Regarding the rules of probability, the correct statement is:
D. The probability of A and its complement will sum to one.
The statement "The probability of A and its complement will sum to one" is a fundamental rule in probability known as the Complement Rule. It states that if A is an event, then the probability of A occurring (denoted as P(A)) plus the probability of A not occurring (denoted as P(A')) is equal to one.
Mathematically, this can be expressed as:
P(A) + P(A') = 1
This rule follows from the fact that the sample space, which includes all possible outcomes, is divided into two mutually exclusive and exhaustive events: A and its complement A'. One of these events must occur, so their probabilities sum to one.
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The sequence (a_n) is defined recursively by a_1 = - 36, a_n+1 = a-n/2 + 72/a-n 1) Find the term a_3 of this sequence. a3 = _________
2) Prove by induction that for all n ∈ N, a_n < 0.
1) The term a_3 of this sequence a3 = -362/37.2
2) By the principle of mathematical induction, for all n ∈ N, aₙ < 0.
1) We are given the recursive formula:
a₁ = -36, aₙ₊₁ = aₙ/2 + 72/aₙ.
We need to find the term a₃ of this sequence. a₂ is given by the recursive formula as:
a₂ = a₁/2 + 72/a₁a₂ = -36/2 + 72/(-36) = -37/2
a₃ is given by the recursive formula as:
a₃ = a₂/2 + 72/a₂= (-37/2)/2 + 72/(-37/2)= -74/37 + (-288/37) = -362/37
Therefore, a₃ = -362/37.2
2) We need to prove by induction that for all n ∈ N, aₙ < 0.
Base case:
For n = 1, we have a₁ = -36 < 0. So, the base case is true.
Inductive step:
Let's assume that for some arbitrary n = k, aₖ < 0.
We need to show that aₖ₊₁ < 0.
Using the recursive formula: aₖ₊₁ = aₖ/2 + 72/aₖ
Since aₖ < 0, -aₖ > 0 and aₖ/2 < 0.Hence, aₖ/2 + 72/aₖ < 0
Therefore, aₖ₊₁ < 0.So, the statement that for all n ∈ N, aₙ < 0 is true for n = 1 and if it's true for n = k, then it's true for n = k + 1.
Therefore, by the principle of mathematical induction, for all n ∈ N, aₙ < 0.
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Find the volume of a prism of altitude "h" with an equilateral triangular base of side "S" BY integration.
The volume of the prism with an equilateral triangular base of side S and altitude h is (S * h^2 * sqrt(3))/4.
To find the volume of a prism with an equilateral triangular base using integration, we can divide the prism into infinitesimally small slices parallel to the base and integrate their volumes.
Consider an infinitesimally thin slice located at a distance "y" from the base. The length of this slice is equal to the length of the base, S. The width of the slice at distance "y" can be determined by considering the height of the equilateral triangle at that distance, which is given by h - (h/S) * y.
The volume of this slice is then given by the product of its length, width, and infinitesimal thickness dy, which is S * [h - (h/S) * y] * dy.
To find the total volume, we integrate this expression from y = 0 to y = h:
V = ∫[0,h] S * [h - (h/S) * y] dy.
Evaluating this integral gives us the volume of the prism:
V = S * [h * y - (h/S) * (y^2/2)] evaluated from y = 0 to y = h.
Simplifying this expression yields:
V = (S * h^2 * sqrt(3))/4.
Therefore, the volume of the prism with an equilateral triangular base of side S and altitude h is (S * h^2 * sqrt(3))/4.
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We formally define the length function f(w) of a string w = ww2...W), (where ne N, and Vi= 1,..., n W; € 2) as 1. if w = €, then f(w) = 0. 2. if w = au for some a € and some string u over 2, then f(x) = 1 + f(u). 1, ..., Prove using proof by induction: For any strings w = wW2...Wy. (where n € N, and Vi W: € 9), f(w) = n.
The length function f(w) of a string w = w₁w₂...W), where n ∈ N and Vi ∈ W: € 9, is equal to n.
The length function f(w) is defined recursively based on the structure of the string w. In the base case, if w is an empty string (ε), the length is defined as 0. In the recursive case, if w can be written as au, where a is a character from the alphabet and u is a string over the alphabet Σ, then the length is defined as 1 plus the length of u.
To prove that for any string w =w₁w₂...wy, where n ∈ N and Vi ∈ W: € 9, the length function f(w) is equal to n, we will use a proof by induction.
Base case:For w = ε (an empty string), we have f(ε) = 0, which satisfies the condition when n = 0.
Inductive step:Assume that for any string w = w₁w₂...wn, where n ∈ N and Vi ∈ W: € 9, the length function f(w) = n.
Now, consider a string w' = w₁w₂...wn+1. By the recursive definition, we can write w' as au, where a is the last character wn+1 and u is the string w₁w₂...wn. From our assumption, we know that f(u) = n.
Therefore, f(w') = 1 + f(u) = 1 + n = n + 1.
Since we have established that for any string w = w₁w₂...wy, where n ∈ N and Vi ∈ W: € 9, the length function f(w) = n, we can conclude that f(w) = n.
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Consider random variables (X, Y ) with joint p.d.f.
fX,Y (x, y) = 1/3 x ≥ 0, y ≥ 0, 2x + 3y ≤ 6
0 otherwise.
(a) Let W = X + Y . Compute FW (w) and fw(w).
(b) Compute E[W] and V ar[W].
(c) Let Z = Y − X. What are the minimum and maximum of Z?
(d) Write FZ(z) in terms of double integral on x and y. You want to consider two separate cases for w ≥ 0 and w < 0.
(e) Find fZ(z).
(f) Compute E[Z] and V ar[Z].
(a) To compute the cumulative distribution function (CDF) of W, denoted as FW(w), we integrate the joint probability density function (PDF) over the appropriate region. The region is defined by the inequalities x ≥ 0, y ≥ 0, and 2x + 3y ≤ 6. The CDF is given by: FW(w) = P(W ≤ w) = ∫∫[fX,Y(x, y)] dy dx
To find the PDF fw(w), we differentiate FW(w) with respect to w.
(b) To compute E[W], we integrate the product of w and the PDF fw(w) over the range of W. The variance V ar[W] is calculated by finding E[W^2] and subtracting (E[W])^2.
(c) To find the minimum and maximum values of Z, we need to determine the range of Y - X. We consider the range of x and y that satisfy the given conditions. By substituting the limits of x and y, we can calculate the minimum and maximum values of Z.
(d) The cumulative distribution function FZ(z) can be written as a double integral over the joint PDF fX,Y(x, y). We consider two cases: w ≥ 0 and w < 0. For each case, we determine the appropriate region and integrate the PDF accordingly.
(e) To find the PDF fZ(z), we differentiate FZ(z) with respect to z.
(f) To calculate E[Z], we integrate the product of z and the PDF fZ(z) over the range of Z. The variance V ar[Z] is computed by finding E[Z^2] and subtracting (E[Z])^2.
Please note that without the specific range or shape of the region defined by the inequalities, it is not possible to provide detailed numerical calculations for each part.
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A quality control company was hired to study the length of meter sticks produced by a certain company. The team carefully measured the length of many meter sticks, and the distribution seems to be severely skewed to the right with a mean of 99.84 cm and a standard deviation of 0.2 cm.
a) What is the probability of finding a meter stick with a length of more than 100.04 cm? ____
b) What is the probability of finding a group of 42 meter sticks with a mean length of less than 99.82 cm?_____
c) What is the probability of finding a group of 50 meter sticks with a mean length of more than 99.87 cm? _____
d) What is the probability of finding a group of 28 meter sticks with a mean length of between 99.82 and 99.86 cm? ______
e) For a random sample of 32 meter sticks, what mean length would be at the 92nd percentile? ______
a) The probability of finding a meter stick with a length of more than 100.04cm is 0.44013.
b) The probability of finding a group of 42 meter sticks with a mean length of less than 99.82 cm is 0.65866.
c) The probability of finding a group of 50 meter sticks with a mean length of more than 99.87 cm is 0.44013.
d) The probability of finding a group of 28 meter sticks with a mean length of between 99.82and 99.86 cm is 0.11974.
e) The mean length that would be at the 92nd percentile for a random sample of 32 meter sticks is 99.89714 cm.
How is this so ?a) The probability of finding a meter stick with a length of more than 100.04cm is
P(X > 100.04 ) =1 - P(X <= 100.04)
= 1- Φ((100.04 - 99.84) / 0.2)
= 1 - Φ(0.12)
= 1 - 0.55987
= 0.44013
b) The probability of finding a group of 42 meter sticks with a mean length of less than 99.82 cm is
P(¯X < 99.82) = 1 - P(X >= 99.82)
= 1 - Φ((99.82 - 99.84) / 0.2 / √42)
= 1 - Φ(-0.1)
= 1 - 0.34134
= 0.65866
c) The probability of finding a group of 50 meter sticks with a mean length of more than 99.87 cm is
P(X > 99.87) = 1 - P(X <= 99.87)
= 1 - Φ((99.87 - 99.84) / 0.2 / √50)
= 1 - Φ(0.15)
= 0.44013
d) The probability of finding a group of 28 meter sticks with a mean length of between 99.82 and 99.86 cm is
P(99.82 < X < 99.86) = Φ ((99.86 - 99.84)/ 0.2 / √28) - Φ((99.82 - 99.84) / 0.2/ √28)
= Φ(0.15) - Φ(0.12)
= 0.55987 - 0.44013
= 0.11974
e) The mean length that would be at the 92nd percentile for a random sample of 32-meter sticks is
X₉₂ = μ + z₉₂ σ / √n
= 99.84 + z₉₂ (0.2) / √32
= 99.84 + 1.85 (0.2) / √32
= 99.84 + 0.05714
= 99.89714
Therefore, the mean length that would be at the 92nd percentile for a random sample of 32 meter sticks is 99.89714 cm.
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Please help me with this I am stuck
Answer:
450 cm ^3
Step-by-step explanation:
Determine the standard error of the estimated slope coefficient for the price of roses (point F) and whether that estimated slope coefficient is statistically significant at the 5 percent level. A. 9.42 and statistically significant since the t-statistic is greater than 2 in absolute value. B. 9.42 and statistically insignificant since the t-statistic is less than 2 in absolute value. C. 4.74 and statistically insignificant since the P-value is greater than 5 percent. D. 4.74 and statistically significant since the P-value is greater than 5 percent.
To determine the standard error of the estimated slope coefficient and its statistical significance, more information is needed, such as the t-statistic or the p-value associated with the estimated slope coefficient. The options provided do not include the necessary details to make a conclusion.
The standard error of the estimated slope coefficient measures the precision or variability of the estimated coefficient. It provides information about how much the estimated slope coefficient could vary across different samples.
The t-statistic and the p-value, on the other hand, are used to assess the statistical significance of the estimated slope coefficient. The t-statistic measures the number of standard errors the estimated coefficient is away from zero, while the p-value indicates the probability of observing a coefficient as extreme as the estimated one under the null hypothesis that the true coefficient is zero.
Without the t-statistic or p-value, it is not possible to determine the statistical significance of the estimated slope coefficient at the 5% level.
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Let v = - [3 1] and u=[2 1]. Write v as the sum of a vector in Span{u} and a vector orthogonal to u. (2) Find the distance from v to the line through u and origin.
The vector v can be written as the sum of a vector in Span{u} and a vector orthogonal to u as follows: v = (1/5)u + (-4/5)[1 -3].
The main answer can be obtained by decomposing the vector v into two components: one component lies in the span of vector u, and the other component is orthogonal to u. To find the vector in the span of u, we scale the vector u by the scalar (1/5) since v = - [3 1] can be written as (-1/5)[2 1]. This scaled vector lies in the span of u and can be denoted as (1/5)u.
To find the vector orthogonal to u, we subtract the vector in the span of u from v. This can be calculated by multiplying the vector u by the scalar (-4/5) and subtracting the result from v. The orthogonal component is obtained as (-4/5)[1 -3].
Thus, we have successfully decomposed vector v as v = (1/5)u + (-4/5)[1 -3], where (1/5)u lies in the span of u and (-4/5)[1 -3] is orthogonal to u.
In linear algebra, vector decomposition is a fundamental concept that allows us to express a given vector as a sum of vectors that have specific properties. The decomposition involves finding a vector in the span of a given vector and another vector that is orthogonal to it. This process enables us to analyze the behavior and properties of vectors more effectively.
In the context of this problem, the vector v is decomposed into two components. The first component, (1/5)u, lies in the span of the vector u. The span of a vector u is the set of all vectors that can be obtained by scaling u by any scalar value. Therefore, (1/5)u represents the part of v that can be expressed as a linear combination of u.
The second component, (-4/5)[1 -3], is orthogonal to u. Two vectors are orthogonal if their dot product is zero. In this case, we subtract the vector in the span of u from v to obtain the orthogonal component. By choosing the scalar (-4/5), we ensure that the resulting vector is orthogonal to u.
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ok soooo
when you rrly think abt it we all have kicked a pregnant lady
Answer:
Step-by-step explanation:
lol true
In a recent year, a research organization found that 517 of 766 surveyed male Internet users use social networking. By contrast 692 of 941 female Internet users use social networking. Let any difference refer to subtracting male values from female values. Complete parts a through d below. Assume that any necessary assumptions and conditions are satisfied. c) What is the standard error of the difference? (Round to four decimal places as needed.) d) Find a 95% confidence interval for the difference between these proportions. 00 (Round to three decimal places as needed.) In a recent year, a research organization found that 517 of 766 surveyed male Internet users use social networking. By contrast 692 of 941 female Internet users use social networking. Let any difference refer to subtracting male values from female values. Complete parts a through d below. Assume that any necessary assumptions and conditions are satisfied. c) What is the standard error of the difference? (Round to four decimal places as needed.) d) Find a 95% confidence interval for the difference between these proportions. 00
a) The proportion of male Internet users who use social networking is approximately 0.6747, and the proportion of female Internet users who use social networking is approximately 0.7358.
b) The difference in proportions is approximately -0.0611.
c) The standard error of the difference is approximately 0.0181.
d) The 95% confidence interval for the difference between these proportions is (-0.096, -0.026).
To calculate the standard error of the difference and find a 95% confidence interval for the difference between the proportions, we can use the formulas for proportions and their differences.
Let [tex]p_1[/tex] be the proportion of male Internet users who use social networking, and [tex]p_2[/tex] be the proportion of female Internet users who use social networking.
a) Proportion for male Internet users: [tex]p_1[/tex] = 517/766 = 0.6747
Proportion for female Internet users: [tex]p_2[/tex] = 692/941 = 0.7358
b) Difference in proportions: [tex]p_1 - p_2[/tex] = 0.6747 - 0.7358 = -0.0611
c) The standard error of the difference (SE) can be calculated using the formula:
[tex]SE = \sqrt{(p_1(1-p_1)/n_1) + (p_2(1-p_2)/n_2)}[/tex]
Where [tex]n_1[/tex] and [tex]n_2[/tex] are the sample sizes for male and female Internet users, respectively.
For male Internet users: [tex]n_1[/tex] = 766
For female Internet users: [tex]n_2[/tex] = 941
Plugging in the values, we have:
[tex]SE = \sqrt{(0.6747(1-0.6747)/766) + (0.7358(1-0.7358)/941)}[/tex]
d) To find the 95% confidence interval for the difference between these proportions, we can use the formula:
[tex]CI = (p_1 - p_2) \± (Z * SE)[/tex]
Where Z is the critical value corresponding to a 95% confidence level. For a large sample size, Z is approximately 1.96.
CI = (-0.0611) ± (1.96 * SE)
CI = -0.0611 ± 0.0355
CI = (-0.096, -0.026)
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COMPUTE THE AREA PARALLELOGRAM DETERMINED BY U= (4,-1) AND V = (-6,-2).
To compute the area of a parallelogram determined by two vectors U = (4, -1) and V = (-6, -2), we can use the formula that states the area of a parallelogram is equal to the magnitude of the cross product of the two vectors. Therefore, the area of the parallelogram determined by U = (4, -1) and V = (-6, -2) is 28 square units.
The formula to compute the area of a parallelogram determined by two vectors U and V is given by:
Area = |U x V|
To calculate the cross product U x V, we can use the following determinant:
| i j k |
| 4 -1 0 |
|-6 -2 0 |
Expanding the determinant, we get:
i * (0 * -2 - 0 * -2) - j * (4 * -2 - 0 * -6) + k * (4 * -2 - (-1) * -6)
= -12i + 24j + 8k
Taking the magnitude of the cross product, we have:
|U x V| = √((-12)^2 + 24^2 + 8^2) = √(144 + 576 + 64) = √784 = 28
Therefore, the area of the parallelogram determined by U = (4, -1) and V = (-6, -2) is 28 square units.
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PLS HELP! NEED TO RAISE GRADE! WILL GIVE BRAINLIEST AND A LOT OF POINTS!
2. A sequence can be generated by using , where and n is a whole number greater than 1.
(a) What are the first five terms in the sequence?
(b) Write an iterative rule for the sequence. Show your work.
{{{ THE BOLDED CHARACTERS SHOULD BE SMALL. }}}
A sequence can be generated by using an = a(n-1) - 5, where a1 = 100 and n is a whole number greater than 1.
a1 = 100 (given)
a2 = a1 - 5 = 100 - 5 = 95
a3 = a2 - 5 = 95 - 5 = 90
a4 = a3 - 5 = 90 - 5 = 85
a5 = a4 - 5 = 85 - 5 = 80
ANSWER for PART (a): 100, 95, 90, 85, 80
-----------------------------------------------------------------------------
an = a1 + d(n - 1)
a1 = 100d is -5 (common difference, and we know it is -5)an = 100 + -5(n - 1)
an = 100 + -5n + 5
an = 105 - 5n
ANSWER for PART (b): an = 105 - 5n
Answer:
I agree with the person above me and here give him brainliest
Step-by-step explanation:
What is the distance to the nearest tenth A unit, between point M (-8,-1) and point N (3,5)?
Answer: 12.5
Step-by-step explanation:
Please Answer This, the question is on the picture. it needs to be a fraction
will mark brainllest if its right, no links!
(for hamster dude)
Answer:
x = 27.2 or 27 1/5
Step-by-step explanation:
cos 54° = 16/x
x = 27.2 or 27 1/5
find the value of a and b.
Answer:
2a°+2a°=180°[opposite angle of cyclic quadrilateralis 180°]
4a°=180
a=180°/4
a=45°
4b+2b=180°[similarly]
6b=180°
b=180°/6
b=30
what is the approximate percentage of a 10c sample left after the time it took a to walk one lap around the gym, where 5 laps takes 200 seconds
The approximate percentage of a 10c sample left after the time it took to walk one lap around the gym is 100 - 25c.
Let x be the time it takes to walk one lap around the gym.
We know that 5 laps take 200 seconds.
Therefore, x can be found by dividing 200 by 5:
x = 200/5 = 40 seconds.
Now, let's find the percentage of the sample left after walking one lap around the gym.
Since x is the time it takes to walk one lap around the gym, we know that the sample decreases at a rate of 10c/x per second.
Therefore, after x seconds, the percentage of the sample remaining is given by: 100(1 - 10c/x)
Substituting x = 40, we get:
100(1 - 10c/40) = 100(1 - 0.25c) = 100 - 25c
So the approximate percentage of a 10c sample left after the time it took to walk one lap around the gym is 100 - 25c.
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what is the sum complete the equation-5 + (20)
Consider a regular surface S in R given by x2 + y2 = 2022. Is S orientable ? Justify your answer.
S is not orientable.
Given a regular surface S in R given by x² + y² = 2022, we need to find out whether S is orientable or not.
The surface is given by, x² + y² = 2022.
Rearranging the terms, we get, y² = 2022 - x²
Let the differentiable function g(x, y) = y, and the set U be the upper hemisphere (upper half) of the surface S.
Then, U = {(x, y, z) : x² + y² = 2022, z ≥ 0}
We know that the partial derivatives of the above function are continuous in U and it follows that U is a regular surface.
We now compute the partial derivatives of g(x, y) :∂g/∂x = 0, and ∂g/∂y = 1
Taking the cross-product of the two partial derivatives, we get : (0i - j + 0k) which is -j.
Now, if we define the positive normal to U to be the upward-pointing unit normal, then we see that (-j) points downward at all points on U.
Thus, U is not orientable.
Therefore, we conclude that the given surface S in R is not orientable.
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Suppose data collected by observers at randomly selected intersections across the country revealed that in a sample of 100 drivers, 30 were using their cell phone. a. Give a point estimate of the true driver cell phone use rate that is, the true proportion-or-population porportion of drivers who are using their cell phone while driving). b. Computea 90% confidence interval for c. Give a practical interpretation of the interval, part b.
a. the point estimate of the true driver cell phone use rate is 0.3 or 30%.
b. the 90% confidence interval for the true driver cell phone use rate is approximately (21.5%, 38.5%).
c. The practical interpretation of the confidence interval is that we are 90% confident that the true driver cell phone use rate falls within the range of 21.5% to 38.5%
a. The point estimate of the true driver cell phone use rate (population proportion) can be calculated by dividing the number of drivers using their cell phone by the total sample size. In this case, the sample size is 100, and 30 drivers were using their cell phone.
Point estimate = Number of drivers using their cell phone / Total sample size
Point estimate = 30/100 = 0.3 (or 30%)
Therefore, the point estimate of the true driver cell phone use rate is 0.3 or 30%.
b. To compute a 90% confidence interval for the true driver cell phone use rate, we can use the formula for a confidence interval for a proportion. The formula is:
Confidence interval = Point estimate ± (Critical value × Standard error)
The critical value depends on the desired level of confidence. For a 90% confidence interval, the critical value is typically obtained from the standard normal distribution and is approximately 1.645.
The standard error can be calculated using the formula:
Standard error = sqrt((point estimate * (1 - point estimate)) / sample size)
In this case, the point estimate is 0.3, and the sample size is 100.
Standard error = sqrt((0.3 * (1 - 0.3)) / 100) ≈ 0.048
Plugging in the values, we can calculate the confidence interval:
Confidence interval = 0.3 ± (1.645 * 0.048)
Confidence interval = (0.215, 0.385)
Therefore, the 90% confidence interval for the true driver cell phone use rate is approximately (21.5%, 38.5%).
c. The practical interpretation of the confidence interval is that we are 90% confident that the true driver cell phone use rate falls within the range of 21.5% to 38.5%. This means that based on the sample data, we can estimate with 90% confidence that the proportion of drivers using their cell phone while driving in the entire population lies between these two percentages. It provides a range of likely values for the true population proportion.
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Find the area of the triangle below.
Answer:
20.25
Step-by-step explanation:
Just multiply the base and height then divide the results in half.
Hope this helps!
First try was incorrect
A circle's diameter is 14 centimeters.
What is the circle's circumference (use 3.14 for pi and round to
nearest tenth)?
Answer:
Circumference = 2(pi)r
C = 2*3.14*4.0= 25.12 round 25.10 m
Step-by-step explanation:
Pls help me the question is in the photo !!
Answer:
14
Step-by-step explanation:
separate the figure into two shapes. On the little one is 2x1 which the answer is 2 then on the larger shape is 6x2 which is 12 then you add 12+2 and then you get your answer
I need help pls and no files ok pls no files
Answer:
$2,725
Step-by-step explanation:
The maximum number of days is 7.
The cost per days is $375.
The cost for 7 days is 7 * $375 = $2,625.
The owner applies a $100 fee for cleaning which must be added to the cost of the days.
$2,625 + $100 = $2,725
The greatest value for the range is the greatest cost there can be which is $2,725.
Please help I need this quickly
Answer:
31 degrees
Step-by-step explanation:
1. since angle x and angle y are congruent, and segment xw and segment yw are congruent, you can assume that angle xzw and angle wzy are also congruent.
2. set angle xzw and angle wzy equal to each other
8x-1 = 5x+11
3. combine like terms
3x = 12
4. simplify to find the value of x
x = 4
5. plug x into the equation to find the degree of angle wzy
5(4)+11
20+11
=31 degrees
Solve the problem. The function D(h) = 5e^-0.4h can be used to determine the milligrams D of a certain drug in a patient's bloodstream h hours after the drug has been given. How many milligrams (to two decimals) will be present after 9 hours? a. 182.99 mg b. 0.14 mg c. 1.22 mg d. 3.35 mg
B. 0.14 mg will be present after 9 hours.
The given function is D(h) = 5e^(-0.4h), which can be used to determine the milligrams D of a certain drug in a patient's bloodstream h hours after the drug has been given.
To find the milligrams after 9 hours, we need to plug in h = 9 in the function D(h) = 5e^(-0.4h).
D(h) = 5e^(-0.4h)
D(9) = 5e^(-0.4(9))
D(9) = 5e^(-3.6)
D(9) = 5 × 0.024419
D(9) = 0.1220 ≈ 0.12 mg
Hence, the answer is option (c) 1.22 mg.
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6.01 x 0.2 =
Can anyone do this plz
Answer:
1.202
Step-by-step explanation:
Calculate (4 + 10i)^2
By applying the the FOIL method, which stands for First, Outer, Inner, Last we obtained the result -84 + 80i for (4 + 10i)^2.
To calculate (4 + 10i)^2, we can:
First, we multiply the first terms of each binomial:
(4 + 10i) * (4 + 10i) = 16 + 40i
Next, we multiply the outer terms of each binomial:
(4 + 10i) * (4 + 10i) = 16 + 40i
Then, we multiply the inner terms of each binomial:
(4 + 10i) * (4 + 10i) = 16 + 40i
Finally, we multiply the last terms of each binomial:
(4 + 10i) * (4 + 10i) = 100i^2
We know that i^2 is equal to -1, so we can substitute that in:
100(-1) = -100
Putting it all together, we get:
(4 + 10i)^2 = 16 + 40i + 40i + (-100)
= -84+80i
Therefore, by applying this method for squaring a complex number, we obtained the result -84 + 80i for (4 + 10i)^2.
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the heights of mature pecan trees are approximately normally distributes with a mean of 42 feet and a standard deviation of 7.5 feet. what proportion are between 43 and 46 feet tall.
The proportion of mature pecan trees between 43 and 46 feet tall can be calculated using the normal distribution with a mean of 42 feet and a standard deviation of 7.5 feet.
To find the proportion, we need to calculate the z-scores corresponding to the given heights and then find the area under the normal curve between those z-scores.
First, we calculate the z-score for 43 feet:
z1 = (43 - 42) / 7.5 = 0.1333
Next, we calculate the z-score for 46 feet:
z2 = (46 - 42) / 7.5 = 0.5333
Using a standard normal distribution table or a calculator, we can find the area between these two z-scores. The area corresponds to the proportion of trees between 43 and 46 feet tall.
The explanation would involve using a standard normal distribution table or a calculator to find the area under the normal curve between the z-scores of 0.1333 and 0.5333. This area represents the proportion of mature pecan trees between 43 and 46 feet tall.
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Find the distance between the points (3, -8) and (8,4).
Answer:
13
Step-by-step explanation:
Answer:
13
Step-by-step explanation:
Hello There!
We can calculate the distance between two points using the distance formula
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
where the x and y values are derived from the coordinates in which you are trying to find the distance between
The points we need to find the distance between is (3,-8) and (8,4)
So we plug the x values and y values into the formula
[tex]d=\sqrt{(8-3)^2+(4-(-8)^2} \\8-3=5\\4-(-8)=12\\d=\sqrt{5^2+12^2} \\5^2=25\\12^2=144\\144+25=169\\\sqrt{169} =13[/tex]
so we can conclude that the distance between the two points is 13 units
please help me solve these equations for geometry :-((
The length M is 12
The measure of the angle BEC is 102
The measure of the arc AB is 128
The length PQ is 18.3
How to calculate the length MThe length M can be calculated using
M² = 8 * (8 + 10)
So, we have
M² = 144
Take the square roots
M = 12
How to calculate the BECThe measure of the angle BEC can be calculated using
BEC = 1/2 * (BC + AD)
So, we have
BEC = 1/2 * (156 + 48)
Evaluate
BEC = 102
How to calculate the ABThe measure of the arc ABcan be calculated using
AB = 180 - 2 * BC
So, we have
AB = 180 - 2 * 26
Evaluate
AB = 128
How to calculate the PQThe length PQ can be calculated using
x² = (12 + 8)² - 8²
So, we have
x² = 336
Take the square roots
x = 18.3
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