranges from the origin to the sphere of radius 4. The range for [tex]\varphi[/tex] starts at the intersection of the cone [tex]z=x^2+y^2[/tex] with the sphere [tex]x^2+y^2+z^2=16[/tex], which gives
[tex]z+z^2=16\implies z^2+z-16=0\implies z=\dfrac{\sqrt{65}-1}2[/tex]
and
[tex]z=4\cos\varphi\implies\varphi=\cos^{-1}\left(\dfrac{\sqrt{65}-1}8\right)[/tex]
and extends to the x-y plane where [tex]\varphi=\frac\pi2[/tex]. The range for [tex]\theta[/tex] is self-evident.
The volume is then
[tex]V=\displaystyle\int_0^{2\pi}\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\int_0^4\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta[/tex]
[tex]V=\displaystyle\left(\int_0^{2\pi}\mathrm d\theta\right)\left(\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\sin\varphi\,\mathrm d\varphi\right)\left(\int_0^4\rho^2\,\mathrm d\rho\right)[/tex]
[tex]V=2\pi\left(\dfrac{\sqrt{65}-1}8\right)\left(\dfrac{64}3\right)=\boxed{\dfrac{16\pi(9-\sqrt{65})}3}[/tex]
A sphere is a three-dimensional object with a round form. The volume of the sphere is [16π(9-√65)]/3 unit³.
What is a sphere?A sphere is a three-dimensional object with a round form. A sphere, unlike other three-dimensional shapes, has no vertices or edges. Its centre is equidistant from all places on its surface. In other words, the distance between the sphere's centre and any point on its surface is the same.
We know that the volume of the given sphere can be given by the integral,
[tex]{\rm Volume} = \int^{2\pi}_0\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} \int_0^4\rho^2sin\varphi\ d\rho\ d\varphi\ d\theta[/tex]
where ρ ranges from the origin of the plot to the sphere of radius 4 while the range of φ starts at the intersection of the cone z=x²+y² with the sphere x²+y²+z²=16.
Now, the value of z and φ can be written as,
[tex]x^2+y^2+z^2 = 16\\\\(x^2+y^2)+z^2 = 16\\\\z+z^2 = 16\\\\z^2+z-16=0 \implies z=\dfrac{\sqrt{65}-1}{2}[/tex]
And
[tex]z =4\ cos\ \varphi \implies \varphi =cos^{-1}(\dfrac{\sqrt{65}-1}{8})[/tex]
Further, the volume of the sphere can be written as,
[tex]{\rm Volume} = \int^{2\pi}_0\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} \int_0^4\rho^2sin\varphi\ d\rho\ d\varphi\ d\theta\\\\\\{\rm Volume} = (\int^{2\pi}_0\ d\theta)(\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} sin\varphi\ d\varphi)(\int_0^4\rho^2 d\rho)\\\\\\V = 2\pi(\dfrac{\sqrt{65}-1}{8})(\dfrac{64}{3}) = \dfrac{16\pi(9-\sqrt{65})}{3}[/tex]
Hence, the volume of the sphere is [tex]\dfrac{16\pi(9-\sqrt{65})}{3}[/tex].
Learn more about Sphere:
https://brainly.com/question/11374994
Suppose the weather forecast calls for a 60% chance of rain each day for the next 3 days. What is the probability that it will NOT rain during the next 3 days
Answer:
Probability that it'll not rain during the next three days = 0.064
Step-by-step explanation:
Given
Let:
P(R) represent the probability that it'll rain each day
P(R') represent the probability that it'll not
[tex]P(R) = 60\%[/tex]
Required
Probability that it'll not rain during the next three days
From concept of probability;
[tex]P(R) + P(R') = 1[/tex]
Substitute 60% for P(R)
[tex]60\% + P(R') = 1[/tex]
Subtract 60% from both sides
[tex]60\% - 60\% + P(R') = 1 - 60\%[/tex]
[tex]P(R') = 1 - 60\%[/tex]
Convert % to decimal
[tex]P(R') = 1 - 0.6[/tex]
[tex]P(R') = 0.4[/tex]
The probability that it'll not rain during the next 3 days is:
[tex]P(R') * P(R') * P(R')[/tex]
[tex]P(R') * P(R') * P(R') =0.4 * 0.4 * 0.4[/tex]
[tex]P(R') * P(R') * P(R') = 0.064[/tex]
Tabitha wants to hang a painting in a gallery. The painting and frame must have an area of 58 square feet. The painting is 7 feet wide by 8 feet long. Which quadratic equation can be used to determine the thickness of the frame, x? (5 points)
Answer:
4x² + 30x - 2 = 0
Step-by-step explanation:
Given:
Area = 58 square feet
Width = 7 feet
Length = 8 feet
Since the area is 58, writing the equation, we have:
(8 + 2x)(7 + 2x) = 58
Now expand the equation:
56 + 16x + 14x + 4x² = 58
56 + 30x + 4x² = 58
Collect like terms:
30x + 4x² + 56 - 58 = 0
30x + 4x² - 2 = 0
Rearrange the equation to a proper quadratic equation:
4x² + 30x - 2 = 0
The quadratic equation that can be used to determine the thickness of the frame, x is 4x² + 30x - 2 = 0
Tristan wants to buy a car and has a choice between two different banks. One bank is offering a simple interest rate of 3% and the other bank is offering a rate of 2.5%
compounded annually. If Tristan decides to deposit $7,000 for 4 years, which bank would be the better deal?
Answer:
The better deal would be simple interest rate of 3%
Step-by-step explanation:
In order to calculate which bank would be the better deal If Trsitam decides to deposit $7,000 for 4 years, we would have to make the following calculation:
simple interest rate of 3%.
Therefore, I= P*r*t
=$7,000*3%*4
I=$840
FV= $7,000+$840
FV=7,840
compound interest rate of 2.5%
Therefore, FV=PV(1+r)∧n
FV=$7,000(1+0.25)∧4
FV=$17,089
The better deal would be simple interest rate of 3%
WILL MARK AS BRAINLIEST!!! 5. A 2011 study by The National Safety Council estimated that there are nearly 5.7 million traffic accidents year. At least 28% of them involved distracted drivers using cell phones or texting. The data showed that 11% of drivers at any time are using cell phones . Car insurance companies base their policy rates on accident data that shows drivers have collisions approximately once every 19 years. That’s a 5.26% chance per year. Given what you know about probability, determine if cell phone use while driving and traffic accidents are related. Step A: Let DC = event that a randomly selected driver is using a cell phone. What is P(DC)? (1 point) Step B: Let TA = event that a randomly selected driver has a traffic accident. What is P(TA)? Hint: What is the probability on any given day? (1 point) Step C: How can you determine if cell phone use while driving and traffic accidents are related? (1 point) Step D: Given that the driver has an accident, what is the probability that the driver was distracted by a cell phone? Write this event with the correct conditional notation. (1 point) Step E: What is the probability that a randomly selected driver will be distracted by using a cell phone and have an accident? (2 points) Step F: For a randomly selected driver, are the events "driving while using a cell phone" and "having a traffic accident" independent events? Explain your answer. (2 points)
Answer:
Step-by-step explanation:
Hello!
Regarding the reasons that traffic accidents occur:
28% are caused by distracted drivers using cell phones or texting
11% of the drivers' user their phones at any time
The probability of a driver having an accident is 5.26%
a)
DC = event that a randomly selected driver is using a cell phone.
P(DC)= 0.11
b)
TA = event that a randomly selected driver has a traffic accident.
P(TA)= 0.0526
c) and f)
If both events are related, i.e. dependent, then you would expect that the occurrence of one of these events will affect the probability of the other one. If they are not related, i.e. independent events, then their probabilities will not be affected by the occurrence of one or another:
If both events are independent P(TA|DC)= P(TA)
If they are dependent, then:
P(TA|DC)≠ P(TA)
P(TA|DC)= 0.28
P(TA)= 0.0526
As you can see the probability of the driver having an accident given that he was using the cell phone is different from the probability of the driver having an accident. This means that both events are related.
d) and e)
You have to calculate the probability that "the driver was distracted with the phone given that he had an accident", symbolically P(DC|TA)
P(DC|TA) = [tex]\frac{P(DCnTA)}{P(TA)}[/tex]
[tex]P(TA|DC)= \frac{P(TAnDC}{P(DC)}[/tex] ⇒ P(DC∩TA)= P(TA|DC)*P(DC)= 0.28 * 0.11= 0.0308
P(DC|TA) = [tex]\frac{0.0308}{0.0526}= 0.585= 0.59[/tex]
I hope this helps!
What is the degree of the polynomial?
Answer: 3rd Degree
Step-by-step explanation:
What is x when: |5x|=3
Answer:
3/5
Step-by-step explanation:
5x= 3
x= 3/5
hope you understand the answer
stay at home stay safe
keep rocking
pls mark me as BRAINLIEST
Change Y - 4X = 0 to the slope-intercept form of the equation of a line.
Answer:
y=4x
Step-by-step explanation:
Add 4x to both sides to get y=mx+b
0 is y-intercept.
4x is the slope.
Select the correct answer from each drop down menu.
The slope of diagonal OA is
A) 4/3
B) 3/4
C) 1
,and it’s equation is
A) 4x-y=0
B) x-3y=0
C) 4x-3y=0
Answer:
(A) [tex]\text{The slope of OA is }\dfrac{4}{3}[/tex]
(C) It’s equation is 4x-3y=0.
Step-by-step explanation:
Point O is at (0,0)
Point A is at (3,4)
[tex]\text{Slope of OA}=\dfrac{4-0}{3-0} \\m=\dfrac{4}{3}[/tex]
The equation of a straight line is in the form: y=mx+b
The y-intercept of the line OA=0
Therefore, we have:
[tex]y=\dfrac{4}{3}x+0\\3y=4x+0\\$Subtract 3y from both sides$\\4x-3y=0[/tex]
The equation of the line is: 4x-3y=0
In a random sample of 400 residents of Boston, 320 residents indicated that they voted for Obama in the last presidential election. Develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.
Answer:
C.I = 0.7608 ≤ p ≤ 0.8392
Step-by-step explanation:
Given that:
Let consider a random sample n = 400 candidates where 320 residents indicated that they voted for Obama
probability [tex]\hat p = \dfrac{320}{400}[/tex]
= 0.8
Level of significance ∝ = 100 -95%
= 5%
= 0.05
The objective is to develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.
The confidence internal can be computed as:
[tex]=\hat p \pm Z_{\alpha/2} \sqrt{\dfrac{ p(1-p)}{n } }[/tex]
where;
[tex]Z_{0.05/2}[/tex] = [tex]Z_{0.025}[/tex] = 1.960
SO;
[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(1-0.8)}{400 } }[/tex]
[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(0.2)}{400 } }[/tex]
[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.16}{400 } }[/tex]
[tex]=0.8 \pm 1.960 \sqrt{4 \times 10^{-4}}[/tex]
[tex]=0.8 \pm 1.960 \times 0.02}[/tex]
[tex]=0.8 \pm 0.0392[/tex]
= 0.8 - 0.0392 OR 0.8 + 0.0392
= 0.7608 OR 0.8392
Thus; C.I = 0.7608 ≤ p ≤ 0.8392
helppppppppp pleasee me give bralienst,stars and thanks
Answer:
(Going from left to right)
Box #1=3
Box #2=5
Box #3=7
Box #4=2
Step-by-step explanation:
For Box #4, there is nothing for the 2 to subtract from so it just goes down
For Box #3, it has to be 7, because nothing can be subtracted from 1 to get 3, so you would have to bring a 1 from the 4 to the left to make the 1 to a 10. 10-7=3
For Box #2 and 1, 3(we changed it in the last step) -9 = a negative number so we have to bring a 1 from the number to the left. This is a hard step but what you have to do it look at the bottom number, which is a 2, so that number had to be a 3 because 3-1=2. 4 becomes 14, and 14-9=5
Hope this helps, if it did, please consider giving me brainliest, it will help me a lot. If you have any questions, feel free to ask.
Have a good day! :)
At noon, ship A is 170 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM
Answer:
The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.
Step-by-step explanation:
Vectorially speaking, let assume that ship A is located at the origin and the relative distance of ship B with regard to ship A at noon is:
[tex]\vec r_{B/A} = \vec r_{B} - \vec r_{A}[/tex]
Where [tex]\vec r_{A}[/tex] and [tex]\vec r_{B}[/tex] are the distances of ships A and B with respect to origin.
By supposing that both ships are travelling at constant speed. The equations of absolute position are described below:
[tex]\vec r_{A} = \left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i[/tex]
[tex]\vec r_{B} = \left(170\,km\right)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j[/tex]
Then,
[tex]\vec r_{B/A} = (170\,km)\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j-\left[\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i[/tex]
[tex]\vec r_{B/A} = \left[170\,km-\left(40\,\frac{km}{h} \right)\cdot t\right]\cdot i +\left[\left(15\,\frac{km}{h} \right)\cdot t\right]\cdot j[/tex]
The rate of change of the distance between the ship is constructed by deriving the previous expression:
[tex]\vec v_{B/A} = -\left(40\,\frac{km}{h} \right)\cdot i + \left(15\,\frac{km}{h} \right)\cdot j[/tex]
Its magnitude is determined by means of the Pythagorean Theorem:
[tex]\|\vec v_{B/A}\| = \sqrt{\left(-40\,\frac{km}{h} \right)^{2}+\left(15\,\frac{km}{h} \right)^{2}}[/tex]
[tex]\|\vec r_{B/A}\| \approx 42.720\,\frac{km}{h}[/tex]
The distance between the ships is changing at 42.720 kilometers per hour at 4:00 PM.
A man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the constant rate of 45 mph . The man can run at a constant rate in either direction to get off the bridge just in time before the train hits him. How fast can the man run?
Answer:
The Man needs to run at 9 mph
Step-by-step explanation:
Let M stand for the man's speed in mph. When the man
runs toward point A, the relative speed of the train with respect
to the man is the train's speed plus the man's speed (45 + M).
When he runs toward point B, the relative speed of the train is the
train's speed minus the man's speed (45 - M).
When he runs toward the train the distance he covers is 2 units.
When he runs in the direction of the train the distance he covers
is 3 units. We can now write that the ratio of the relative speed
of the train when he is running toward point A to the relative speed
of the train when he is running toward point B, is equal to the
inverse ratio of the two distance units or
(45 + M) 3
----------- = ---
(45 - M) 2
90+2 M=135-3 M
⇒5 M = 45
⇒ M = 9 mph
The Man needs to run at 9 mph
Answer: 9 mph
Step-by-step explanation:
Given that a man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the constant rate of 45 mph .
If the man tend to run in the forward direction, he will cover another 2/5 before the train reaches his initial position. The distance covered by the man will be 2/5 + 2/5 = 4/5
The remaining distance = 1 - 4/5 = 1/5
If the man can run at a constant rate in either direction to get off the bridge just in time before the train hits him, the time it will take the man will be
Speed = distance/time
Time = 1/5d ÷ speed
The time it will take the train to cover the entire distance d will be
Time = d ÷ 45
Equate the two time
1/5d ÷ speed = d ÷ 45
Speed = d/5 × 45/d
Speed = 9 mph
Can someone help me with this question
Answer:
Product of given question is 6√5
Step-by-step explanation:
Given:
3√2(√10)
Find:
Product.
Computation:
⇒ 3√2(√10)
⇒ 3√20
⇒ 3√4×5
⇒ 3√2×2×5
⇒ 3×2√5
⇒ 6√5
Product of given question is 6√5
According to a study done by the Gallup organization, the proportion of Americans who are satisfied with the way things are going in their lives is 0.82. What is the probability the sample proportion who are satisfied with the way things are going in their life is greater than 0.85
Complete Question
According to a study done by the Gallup organization, the proportion of Americans who are satisfied with the way things are going in their lives is 0.82. Suppose a random sample of 100 Americans is asked "Are you satisfied with the way things are going in your life?"
What is the probability the sample proportion who are satisfied with the way things are going in their life is greater than 0.85
Answer:
The probability is [tex]P(X > 0.85 ) = 0.21745[/tex]
Step-by-step explanation:
From the question we are told that
The population proportion is [tex]p = 0.82[/tex]
The value considered is x = 0.85
The sample size is n = 100
The standard deviation for this population proportion is evaluated as
[tex]\sigma = \sqrt{\frac{p(1-p)}{n} }[/tex]
substituting values
[tex]\sigma = \sqrt{\frac{0.82(1-0.82)}{100} }[/tex]
[tex]\sigma = 0.03842[/tex]
Generally the probability that probability the sample proportion who are satisfied with the way things are going in their life is greater than x is mathematically represented as
[tex]P(X > x ) = P( \frac{X - p }{ \sigma } > \frac{x - p }{ \sigma } )[/tex]
Where [tex]\frac{X - p }{ \sigma }[/tex] is equal to Z (the standardized value of X ) so
[tex]P(X > x ) = P( Z> \frac{x - p }{ \sigma } )[/tex]
substituting values
[tex]P(X > 0.85 ) = P( Z> \frac{ 0.85 - 0.82 }{ 0.03842 } )[/tex]
[tex]P(X > 0.85 ) = P( Z> 0.78084)[/tex]
from the standardized normal distribution table [tex]P( Z> 0.78084)[/tex] is 0.21745
So
[tex]P(X > 0.85 ) = 0.21745[/tex]
Farmers Jay, Peter, and Sam own rectangular farms, as indicated in the figure. Jay owns 2 acres of land, Peter owns 4 acres and Sam owns 6 acres. Find the area of the common pasture. PLEASE HELP ASAP!
Answer:
Area of the common pasture = 12 acres
Step-by-step explanation:
Let the dimensions of the farm owned by Jay are 'a' units and 'b' units.
Area of the farm = ab = 2 acres
Similarly, areas of the farm owned by Peter with dimensions 'a' unit and 'c' unit = ac = 4 acres
And area of the farm owned by Sam with dimensions 'b' and 'd' units = bd = 6 acres
Now, [tex]\frac{ab}{ac}=\frac{2}{4}[/tex]
[tex]\frac{b}{c}=\frac{1}{2}[/tex] ---------(1)
[tex]\frac{ab}{bd}=\frac{2}{6}[/tex]
[tex]\frac{a}{d}=\frac{1}{3}[/tex] ---------(2)
[tex]\frac{b}{c}\times \frac{a}{d}=\frac{1}{2}\times \frac{1}{3}[/tex]
[tex]\frac{ab}{cd}=\frac{1}{6}[/tex]
cd = 6(ab)
cd = 6 × 2 [Since ab = 2 acres]
= 12 acres
Therefore, area of the common pasture will be 12 acres.
Find m<1 .Triangle Angle-Sum Theorem.
Answer:
m<1 = 30
Step-by-step explanation:
To find m<1, we can do 180 - 75 - 75, which will give us 30 degrees, so m<1 = 30
A rectangle has length 4 inches and width 2 inches. If the length and width of the rectangle are
reduced by 50 percent, by what percent will the area of the rectangle be reduced?
40 percent
50 percent
60 percent
75 percent
Answer:
75%
Step-by-step explanation:
First we can solve the area of the rectangle originally the answer is:
4 × 2 = 8
Then we decrease both measurements by 50% to get the dimensions 1 and 2. The new area will be 1 × 2 which is 2.
2 is 25% of 8 which means that the area of the rectangle has been reduced by 75%.
helpppppp pleaseeee me helpppp
Answer:
$10 + $10 + $1 + 25¢ + 5¢
or
$20 + $1 + 25¢ + 5¢
Step-by-step explanation:
Each one must pay $21.30
Section 8
Find the mean of these numbers:
24 18
37
82 17
26
Answer:
[tex]\boxed{Mean = 34.33}[/tex]
Step-by-step explanation:
Mean = Sum of Observations / No. Of Observations
Mean = (24+18+37+82+17+26)/6
Mean = 206 / 6
Mean = 34.33
um aluno tinha uma quantidade de questoes para resolver se ja resolveu a quinta parte da sua tarefa então a razão entre o numero de questoes resolvidas e o numero restante de questoes nessa ordem é a) 1/20 b)1/5 c)1/4 d)4 e)5
Answer:
can you please simplify that in English?
1. What is the length of the shortest side if the perimeter of the rectangle is
56 inches?
3х
5х – 4
Answer:
Length of Shortest Side = 12 inches
Step-by-step explanation:
Length of Shortest Side = L = 3x
Length of Longest Side = W = 5x-4
Condition:
2L+2W = Perimeter
2(3x)+2(5x-4) = 56
6x+10x-8 = 56
16x-8 = 56
Adding 8 to both sides
16x = 56+8
16x = 64
Dividing both sides by 14
=> x = 4
Now,
Length of the Shortest Side = L = 3(4) = 12 inches
Length of the Longest Side = W = 5(4)-4 = 16 inches
Answer:
12 inches
Step-by-step explanation:
The length is the longest side.
The width is the shortest side.
Length : [tex]l=5x-4[/tex]
Width : [tex]w=3x[/tex]
Apply formula for the perimeter of a rectangle.
[tex]P=2l+2w[/tex]
[tex]P=perimeter\\l=length\\w=width[/tex]
Plug in the values.
[tex]56=2(5x-4)+2(3x)[/tex]
[tex]56=10x-8+6x[/tex]
[tex]56=16x-8[/tex]
[tex]64=16x[/tex]
[tex]4=x[/tex]
The shortest side is the width.
[tex]w=3x[/tex]
Plug in the value for x.
[tex]w=3(4)[/tex]
[tex]w=12[/tex]
In a study of the accuracy of fast food drive-through orders, Restaurant A had 302accurate orders and 59that were not accurate.a. Construct a 95%confidence interval estimate of the percentage of orders that are not accurate.b. Compare the results from part (a) to this 95%confidence interval for the percentage of orders that are not accurate at Restaurant B: 0.143less thanpless than0.219.What do you conclude?
Answer:
(a) A 95% confidence interval estimate of the percentage of orders that are not accurate is [0.125, 0.201].
(b) We can conclude that both restaurants can have the same inaccuracy rate due to the overlap of interval areas.
Step-by-step explanation:
We are given that in a study of the accuracy of fast food drive-through orders, Restaurant A had 302 accurate orders and 59 orders that were not accurate.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of orders that were not accurate = [tex]\frac{59}{361}[/tex] = 0.163
n = sample of total orders = 302 + 59 = 361
p = population proportion of orders that are not accurate
Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.
So, 95% confidence interval for the population proportion, p is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95
P( [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95
95% confidence interval for p = [ [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]
= [ [tex]0.163 -1.96 \times {\sqrt{\frac{0.163(1-0.163)}{361} } }[/tex] , [tex]0.163 +1.96 \times {\sqrt{\frac{0.163(1-0.163)}{361} } }[/tex] ]
= [0.125, 0.201]
(a) Therefore, a 95% confidence interval estimate of the percentage of orders that are not accurate is [0.125, 0.201].
(b) We are given that the 95% confidence interval for the percentage of orders that are not accurate at Restaurant B is [0.143 < p < 0.219].
Here we can observe that there is a common area of inaccurate order of 0.058 or 5.85% for both the restaurants.
So, we can conclude that both restaurants can have the same inaccuracy rate due to the overlap of interval areas.
What is the rate of change of the function
The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.
The following data values represent a sample. What is the variance of the
sample? X = 8. Use the information in the table to help you.
х
12
9
11
5
3
(x; - x)²
16
1
9
9
25
Answer:
The variance of the data is 15.
σ² = 15
Step-by-step explanation:
The mean is given as
X = 8
х | (x - X) | (x - X) ²
12 | 4 | 16
9 | 1 | 1
11 | 3 | 9
5 | -3 | 9
3 | -5 | 25
The variance is given by
[tex]\sigma^2 = \frac{1}{n-1} \sum (x - X)^2[/tex]
[tex]\sigma^2 = \frac{1}{5 - 1} (16 + 1 + 9 + 9 +25) \\\\\sigma^2 = \frac{1}{4} ( 16 + 1 + 9 + 9 +25) \\\\\sigma^2 = \frac{1}{4} (60) \\\\\sigma^2 = 15[/tex]
Therefore, the variance of the data is 15.
1/3 x =6 What would like match this answer
Answer:
x=18
Step-by-step explanation:
Answer:
x = 18
Step-by-step explanation:
Since 1/3x = 6,
x = 6 x 3
Thus, x = 18
Find all of the angle measures in the image.
Answer:
Angle 2= 45
Angle 3= 45
Angle 4= 135
Angle 5= 135
Angle 6= 45
Angle 7= 45
Angle 8= 135