Answer:
i dont see no angle
Step-by-step explanation:
Angles can be named according to their measure in degrees or position on a plane. Angles are always counted to be positive in the counter-clockwise direction and negative in the clockwise direction.
Explanation:The name of an angle in Mathematics can vary based on its measure and location. Angles are generally named according to their measure in degrees, and they can be categorized as acute (less than 90 degrees), right (90 degrees), obtuse (greater than 90 degrees but less than 180 degrees), straight (180 degrees), reflex (greater than 180 degrees), or full (360 degrees).
Angles can also be referred to in terms of their position on a Cartesian plane: angles in the first quadrant are positive and less than 90; in the second quadrant, they are more than 90 but less than 180; in the third quadrant, they exceed 180 but are less than 270; and in the fourth quadrant, they are greater than 270 but less than 360 degrees.
This consistently maintains the rule that angles are defined as positive in the counter-clockwise direction, and negative in the clockwise direction. For example, an angle of 30° south of west is the same as the global angle 210°, or it can also be expressed as −150° relative to the positive x-axis.
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ΔWXY, the measure of ∠Y=90°, WY = 8, YX = 15, and XW = 17. What ratio represents the tangent of ∠X?
In a right triangle, the tangent of an angle is defined as the ratio of the opposite side to the adjacent side. Therefore, for triangle ΔWXY, the tangent of ∠X is the ratio of side WY to YX, which is 8/15.
Explanation:To understand this question, we need to know that in the context of a right triangle, the tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side. In ΔWXY, where the measure of ∠Y=90°, ∠X is the angle we are considering. The side opposite to ∠X is WY and the side adjacent to ∠X is YX. Therefore, the tangent of ∠X can be calculated using the formula: tan(X) = WY / YX.
In this scenario, we know that WY = 8 and YX = 15. So, the tangent of ∠X is given by: tan(X) = WY / YX = 8 / 15. Hence, the ratio that represents the tangent of ∠X is 8 / 15.
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The area of a triangle that is similar to the one below is the area of this triangle. What is the length of the base of the similar triangle? 2.3 ft 3.3 ft 7 ft 63 ft
The answer is 7 feet
Answer:
c. 7ft
good luck, i hope this helps :)
Researchers are studying two populations of sea turtles. In population D, 30 percent of the turtles have a shell length greater than 2 feet. In population E, 20 percent of the turtles have a shell length greater than 2 feet. From a random sample of 40 turtles selected from D, 15 had a shell length greater than 2 feet. From a random sample of 60 turtles selected from E, 11 had a shell length greater than 2 feet. Let pˆD represent the sample proportion for D, and let pˆE represent the sample proportion for E.
Complete question:
Researchers are studying two populations of sea turtles. In population D, 30 percent of the turtles have a shell length greater than 2 feet. In population E, 20 percent of the turtles have a shell length greater than 2 feet. From a random sample of 40 turtles selected from D, 15 had a shell length greater than 2 feet. From a random sample of 60 turtles selected from E, 11 had a shell length greater than 2 feet. Let pˆD represent the sample proportion for D, and let pˆE represent the sample proportion for E.
(a) What is the value of the difference pˆD−pˆE? Show your work.
(b) What are the mean and standard deviation of the sampling distribution of the difference in sample proportions pˆD−pˆE? Show your work and label each value.
(c) Can it be assumed that the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal? Justify your answer.
(d) Consider your answer in part (a). What is the probability that pˆD−pˆE is greater than the value found in part (a)?
Answer:
a) 0.1917
b) The mean is 0.1917 and the standard deviation is 0.0914.
c)Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal
d) 0.1580
Step-by-step explanation:
a) for p`D we have:
[tex] \frac{15}{40} [/tex]
= 0.375
For p`E we have:
[tex] \frac{11}{60} [/tex]
= 0.1833
Therefore, p`D - p`E, we have:
0.375 - 0.1833
=0.1917
b) The Mean can be calculated as p`D - p`E =
0.375 - 0.1833
=0.1917
For standard deviation:
[tex] s.d = \sqrt{\frac{p`D (1-p`D)}{N_D} + \frac{p`E(1-p`E)}{N_E}}[/tex]
[tex] s.d = \sqrt{\frac{0.375(1 - 0.375)}{40} + \frac{0.1833(1 - 0.1833)}{60}}= 0.0914[/tex]
The mean is 0.1917 and the standard deviation is 0.0914.
c) Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal, because for normal condition, we have:
i) np ≥ 10
ii) n(1-p) ≥ 10.
From the expressions, we can see the samples satisfy the condition for normality.
d) To get the probability, wen need to find the Z score.
The Z score can be calculated using the formula:
[tex] Z = \frac{(p`D - p`E) -(pD - pE)}{s.d}[/tex]
[tex] = \frac{(0.1917) -(0.1)}{0.0914}[/tex]
= 1.0029
Therefore,
P(Z > 1.0029) = 1 - P(Z ≤ 1.0029)
From the z distribution table, we have:
P (Z > 1.0029) = 1 - 0.8420 = 0.1580
The probability is 0.1580
The value of the difference pˆD−pˆE is 0.1917. The mean and standard deviation is 0.1917 and 0.0914 respectively and the probability that pˆD−pˆE is greater than the 0.1917 is 0.1580.
Given :
In population D, 30 percent of the turtles have a shell length greater than 2 feet.In population E, 20 percent of the turtles have a shell length greater than 2 feet.From a random sample of 40 turtles selected from D, 15 had a shell length greater than 2 feet.From a random sample of 60 turtles selected from E, 11 had a shell length greater than 2 feet.a) The value of pˆD is:
[tex]=\dfrac{15}{40}[/tex]
The value of pˆE is:
[tex]=\dfrac{11}{60}[/tex]
So, the value of (pˆD - pˆE) is:
[tex]=\dfrac{15}{40}-\dfrac{11}{60}[/tex]
= 0.375 - 0.1833
= 0.1917
b) Mean is given by the formula:
pˆD - pˆE = 0.1917
For standard deviation using the formula:
[tex]\rm SD =\sqrt{ \dfrac{p\hat{}D(1-p\hat{}D)}{N_D}+\dfrac{p\hat{}E(1-p\hat{}E)}{N_E}}[/tex]
[tex]\rm SD = \sqrt{\dfrac{0.375(1-0.375)}{40}+\dfrac{0.1833(1-0.1833)}{60}}[/tex]
SD = 0.0914
c). Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal.
d). To determine the probability, first evaluate the z-score.
[tex]\rm Z=\dfrac{(p\hat{}D-p\hat{}E)-(pD-pE)}{SD}[/tex]
[tex]\rm Z = \dfrac{0.1917-0.1}{0.0914}[/tex]
Z = 1.0029
Now, P(Z>1.0029) = 1 - P(Z [tex]\leq[/tex] 1.0029)
= 1 - 0.8420
= 0.1580
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A multiple-choice test contains 10 questions. There are four possible answers for each question. a) In how many ways can a student answer the questions on the test if the student answers every question
Answer :
[tex]4^{10}[/tex] =1048576 ways a student can answer the questions on the test if the student answers every question.Step-by-step explanation:
Given that a multiple-choice test contains 10 questions and there are 4 possible answers for each question.
∴ Answers=4 options for each question.
To find how many ways a student can answer the given questions on the test if the student answers every question :Solving this by product rule
Product rule :
If one event can occur in m ways and a second event occur in n ways, the number of ways of two events can occur in sequence is then m.n
From the given the event of choosing the answer of each question having 4 options is given by
The 1st event of picking the answer of the 1st question=4 ,
2nd event of picking the answer of the 2nd question=4 ,
3rd event of picking the answer of the 3rd question=4
,....,
10th event of picking the answer of the 10th question=4.
It can be written as by using the product rule
[tex]=4.4.4.4.4.4.4.4.4.4[/tex]
[tex]=4^{10}[/tex]
[tex]=1048576[/tex]
∴ there are 1048576 ways a student can answer the questions on the test if the student answers every question.The price of a stock is uniformly distributed between $30 and $40. a. Write the probability density function, f(x), for the price of the stock. b. Determine the expected price of the stock. c. Determine the standard deviation for the stock. d. What is the probability that the stock price will be between $34 and $38
Answer:
a) [tex] f(x) = \frac{1}{40-30}, 30 \leq x \leq 40[/tex]
b) [tex] E(X) = \frac{a+b}{2}= \frac{30+40}{2}=35[/tex]
c) [tex] Var(X) = \frac{(b-a)^2}{12}= \frac{(40-30)^2}{12}= 8.33[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{8.33}= 2.887[/tex]
d) [tex]P(34< X <38) = F(38) -F(34)= \frac{38-30}{10} -\frac{34-30}{10}= 0.8-0.4=0.4[/tex]
Step-by-step explanation:
For this case we define the random variable X with this distribution:
[tex] X \sim Unif (a=30, b=40)[/tex]
Part a
The density function since is an uniform distribution is given by:
[tex] f(x) = \frac{1}{40-30}, 30 \leq x \leq 40[/tex]
Part b
The expected value is given by:
[tex] E(X) = \frac{a+b}{2}= \frac{30+40}{2}=35[/tex]
Part c
The variance is given by:
[tex] Var(X) = \frac{(b-a)^2}{12}= \frac{(40-30)^2}{12}= 8.33[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{8.33}= 2.887[/tex]
Part d
For this case we want this probability:
[tex] P(34< X <38)[/tex]
And we can use the cumulative distribution function given by:
[tex] F(x)= \frac{x-30}{40-30}, 30 \leq X \leq 40[/tex]
And using this we got:
[tex]P(34< X <38) = F(38) -F(34)= \frac{38-30}{10} -\frac{34-30}{10}= 0.8-0.4=0.4[/tex]
Final answer:
The probability density function, expected price, standard deviation, and the probability of the stock price being between $34 and $38 are calculated.
Explanation:
To find the probability density function, f(x), for the price of the stock, we need to determine the range of the variable x, and then divide the range by the total probability. In this case, the range of x is from $30 to $40, and the total probability is equal to 1. So, the probability density function is:
f(x) = 1 / (40 - 30) = 1/10 = 0.1
To determine the expected price of the stock, we can calculate the average of the range of x:
Expected price = (30 + 40) / 2 = $35
To determine the standard deviation for the stock, we can use the formula:
Standard deviation = (40 - 30) / sqrt(12) = 10 / sqrt(12) ≈ 2.8879
To find the probability that the stock price will be between $34 and $38, we need to find the area under the probability density function curve between these two prices. Since the probability density function is a uniform distribution, the probability is equal to the width of the range divided by the total width of the distribution:
Probability = (38 - 34) / (40 - 30) = 4 / 10 = 0.4
In repeated samples, approximately 99% of all differences in sample means will fall within the bounds of the interval already computed.
a. True
b. False
Answer:
a) True
Step-by-step explanation:
Repeated samples are a type of samples that are used to determine the features or characteristics or a given set of data.
In repeated samples, statistical techniques are applied whereby two samples that have similar characteristics are tested or analysed under different conditions.
Repeated samples can also be called matched or paired samples.
In repeated samples , we have what we refer to as confidence intervals. These are intervals whereby the true and correct value of certain parameters such as mean, the standard deviation of a given data or distribution is determined. We have confidence interval levels of 90%, 95% and 99%.
In repeated samples, approximately 99% of all differences in sample means will fall within the bounds of the interval already computed.
A softball pitcher has a 0.487 probability of throwing a strike for each pitch. If the softball pitcher throws 29 pitches, what is the probability that no more than 14 of them are strikes?
Answer:
0.4801
Step-by-step explanation:
This is a binomial distribution question.
It can be approximated using normal distribution if the following conditions are met:
np > 10
n(1-p) > 10
Here,
n = 29
p = 0.487
So,
np = 14.12
n(1-p) = 14.88
So, we can use normal approximation here:
Binomial: X ~ B(n,p) becomes
Normal Approx: X~ N([tex]np,\sqrt{np(1-p)}[/tex])
Mean is:
[tex]\mu=np=14.123[/tex]
Standard Deviation is:
[tex]\sigma=\sqrt{np(1-p)} =2.69[/tex]
We need probability of less than or equal to 14, so we can say:
P(x ≤ 14)
Using [tex]z=\frac{x-\mu}{\sigma}[/tex], we have:
P(x ≤ 14) = [tex]P(\frac{x-\mu}{\sigma} \leq \frac{14-14.123}{2.69})\\=P(z \leq -0.05)\\=0.4801[/tex]
Note: We used z table in the last line
So the probability is 0.4801
PLEASE CALCULUS HELP!!!!!!
Answer:
work and answer are shown in the picture
Step-by-step explanation:
if you have any questions about my work please let me know
Hi please help I keep getting the anwser wrong and really need to get at least 1/2 the two right or I’ll get a zero!!!pls
Answer:
x = [tex]14\frac{7}{9}[/tex] or 14.78
Step-by-step explanation:
The lines are parallel, therefore, k(18) acts same as g(x)
That means that:
k(18) = g(x)
- 14 = [tex]-\frac{18}{7}x[/tex] + 24
- 14 - 24 = [tex]-\frac{18}{7}x[/tex]
- 38 = [tex]-\frac{18}{7}x[/tex]
38(7) = 18x
266 = 18x
266 / 18 = x
133 / 9 = x
x = [tex]14\frac{7}{9}[/tex] or 14.78
Help Fast Which transformations could have occurred to map △ABC to △A"B"C"? a rotation and a dilation a rotation and a reflection a reflection and a dilation a translation and a dilation
Answer:
its A
Step-by-step explanation:
A toolbox has 10 screwdrivers Sid 6 wrenches.
Bella puts 8 more wrenches in the toolbox.
*) How many more wrenches are in the toolbox
than screwdrivers?
Answer: There are 4 more wrenches in the toolbox then the screwdrivers.
Step-by-step explanation: Add the 6 wrenches Sid put in the toolbox with the 8 wrenches Bella added to get 14 wrenches in total. Then, subtract the 10 screwdrivers from the 14 wrenches to get 4 wrenches.
Final answer:
Bella added 8 wrenches to the toolbox, making a total of 14 wrenches. There were initially 10 screwdrivers, so there are now 4 more wrenches than screwdrivers.
Explanation:
Calculating the Difference Between Wrenches and Screwdrivers in a Toolbox
Initially, there are 10 screwdrivers and 6 wrenches in the toolbox. Bella adds 8 more wrenches, which brings the total number of wrenches to 6 + 8, which equals 14 wrenches. The question asks how many more wrenches there are than screwdrivers. To find this, we subtract the number of screwdrivers from the number of wrenches:
14 wrenches - 10 screwdrivers = 4 more wrenches than screwdrivers in the toolbox.
What is equivalent to 16 3/4x
Answer:
⁴ˣ√16³
Step-by-step explanation:
The equivalent to 16^(3/4x) is ⁴ˣ√16³. It reads, 4x root of 16 raised to the power of 3. 1/4x as an exponent means the 4x root of the base number. 3 as an exponent simply means that the base number is raised to the third power.
A scatterplot shows a strong, positive, linear relationship between the number of rebounds a basketball team averages and the number of wins that team records in a season. Which conclusion is most appropriate?
Answer:
The correct answer is wins and rebounds are correlated positively ,but we cannot decided that having more rebounds leads to more wins,on average.
Step-by-step explanation:
From the example given, the most appropriate conclusion is that, because causation is not the same as correlation, If two variables are compared,this does not mean that one leads to the other.
An observed data is based on correlation,but for description of causation ,we need to make experiments,as we update the variable treatment regarding to the changes in response variable.
When a car is first observed it has a speed of 20 ms-1. after a time of 10 S it is observed that the speed is 50 MS-1
Answer:
i need points.
Step-by-step explanation:
Over 10 minutes ,how far on a clock does the tip of a 12 inch minute hand move ?
A: 2.09inches
B: 6.28 inches
C: 12.56 inches
D: 75.36 inches
Need help please anyone
Answer:
C: 12.56 inches
Step-by-step explanation:
We know that the minute hand can move an equivalent of 60 minutes in any one revolution.
-10 minutes movement is equal to 1/6 the total distance and the circumference covered in that time is calculated as:
[tex]C=\pi D\\\\=\frac{1}{6}\pi \times (12\times 2)\\\\\\=12.56\ in[/tex]
Hence, over 10 minutes the minutes hand moves 12.56 inches away.
The tip of a 12 inch minute hand will move approximately 12.56 inches over the course of 10 minutes, which aligns with option C in your given choices.
Explanation:The subject of this question is Mathematics, specifically geometry and involves calculating the length of an arc within a circle. The minute hand of a clock can be thought of as the radius of a circle, with a full rotation of the hand representing a complete circle. The minute hand moves 360 degrees in 60 minutes (or 6 degrees per minute), so over 10 minutes, the minute hand will move 60 degrees.
Now, the length of that portion of the circle (the arc length) is calculated using the formula: (2πr)(θ/360), where r is the radius (half of the diameter, or 12 inches in this case), and θ is the angle in degrees. When you plug in the respective values, you will find that the minute hand of the clock moves an approximate distance of 12.56 inches, which corresponds to option C in your given choices.
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Suppose ACT Reading scores are normally distributed with a mean of 21.3 and a standard deviation of 5.9. A university plans to award scholarships to students whose scores are in the top 7%. What is the minimum score required for the scholarship? Round your answer to the nearest tenth, if necessary.
Answer:
30.0
Step-by-step explanation:
Given our data is normally distribute with [tex]\mu=21.3[/tex] and [tex]\sigma=5.9[/tex]
-Top 7% is given by find the z-value corresponding to p=(1-0.07)=0.93
-We substitute our values in the equation below;
[tex]z=\frac{\bar X-\mu}{\sigma}\\\\\\=\frac{X-21.3}{5.9}, z_{0.035}=1.476\\\\\therefore 1.476=\frac{X-21.3}{5.9}\\\\X=5.9\times 1.476+21.3\\\\=30.0084\approx30.0[/tex]
Hence, the minimum score required for the scholarship is 30.0
The minimum ACT Reading score required for a university scholarship awarded to the top 7% is approximately 30.0.
To find the minimum ACT Reading score required for a scholarship awarded to students in the top 7%, we need to determine the z-score that corresponds to the top 7% of a normal distribution. We can then use this z-score to find the corresponding ACT score.
The z-score for the top 7% of a standard normal distribution is approximately 1.475. Since the ACT Reading scores have a mean (μ) of 21.3 and a standard deviation (σ) of 5.9, we can use the z-score formula to find the minimum score 'x' required for the scholarship: z = (x - μ) / σ.
Solving for 'x', we get: x = zσ + μ = 1.475(5.9) + 21.3 ≈ 30.0. Therefore, the minimum ACT Reading score required for the scholarship is approximately 30.0.
A guy wire 17 feet long runs from the top of a pole to a spot on the ground. If the height of the pole is 7 feet more than the distance from the base of the pole to the spot where the guy wire is anchored, how tall is the pole?
Answer:
The pole is 15 feet tall
Step-by-step explanation:
Pythagora's Theorem
Let's call x the distance from the base of the pole to the spot where the guy wire is anchored. The height of the pole is 7 feet more, i.e. x+7.
The guy wire is 17 feet long. These dimensions form the sides of a right triangle where the guy wire is the hypotenuse.
Applying Pythagora's Theorem
[tex]x^2+(x+7)^2=17^2[/tex]
Operating
[tex]x^2+x^2+14x+49=289[/tex]
Rearranging and simplifying by 2
[tex]x^2+7x-120=0[/tex]
Factoring
[tex](x-8)(x+15)=0[/tex]
Solving
[tex]x=8,\ x=-15[/tex]
Only the positive solution is valid, thus x=8
The height of the pole is x+7=15 feet
The pole is 15 feet tall
Final answer:
Using the Pythagorean theorem, we find that the height of the pole is approximately 15 feet when solving the derived quadratic equation.
Explanation:
The question involves applying the Pythagorean theorem to solve for the height of the pole. Let's denote the distance from the base of the pole to the spot where the guy wire is anchored as x. Then, the height of the pole is x + 7 feet. Since the guy wire creates a right triangle with the pole and the ground, the Pythagorean theorem states that the square of the hypotenuse (guy wire) is equal to the sum of the squares of the other two sides (height of the pole and the distance from the base). Therefore, we can write the equation 17^2 = (x + 7)^2 + x^2. By solving this equation, we will find the value of x and then determine the height of the pole.
Step by step, we solve the equation: 289 = (x + 7)^2 + x^2, which simplifies to 289 = 2x^2 + 14x + 49. Subtracting 289 from both sides results in 0 = 2x^2 + 14x - 240. By factoring or using the quadratic formula, we find that x is approximately 8 feet. Therefore, the height of the pole is x + 7 feet, which is 15 feet tall.
A coin is tossed and a number cube is rolled what is the probability that the coin shows heads and the number cube shows six
Answer:
There is a 1/2 chance the coin will land on heads and there is a 1/6 chance that the number cube will land on 6. hope this helps
An important problem in thermodynamics is to find the work done by an ideal Carnot engine. A cycle consists of alternating expansion and compression of gas in a piston. The work done by the engine is euqal to the area of the region R enclosed by two isothermal curves xy=a, xy=b and two adiabatic curves xy^1.4=c, xy^1.4=d, where 0
Answer:
The work done is 2.5(b-a)* ln(d/c).
Step-by-step explanation:
Steps are in the following attachments
The work done by an ideal Carnot engine is equal to the area enclosed by the region in the pV diagram.
Explanation:The work done by an ideal Carnot engine is equal to the area enclosed by the region in the pV diagram. This region is bounded by two isothermal curves and two adiabatic curves. The work done by the engine can be calculated by finding the area under the isothermal curves and subtracting the area under the adiabatic curves.
To find the work done, you can divide the region into smaller shapes, such as rectangles or triangles, and calculate the area of each shape. Then, sum up the areas of all the shapes to get the total work done by the engine.
Remember to use the equations for the isothermal and adiabatic processes to relate the pressure and volume of the gas at different points in the cycle.
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g Consider the following statement. For all sets A and B, (A − B) ∪ (A ∩ B) = A. Construct an algebraic proof for the statement. Cite a property from Theorem 6.2.2 for every step.
To prove the statement (A − B) ∪ (A ∩ B) = A, we can use the property of set difference, distribution, and identity from Theorem 6.2.2.
Explanation:To construct an algebraic proof for the statement (A − B) ∪ (A ∩ B) = A, we can use the property of set difference, distribution, and identity from Theorem 6.2.2.
Start with the left side of the equation: (A − B) ∪ (A ∩ B)Apply the property of set difference: (A − B) = A ∩ B'. Now the equation becomes (A ∩ B') ∪ (A ∩ B).Use the property of distribution: A ∩ (B' ∪ B) = A ∩ U = A, where U represents the universal set. Therefore, (A − B) ∪ (A ∩ B) = A. Learn more about Set theory here:https://brainly.com/question/27333813
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If 10 pounds of ice cream are separated into 15 bowls, how much ice cream would be in each bowl?
Answer:
2/3 of a pound.
Step-by-step explanation:
10 pounds per 15 bowls = 2 pounds per 3 bowls, this is equal to 2/3 of ice cream a pound in a single bowl.
Last month, Bethany sent 5,450 texts. This month she sent 7,085 texts. What was the percent increase in her texting from last month to this month?
Answer:
Hello
The answer is 30% increase in texts since last month.
IF you feel any problem in understanding , do comment pls.
Step-by-step explanation:
Let
X = last month sent texts
y = this month sent texts
First of all find the no. of increased texts,
by subtracting x from y
=> y-x= 7085- 5450
= 1635
We want to find these 15 texts % with respect to 5450 texts
i.e. 1635/X
=0.30
for answer in % multiply with 100
i.e. 30%
Need to solve
15,000,000 = 4700e 0.154t
Answer:
[tex]t=52.39[/tex]
Step-by-step explanation:
Figure ABCD is a square. Prove BD ≅ AC. Square A B C D with diagonals is shown. Statements Reasons 1. ABCD is a square 1. given 2. ∠DAB, ∠ABC, ∠BCD, and ∠CDA are right angles 2. definition of a square 3. ∠DAB ≅ ∠ABC ≅ ∠BCD ≅ ∠CDA 3. right angles are congruent 4. AB ≅ BC ≅ CD ≅ DA 4. ? 5. △BAD ≅ △ABC 5. SAS 6. BD ≅ AC 6. CPCTC What is the missing reason in the proof?
all sides of a square are congruent
all right angles measure 90°
definition of diagonal
definition of perpendicular
Answer:
all sides are congruent
Step-by-step explanation:
its talking about sides
I believe A is correct
Good luck!
1. How are systems of equations used in the real world?
Help me plzzzz
Answer:
Systems of linear equations are used in the real world by economists and entrepreneurs to find out when supply equals demand.It's all about the mullah, and if you don't know the numbers when you have a business, it might fail.
Step-by-step explanation:
Find unknown number lb=oz
Answer:
72
Step-by-step explanation:
multiply by 16
Answer:
4 1/2 lb = 72 oz
brainliest?
Use the confidence level and sample data to find a confidence interval for estimating the population muμ. Round your answer to the same number of decimal places as the sample mean. A random sample of 9595 light bulbs had a mean life of x overbar equals 510x=510 hours with a standard deviation of sigma equals 37 hours.σ=37 hours. Construct a 90% confidence interval for the mean life, muμ, of all light bulbs of this type.
Answer:= (504, 516)
Therefore, the 90% confidence interval (a,b) = ( 504, 516)
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
Given that;
Mean gain x = 510
Standard deviation r = 37
Number of samples n = 95
Confidence interval = 90%
z(at 90% confidence) = 1.645
Substituting the values we have;
510+/-1.645(37/√95)
510+/-1.645(3.796)
510+/-6.24
510+/-6
= (504, 516)
Therefore at 90% confidence interval (a,b) = ( 504, 516)
At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?
The distance between the two ships is changing at approximately 26.87 km/h at 4:00 PM, calculated using the Pythagorean theorem and the differentiation principles of related rates in calculus.
To determine how fast the distance between the two ships is changing at 4:00 PM, we need to apply the concept of related rates in calculus. Since at noon ship A is 130 km west of ship B and moving east, and ship B is moving north, we can imagine their paths as legs of a right triangle, where the hypotenuse represents the distance between the two ships.
Let's denote the distance that ship A has traveled east as x (in kilometers), the distance that ship B has traveled north as y (in kilometers), and the distance between the two ships as z (in kilometers). At 4:00 PM, ship A has been sailing east for 4 hours at 25 km/h, so x = 25 km/h * 4 h = 100 km. Ship B has sailed north for the same amount of time at 20 km/h, so y = 20 km/h * 4 h = 80 km.
To find the rate at which the distance z is changing, we use the Pythagorean theorem z = x^2 + y^2.
Differentiating both sides with respect to time t, we get 2zdz/dt = 2xdx/dt + 2ydy/dt. We can cancel the 2's and plug in the values for x, y, dx/dt (25 km/h) and dy/dt (20 km/h) to find dz/dt, which represents the rate at which the distance between the ships is changing.
Solving for dz/dt, we have:
z dz/dt = x dx/dt + y dy/dt
z dz/dt = 100 km * 25 km/h + 80 km * 20 km/h
First, we must find z, which is the distance between the ships at 4:00 PM:
z =
sqrt{130^2 + 80^2} =
sqrt{16900 + 6400} =
sqrt{23300} ≈ 152.65 km
Now, we solve for dz/dt:
152.65 km * dz/dt = 100 km * 25 km/h + 80 km * 20 km/h
dz/dt ≈ (2500 km²/h + 1600 km²/h) / 152.65 km
dz/dt ≈ 4100 km²/h / 152.65 km
dz/dt ≈ 26.87 km/h
Therefore, the distance between the ships is changing at approximately 26.87 km/h at 4:00 PM.
at 4:00 PM, the rate of change of the distance between the ships is[tex]\( \frac{7350}{10\sqrt{593}} \)[/tex] km/h.
To find the rate of change of the distance between the ships at 4:00 PM, we'll first find expressions for the positions of each ship at that time. Then, we'll differentiate the distance formula with respect to time and evaluate it at 4:00 PM.
Let ( t ) be the time in hours since noon.
Ship A's position [tex]\( x_A \)[/tex] at time \( t \) is given by:
[tex]\[ x_A = 130 + 25t \][/tex]
And ship B's position [tex]\( y_B \)[/tex] at time \( t \) is given by:
[tex]\[ y_B = 20t \][/tex]
Using these positions, the distance between the ships ( D ) at time ( t ) is given by the distance formula:
[tex]\[ D(t) = \sqrt{(130 + 25t)^2 + (20t)^2} \][/tex]
Now, let's differentiate [tex]\( D(t) \)[/tex] with respect to time [tex]\( t \)[/tex] using the chain rule:
[tex]\[ \frac{dD}{dt} = \frac{1}{2\sqrt{(130 + 25t)^2 + (20t)^2}} \times \left(2(130 + 25t)(25) + 2(20t)(20)\right) \][/tex]
Now, plug in ( t = 4 ) to find the rate of change of the distance between the ships at 4:00 PM:
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{(130 + 25(4))^2 + (20(4))^2}} \times \left(2(130 + 25(4))(25) + 2(20(4))(20)\right) \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{(130 + 100)^2 + (80)^2}} \times \left(2(130 + 100)(25) + 2(80)(20)\right) \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{230^2 + 80^2}} \times \left(2(230)(25) + 2(80)(20)\right) \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{52900 + 6400}} \times \left(2(5750) + 2(1600)\right) \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{59300}} \times (11500 + 3200) \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{59300}} \times 14700 \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{14700}{2\sqrt{59300}} \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{7350}{\sqrt{59300}} \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{7350}{\sqrt{100 \times 593}} \][/tex]
[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{7350}{10\sqrt{593}} \][/tex]
So, at 4:00 PM, the rate of change of the distance between the ships is[tex]\( \frac{7350}{10\sqrt{593}} \)[/tex] km/h. This can be simplified further, but it's best to leave it in this form to retain the exact value.
5. Oscar needs to fill a sphere-shaped balloon with
helium. If the balloon has a diameter of 8 inches, what is
the total amount of helium that the balloon will hold to
the nearest tenth?
A. 2,143.6 in.3
B. 714.5 in.
C. 268.1 in.3
D. 150.7 in.
Final answer:
Oscar's balloon, which has an 8-inch diameter, will hold approximately 268.1 cubic inches of helium, calculated using the volume formula for a sphere.
Explanation:
Oscar needs to calculate the volume of a sphere-shaped balloon to determine how much helium it can hold. To find the balloon's volume, we use the formula for the volume of a sphere, which is V = ⅓πd³, where V is the volume, π is approximately 3.14159, and d is the diameter of the sphere. Since the balloon has a diameter of 8 inches, its radius r is 4 inches (which is half of the diameter).
Plugging the radius into the formula, we get: V = ⅓π(4 inches)³ = ⅓π(64 inches³) = 268.0826 inches³. Therefore, Oscar's balloon will hold approximately 268.1 cubic inches of helium to the nearest tenth, making the correct answer C. 268.1 in.³
Suppose we want to assess the effect of a one-day SAT prep class at a 5% level of significance. Scores on the SAT writing exam can range from 200 to 800. A random sample of 50 students takes the SAT writing test before and after a prep class. We test the hypotheses: H 0: μ = 0 H a: μ > 0 where μ is the mean of the difference in SAT writing scores (after minus before) for all students who take the SAT prep class. The sample mean is 5 with a standard deviation of 18. Since the sample size is large, we are able to conduct the T-Test. The T-test statistic is approximately 1.96 with a P-value of approximately 0.028. What can we conclude? Group of answer choices The one-day SAT prep class is associated with statistically significant improvements in SAT writing performance. Students taking a one-day SAT prep class performed significantly better on the SAT writing exam than students who did not take the class. Students taking a one-day SAT prep class do not show statistically significant improvements in their SAT writing performance. Scores only increased by 5 points, which is not significant on an exam where scores can range from 200 to 800. The one-day SAT prep class produces statistically significant improvements in SAT writing performance.
Answer: The one-day SAT prep class is associated with statistically significant improvements in SAT writing performance.
Step-by-step explanation: just took the quiz
The correct conclusion about the situation is, the one-day SAT prep class produces statistically significant improvements in SAT writing performance, which is option (e).
Given that:
It is assessing the performance of the students in the SAT writing exam before and after SAT prep class.
The hypothesis is:
H₀: μ = 0
H₁: μ > 0
This is a one-tailed test.
Here, the T-test is used.
Now, the significance level is, α = 0.05
p-value = 0.028
Since, the p-value, 0.028 is less than the significance level 0.05, the null hypothesis is rejected.
So, the mean of the difference in SAT scores is greater than 0.
That is, there is a significant effect in SAT exam by the prep class.
Hence, the correct conclusion is, The one-day SAT prep class produces statistically significant improvements in SAT writing performance, which is option (e).
Learn more about T-test here :
https://brainly.com/question/26462012
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