The 99% confidence interval for the difference in the proportions of patients needing pain medication between the old and new procedures is given as follows:
(0.047, 0.443).
How to obtain the confidence interval?The sample proportion for each case is given as follows:
[tex]p_1 = \frac{24}{58} = 0.414[/tex][tex]p_2 = \frac{14}{83} = 0.169[/tex]Hence the difference is given as follows:
0.414 - 0.169 = 0.245.
The standard error for each sample is given as follows:
[tex]s_1 = \sqrt{\frac{0.414(0.586)}{58}} = 0.065[/tex][tex]s_2 = \sqrt{\frac{0.169(0.831)}{83}} = 0.041[/tex]Hence the standard error for the distribution of differences is given as follows:
[tex]s = \sqrt{0.065^2 + 0.041^2}[/tex]
s = 0.077[/tex]
The confidence level is of 99%, hence the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.
The lower bound of the interval is:
0.245 - 2.575 x 0.077 = 0.047.
The upper bound of the interval is:
0.245 + 2.575 x 0.077 = 0.443.
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we need to calculate a) mean b) variance c) standard
deviation
(2) clarining cinif requang, For the frequency: table on the left, compete (as the main (8), 4) the variance [5] and w the standard deviation 8]. 2 3 6 9. 7 12 4 Sum=20
The mean is ≈ 8.793. The variance is approximately 9.641. The standard deviation is approximately 2.964
Given frequency table:
Value: 2 3 6 9 12
Frequency: 3 6 9 7 4
a) Mean:
[tex]\[\text{{Mean}} = \frac{{\text{{Sum of (Value * Frequency)}}}}{{\text{{Total number of observations}}}}\]\[\text{{Mean}} = \frac{{(2 \times 3) + (3 \times 6) + (6 \times 9) + (9 \times 7) + (12 \times 4)}}{{3 + 6 + 9 + 7 + 4}}\]\[\text{{Mean}} = \frac{{189}}{{29}}\][/tex]
≈ 8.793
b) Variance:[tex]\[\text{{Variance}} = \frac{{(3 \times (2 - \text{{Mean}})^2) + (6 \times (3 - \text{{Mean}})^2) + (9 \times (6 - \text{{Mean}})^2) + (7 \times (9 - \text{{Mean}})^2) + (4 \times (12 - \text{{Mean}})^2)}}{{29}}\][/tex]
≈ 8.793
c) Standard Deviation:
[tex]\[\text{{Standard Deviation}} = \sqrt{{\text{{Variance}}}}\][/tex]
Therefore, the standard deviation is approximately [tex]\sqrt{8.793} \approx 2.964[/tex]
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PLISSSS HELP 20 POINTS
Answer:
x=8
Step-by-step explanation:
Because you are solving for x, you want to cancel out the y terms. You can do this by multiplying the entire equations by numbers that will make the y terms have equal numbers but opposite signs.
2(2x-5y=1)
5(-3x+2y=-18)
This turns into
4x-10y=2
-15x+10y=-90
The y terms cancel out, and the other terms can be added together.
-11x=-88
x=8
Find the area of the figure.
HELP PLZZ
Answer:
159.25 ft²
I hope this helps! :)
Step-by-step explanation:
Formulas:
For the Rectangle... bh = a
For the Semicircle... 1/2 × πr²
Step 1:
Solve the area for the rectangle:
bh = a
10 × 12 = 120
a = 120 ft²
Step 2:
Solve the Area for the Semicircle:
1/2 × πr²
1/2 × 3.14 = 1.57
Radius = Diameter ÷ 2
10 ÷ 2 = 5
Radius = 5
1.57 × 5²
1.57 × 5 × 5
= 39.25 ft²
Step 3:
Add the two areas together:
120 + 39.25 = 159.25 ft²
Can someone plz help me with #2 and #6 plz thank oyu
Answer:
njbrdjbbgdbjbfjfj
Step-by-step explanation:
A random variable X has density function fx(x) e*, x<0, 0, otherwise. The moment generating function My(t)= Use My(t) to compute E(X)= and Var(x)= Use My(t) to compute the compute the mgf for 3 Y= X-2. That is My(t)= = 2
To compute the moment generating function (MGF) for the random variable X, we need to use the formula:
[tex]My(t) = E(e^(tx))[/tex]
Given that the density function for X is fx(x) = e^(-x), x < 0, and 0 otherwise, we can write the MGF as follows:
[tex]My(t) = ∫[from -∞ to ∞] e^(tx) * fx(x) dx[/tex]
Since the density function fx(x) is non-zero only for x < 0, we can rewrite the integral accordingly:
[tex]My(t) = ∫[from -∞ to 0] e^(tx) * e^x dx + ∫[from 0 to ∞] e^(tx) * 0 dx[/tex]
The second integral is zero because the density function is zero for x ≥ 0. We can simplify the expression:
[tex]My(t) = ∫[from -∞ to 0] e^(x(1+t)) dx[/tex]
Using the properties of exponents, we can simplify further:
[tex]My(t) = ∫[from -∞ to 0] e^((1+t)x) dx[/tex]
Now we can evaluate this integral:
[tex]My(t) = [1 / (1+t)] * e^((1+t)x) | [from -∞ to 0)[/tex]
= [tex][1 / (1+t)] * (e^((1+t)(0)) - e^((1+t)(-∞)))[/tex]
= [tex][1 / (1+t)] * (1 - 0)[/tex]
= [tex]1 / (1+t)[/tex]
The moment generating function My(t) simplifies to 1 / (1+t).
To compute the expected value (E(X)) and variance (Var(X)), we can differentiate the MGF with respect to t:
E(X) = My'(t) evaluated at t=0
Var(X) = My''(t) evaluated at t=0
Taking the derivative of My(t) = 1 / (1+t) with respect to t, we get:
[tex]My'(t) = -1 / (1+t)^2[/tex]
Evaluating My'(t) at t=0:
E(X) = [tex]My'(0) = -1 / (1+0)^2 = -1[/tex]
Thus, the expected value of X is -1.
To compute the second derivative, we differentiate My'(t) =[tex]-1 / (1+t)^2[/tex]again:
[tex]My''(t) = 2 / (1+t)^3[/tex]
Evaluating My''(t) at t=0:
Var(X) =[tex]My''(0) = 2 / (1+0)^3 = 2[/tex]
Thus, the variance of X is 2.
Now, let's compute the MGF for the random variable Y = X - 2:
[tex]My_Y(t) = E(e^(t(Y)))= E(e^(t(X - 2)))= E(e^(tX - 2t))[/tex]
Using the properties of the MGF, we know that if X is a random variable with MGF My(t), then e^(cX) has MGF My(ct), where c is a constant. Therefore, we can rewrite the MGF for Y as:
[tex]My_Y(t) = e^(-2t) * My(t)[/tex]
Substituting My(t) = 1 / (1+t) from the previous calculation, we get:
[tex]My_Y(t) = e^(-2t) * (1 / (1+t))[/tex]
Simplifying further:
[tex]My_Y(t) = e^(-2t) / (1+t)[/tex]
Thus, the MGF for Y = X
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Jen has 3 bags of pears. Each Bag has 5 pears. If Jen gives the same number of pears to 4 friends, how many pears will each friend get?
3bags x 5 pearls per bag=15 bags
15 / (4 friends+you)=5
15/5=3 pearls per friend.
Answer:3 pearls per person.
Answer:
It is 3x5 is 15 the you divide 15 by 4 which is 3.75 per person
Step-by-step explanation:
3x5=15
15\4=3.75
The mean score of a competency test is 64, with a standard deviation of 4. Between what two values do about 99.7% of the values lie? (Assume the data set has a bell-shaped distribution.) Between 56 and 72 Between 60 and 68 O Between 52 and 76 Between 48 and 80
In a dataset with a bell-shaped distribution, approximately 99.7% of the values lie within three standard deviations of the mean. Given a mean score of 64 and a standard deviation of 4 on a competency test, we can determine the range within which about 99.7% of the values will fall. The correct range is between 56 and 72.
To calculate the range, we need to consider three standard deviations above and below the mean. Three standard deviations from the mean account for approximately 99.7% of the data in a bell-shaped distribution.
Lower limit: Mean - (3 * Standard Deviation)
= 64 - (3 * 4)
= 64 - 12
= 52
Upper limit: Mean + (3 * Standard Deviation)
= 64 + (3 * 4)
= 64 + 12
= 76
Therefore, about 99.7% of the values lie between 52 and 76.
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PROCESS A: "Driftless" geometric Brownian motion (GBM). "Driftless" means no "dt" term. So it's our familiar process: ds = o S dw with S(O) = 1. o is the volatility. PROCESS B: ds = a S2 dw for some constant a, with S(0) = 1 As we've said in class, for any process the instantaneous return is the random variable: dS/S = (S(t + dt) - S(t)/S(t) = [1] Explain why, for PROCESS A, the variance of this instantaneous return (VAR[ds/S]) is constant (per unit time). Hint: What's the variance of dw? The rest of this problem involves PROCESS B. [2] For PROCESS B, the statement in [1] is not true. Explain why PROCESS B's variance of the instantaneous return (per unit time) depends on the value s(t).
In Process A, which is a driftless geometric Brownian motion, the variance of the instantaneous return (VAR[ds/S]) is constant per unit time. However, in Process B, where ds = aS^2dw, the variance of the instantaneous return depends on the value of S(t).
In Process A, since there is no drift term (dt), the random variable dw follows a standard normal distribution with a constant variance of dt. When calculating the instantaneous return dS/S, we can see that the dt terms cancel out, resulting in a constant variance of the instantaneous return (VAR[ds/S]) per unit time. This is because the volatility o remains constant, and the random variable dw has a constant variance.
In Process B, the equation ds = aS^2dw suggests that the variance of the instantaneous return is proportional to the value of S(t). As S(t) increases, the magnitude of ds also increases, leading to a larger variance of the instantaneous return. In other words, the volatility in Process B depends on the level of the underlying process S(t). This is different from Process A, where the variance of the instantaneous return is constant regardless of the value of S(t). Hence, the variance of the instantaneous return in Process B is not constant per unit time, but rather dependent on the value of S(t).
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Please help me!! No files allowed. I need the answer and an explanation!
Answer:
1/324
Step-by-step explanation:
A sample of 1 can be drawn from an automatic storage and retrieval rack with 9 different storage racks and 6 different trays in each rack, find the number of different ways of obtaining a tray?
There are 54 different ways of obtaining a tray from the automatic storage and retrieval rack.
To find the number of different ways of obtaining a tray from an automatic storage and retrieval rack, we can use the concept of the multiplication principle.
The multiplication principle states that if there are m ways to do one thing and n ways to do another thing, then there are m × n ways to do both things together.
In this case, we have two steps involved in obtaining a tray:
Selecting a storage rack: There are 9 different storage racks available. We can choose any one of them.
Selecting a tray within the chosen storage rack: Once we have chosen a storage rack, there are 6 different trays in each rack. We can select any one of these trays.
According to the multiplication principle, the total number of ways to perform both steps is the product of the number of options for each step.
Number of ways = Number of options for Step 1 × Number of options for Step 2
= 9 × 6
= 54
Therefore, there are 54 different ways of obtaining a tray from the automatic storage and retrieval rack.
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If m∠4 = 35°, find m∠2 and m∠3.
There are two parallel lines. Point A and B are on the upper line and point C and D are same are A and B on the lower line. The segment AC is perpendicular to both parallel lines. There is another segment CB. The angle CAB is denoted by 1. The angle ABC is denoted by 2. The angle ACB is denoted by 3. The complementary angle of angle 3 is denoted by 4.
m∠2 =
°
m∠3 =
°
Answer:
Step-by-step explanation:
yes
The ∠3 is an alternate angle of ∠4 so it is 35° while ∠3 is a complimentary angle of ∠4 so it will be 55°.
What is an angle?An angle is a geometry in plane geometry that is created by 2 rays or lines that have an identical terminus.
Since we lack a measurement for angular rotation, the angle is a valuable tool for measuring angular distance. For instance, a meter or an inch is a unit of measurement for linear motion.
Given that AB and CD are parallel and AC is perpendicular to both.
So,
∠ACD = 90°
∠3 + ∠4 = 90°
∠3 = 90 - 35 = 55°.
Now,
Since ∠4 and ∠2 are alternative angles so both will be the same.
So,
∠2 = 35°
Hence "The ∠3 is an alternate angle of ∠4 so it is 35° while ∠3 is a complimentary angle of ∠4 so it will be 55°".
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11 - x when x= -4 how do you solve this
Answer:
15 is the answer
Step-by-step explanation:
We know that x = -4, so substitute x for -4 in the problem
11 - (-4)
2 negative signs make a positive sign
11 + 4
=15
Answer:
Hi! The answer to your question is [tex]15[/tex]
How to solve is whenever there is an x, replace it with a -4 so the problem would be set up like this 11-(-4) and at that point you can just solve it in a calculator
Step-by-step explanation:
☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆
☁Brainliest is greatly appreciated!!☁
Hope this helps!!
- Brooklynn Deka
Let Z= max (X, Y) and W = min (X, Y) are two new random variables as functions of old random variables X and Y. (a). Determine fz (z) and fw (w) in terms of marginal CDFs of X and Y random variables, by first drawing the region of interest on X and Y plane. (b). Let x and y be independent exponential random variables with common parameter A. Define W = min (X, Y). Find fw (w).
(a) fz (z) and fw (w) in terms of cumulative distribution functions (CDFs) are:
fz(z) = Fx(z) * (1 - Fy(z)) + Fy(z) * (1 - Fx(z))
fw(w) = 1 - fz(w)
(b) If X and Y are independent exponential random variables with parameter λ, then fw(w) = [tex]1 - e^{-2\lambda w}[/tex] for w ≥ 0.
To determine fz(z) and fw(w) in terms of the marginal cumulative distribution functions (CDFs) of X and Y random variables, we need to consider the region of interest on the X-Y plane.
(a) Drawing the region of interest on the X-Y plane:
The region of interest can be visualized as the area where Z = max(X, Y) and W = min(X, Y) take specific values. This region is bounded by the line y = x (diagonal line) and the lines x = z (vertical line) and y = w (horizontal line).
Determining fz(z):
To find fz(z), we need to consider the cumulative probability that Z takes a value less than or equal to z. This can be expressed as:
fz(z) = P(Z ≤ z) = P(max(X, Y) ≤ z)
Since X and Y are independent random variables, the probability can be calculated using the joint CDF of X and Y:
fz(z) = P(max(X, Y) ≤ z) = P(X ≤ z, Y ≤ z)
Using the marginal CDFs of X and Y, denoted as FX(x) and FY(y), respectively, we can express fz(z) as:
fz(z) = P(X ≤ z, Y ≤ z) = P(X ≤ z) * P(Y ≤ z) = FX(z) * FY(z)
Determining fw(w):
To find fw(w), we need to consider the cumulative probability that W takes a value less than or equal to w. This can be expressed as:
fw(w) = P(W ≤ w) = P(min(X, Y) ≤ w)
Since X and Y are independent random variables, the probability can be calculated using the joint CDF of X and Y:
fw(w) = P(min(X, Y) ≤ w) = 1 - P(X > w, Y > w)
Using the marginal CDFs of X and Y, denoted as FX(x) and FY(y), respectively, we can express fw(w) as:
fw(w) = 1 - P(X > w, Y > w) = 1 - [1 - FX(w)][1 - FY(w)]
Special case when X and Y are independent exponential random variables with parameter A:
If X and Y are independent exponential random variables with a common parameter A, their marginal CDFs can be expressed as:
[tex]FX(x) = 1 - e^{-Ax}\\FY(y) = 1 - e^{-Ay}[/tex]
Using these marginal CDFs, we can substitute them into the formulas for fz(z) and fw(w) to obtain the specific expressions for the random variables Z and W.
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A rectangle’s length is three times its width, w. Its area is 243 square units. Which equation can be used to find the width of the rectangle?
a. w2 = 3(243)
b. 3w2 = 243
c. 4w2 = 243
d. 3w2 = 3(243)
please im on a timed test
Answer:
b) 3w² = 243
Step-by-step explanation:
area = L x w
L = 3w
substitute for L:
243 = 3w x w = 3w²
3w² = 243
class 9 help who are clever will get a brainlist
8 ft
Find the area of the figure.
Answer:
Area of a rectangle is length multiplied by the width. In this case, length is equal to width. So, Area is 8 ft * 8 ft which is 64 ft2.
If today is Wednesday what is it like hood tomorrow will be Saturday?
Answer:
zero
Step-by-step explanation:
If today is Wednesday, the probability of tomorrow being Saturday is zero.
A company rents out 18 food booths and 26 game booths at the county fair. The fee for a food booth is $100 plus $10 per day. The fee for a game booth is $95 plus $5 per day. The fair lasts for d days, and all the booths are rented for the entire time. Enter a simplified expression for the amount, in dollars, that the company is paid.
Answer:
4,270 + 310d
Step-by-step explanation:
Let
d = number of days
Each food booth = 100 + 10d
18 food booth = 18(100 + 10d)
= 1,800 + 180d
Each game booth = 95 + 5d
26 game booth = 26(95 + 5d)
= 2,470 + 130d
Total amount the company is paid = total cost of food booth + total cost of game booth
= (1,800 + 180d) + (2,470 + 130d)
= 1,800 + 180d + 2,470 + 130d
= 1,800 + 2,470 + 180d + 130d
= 4,270 + 310d
Total amount the company is paid = 4,270 + 310d
Linear programming can be used to find the optimal solution for profit, but cannot be used for nonprofit organizations. False True
The statement "Linear programming can be used to find the optimal solution for profit, but cannot be used for nonprofit organizations" is False.
Linear programming can be used to find the optimal solution for profit as well as for non-profit organizations. Linear programming is a method of optimization that aids in determining the best outcome in a mathematical model where the model's requirements can be expressed as linear relationships. Linear programming can be used to solve optimization problems that require maximizing or minimizing a linear objective function, subject to a set of linear constraints.
Linear programming can be used in a variety of applications, including finance, engineering, manufacturing, transportation, and resource allocation. Linear programming is concerned with determining the values of decision variables that will maximize or minimize the objective function while meeting all of the constraints. It is used to find the optimal solution that maximizes profits for for-profit organizations or minimizes costs for non-profit organizations.
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Can somebody plz help answer these questions correctly (only if u know how to do them) thx sm! :3
WILL MARK BRAINLIEST WHOEVER ANASWERS FIRST :DDDD
Answer:
s= 100
r= 130
x= 38
y= 30
Step-by-step explanation:
A study at the University of Illinois found that young men who drank two pints of beer first were better able to solve certain word puzzles than sober men. Design an experiment that could attempt to verify this result. Describe the population, how you’d collect your sample, how you’d execute the experiment, and what data you’d collect .
Experiment to verify the result of the study:A study at the University of Illinois found that young men who drank two pints of beer first were better able to solve certain word puzzles than sober men. An experiment to verify this result should be designed with the following steps:
Population: The population in this experiment would be young men who are eligible to consume beer legally.
Sampling: The sampling method will be convenient sampling. In this type of sampling, participants will be selected based on their availability to participate. Any participant that is within the age range of eligibility and is willing to participate can be considered for the study.
The participants will be divided into two groups, one group will drink two pints of beer while the other group will not drink any beer.
Executing the experiment: Both groups will be given word puzzles to solve after the beer is consumed by the test group and given to the control group directly.
The participants will not be given any hints on how to solve the puzzle to keep it fair. Data Collection: Both groups will be timed to solve the puzzle.
The group that solves the puzzle faster will be regarded as the winner. The number of people in each group that solve the puzzle will be recorded.
A correlation test would be performed to determine if the solution time is related to the consumption of beer.
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Data will be collected and analyzed using statistical tools such as the t-test to determine if there is a significant difference in performance between the two groups.
The experiment is designed to verify whether young men who drank two pints of beer first were better able to solve certain word puzzles than sober men. This question requires a well-planned experimental design. The experiment requires a hypothesis and a null hypothesis.
Hypothesis
Drinking two pints of beer can improve the performance of young men in word puzzles than sober men.
Null Hypothesis
Drinking two pints of beer cannot improve the performance of young men in word puzzles than sober men.
Population
The target population of the study is young men aged between 18 to 30 years.
Sample collection
To collect the sample, we will identify potential participants based on the age range of 18-30 years. The study will recruit volunteers who drink alcohol regularly and those who don't. Participants who have consumed alcohol before the study will be required to take a breathalyzer test to ensure they are within the recommended limits. Only those with a blood alcohol concentration of 0.08% and below will be included in the study. Participants will also be required to sign informed consent to participate in the study.
Execute the experiment
Participants will be randomly assigned into two groups: the control group and the experimental group. The control group will be given water to drink while the experimental group will be given two pints of beer. Participants will then be given a set of word puzzles to solve, and their performance will be recorded. Each group will be given an equal time limit to solve the word puzzles.
Data Collection
The data collected will include the number of word puzzles solved by each group, the time taken to solve the word puzzles, and the number of incorrect answers. The data collected will be analyzed using statistical tools such as the t-test to determine if the difference in performance between the two groups is statistically significant. ConclusionThe experiment is designed to verify if drinking two pints of beer can improve the performance of young men in solving certain word puzzles than sober men. The experiment involves a sample size of young men aged 18-30 years who will be randomly assigned to two groups; the experimental group and the control group. Data will be collected and analyzed using statistical tools such as the t-test to determine if there is a significant difference in performance between the two groups.
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(a-1)+(b+3)i = 5+8i
please answer me quickly i need it please
Answer:
a = 6, b = 5
Step-by-step explanation:
Assuming you require to find the values of a and b
Given
(a - 1) + (b + 3)i = 5 + 8i
Equate the real and imaginary parts on both sides , that is
a - 1 = 5 ( add 1 to both sides )
a = 6
and
b + 3 = 8 ( subtract 3 from both sides )
b = 5
My friend Yoy purchased some rews for $3 each and some jooghs for
$5 each. The total cost was about $60. Altogether, he purchased 18
items.
Write a system of equations, in standard form, to model the
relationship between Yoy's rews (x) and jooghs (y).
Answer:
x+Y =x68 i thinkStep-by-step explanation:
Answer:
86
Step-by-step explanation:
Example 1
Make a graph for the table in the Opening Exercise.
Example 2
Use the graph to determine which variable is the independent variable and which is the dependent variable. Then state the relationship between the quantities represented by the variables
The base angles of an isosceles trapezoid are__________?
Answer:
Base angles of an isosceles triangle are always congruent.
will give 20 brainly PLEASE NEED HELP NOW
plz put the answer as simple as a b c or d
Answer:
1. A
2. C
Step-by-step explanation:
Please help! This is my last question and I can’t get it... I gave the most points I could. (I will also mark you the brainliest if it works)
Answer:
18.
Step-by-step explanation:
hey hottie pls help me! thank u ;)
Select the context(s) that could be modeled by a linear function.
A)The amount Ms. Ji Woo and Mrs. Rose pay to rent a car is $50 per day.
B) Ms. Jisoo adds $20 to a savings account each month.
C)The value of Principal Kai's car decreases by 15.5% each year.
D)The price of a stock on Robinhood each year is 110% of its price from the previous year.
E)Mrs. Manoban pays $1000 for car insurance the first year and pays an additional $25 per year
Answer: A,B, and E for sure. I'm not sure about C though.
Step-by-step explanation:
can someone please help me out its important please.
encontre as raízes quadradas dos números:
a)²√625
b)²√100
c)²√81
Answer:
a.) 25, b.)10, c.)9
Step-by-step explanation:
a.) 25x25=625
b.)10x10=100
c.) 9x9=81
Find the value of the variables in the simplest form
Answer:
Step-by-step explanation:
Answer:
x = 15[tex]\sqrt{3}[/tex] , y = 15
Step-by-step explanation:
Using the sine and cosine ratios in the right triangle and the exact values
sin60° = [tex]\frac{\sqrt{3} }{2}[/tex] , cos60° = [tex]\frac{1}{2}[/tex] , then
sin60° = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{x}{30}[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex] ( cross- multiply )
2x = 30[tex]\sqrt{3}[/tex] ( divide both sides by 2 )
x = 15[tex]\sqrt{3}[/tex]
---------------------------------------------------------
cos60° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{y}{30}[/tex] = [tex]\frac{1}{2}[/tex] ( cross- multiply )
2y = 30 ( divide both sides by 2 )
y = 15