is csc2 x + sec2x=1 an identity
Answer:
Step-by-step explanation:
=
sin
2
x
+
cos
2
x
sin
2
x
cos
2
x
=
1
Answer:
explanation is below happy to help ya!
Step-by-step explanation:
Petra jogs 5 miles in 40 minutes. At this rate, how long would it take her to jog 11 miles?
AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
Answer:
88 mins
Step-by-step explanation:
Divide 40 by 5 to get 8. That is your mile per minute ratio. Then multiply 11 by 8 to get the time of 11 miles!
What is 2 3/4 in a decimal format
Answer:
2.5
Step-by-step explanation:
Answer:
2.75
Step-by-step explanation:
2 x 4=8 + 3=11/4=2.75
A submarine is positioned at 300 feet below sea level. A bird is flying 100 foot above sea
level. How many feet above the submarine is the bird?
-300-200-1000
100 200 300
Answer:
h = 400 ft
The bird is 400 ft above the submarine
Step-by-step explanation:
Given;
The submarine is 300 feet below sea level
Vertical distance between submarine and sea level h1 = 300 feet
The bird is flying at 100 feet above sea level
Vertical distance between the sea level and the bird h2 = 100 feet
The total vertical distance between the bird and the submarine is the sum of the vertical distance between submarine and sea level and the vertical distance between the sea level and the bird.
h = h1 + h2
h = 300 ft + 100 ft
h = 400 ft
The bird is 400 ft above the submarine
Answer:
The bird is 400 ft above the submarine
Step-by-step explanation:
The reading speed of second grade students in a large city is approximately normal, with a mean of 90 words per
minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (f).
-
(b) What is the probability that a random sample of 10 second grade students from the city results in a mean
reading rate of more than 96 words per minute?
The probability is ___
(Round to four decimal places as needed.)
Answer:
The probability that a random sample of 10 second grade students from the city results in a mean reading rate of more than 96 words per minute
P(x⁻>96) =0.0359
Step-by-step explanation:
Explanation:-
Given sample size 'n' =10
mean of the Population = 90 words per minute
standard deviation of the Population =10 wpm
we will use formula
[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]
Let X⁻ = 96
[tex]Z = \frac{96-90 }{\frac{10}{\sqrt{10} } }[/tex]
Z = 1.898
The probability that a random sample of 10 second grade students from the city results in a mean reading rate of more than 96 words per minute
[tex]P(X^{-}>x^{-} ) = P(Z > z^{-} )[/tex]
= 1- P( Z ≤z⁻)
= 1- P(Z<1.898)
= 1-(0.5 +A(1.898)
= 0.5 - A(1.898)
= 0.5 -0.4641 (From Normal table)
= 0.0359
Final answer:-
The probability that a random sample of 10 second grade students from
= 0.0359
What is the sum of 8 and 10?
Answer:
18
Step-by-step explanation:
pls mark brainliest
idk if this is a joke
What is the mean for the data set? 138,142,105,112,108,134,106,181,164,105
during each cycle, the velocity v (in meters per second) of a robotic welding device is given by v=9t-2/9+t^2, where t is time in seconds. find the expression for the displacement s (in meters) as a function of t if s=0 when t=0.
Answer:
[tex]d = \dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]
Step-by-step explanation:
Given the equation of velocity w.r.to time 't':
[tex]v=9t-\dfrac{2}{9}+t^2 ...... (1)[/tex]
Formula for Displacement:
[tex]Displacement = \text{velocity} \times \text{time}[/tex]
So, if we find integral of velocity w.r.to time, we will get displacement.
[tex]\Rightarrow \text{Displacement}=\int {v} \, dt[/tex]
[tex]\Rightarrow \int {v} \, dt = \int ({9t-\dfrac{2}{9}+t^2}) \, dt \\\Rightarrow \int{9t} \, dt - \int{\dfrac{2}{9}} \, dt + \int{t^2} \, dt\\\Rightarrow s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3} + C ....... (1)[/tex]
Here, C is constant (because it is indefinite integral)
Formula for integration used:
[tex]1.\ \int({A+B}) \, dx = \int {A} \, dx + \int{B} \, dx \\2.\ \int({A-B}) \, dx = \int {A} \, dx - \int{B} \, dx \\3.\ \int{x^{n} } \, dx = \dfrac{x^{n+1}}{n+1}\\4.\ \int{C } \, dx = Cx\ \{\text{C is a constant}\}[/tex]
Now, it is given that s = 0, when t = 0.
Putting the values in equation (1):
[tex]0=\dfrac{9\times 0^{2} }{2} - \dfrac{2}{9}\times 0 + \dfrac{0^3}{3} + C\\\Rightarrow C = 0[/tex]
So, the equation for displacement becomes:
[tex]s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]
Which of the following are solutions to the quadratic equation x squared plus 8x=9
Answer:
X=1 x=-9
Step-by-step explanation:
(x^2)+8x=9
(x^2)+8x-9=0
(x-1)(x+9)=0
x=1 x=-9
Hallar el área y el perímetro de un rombo cuyas diagonales menor y mayor miden, respectivamente,10 cm y 24 cm
Answer:
The area is 120 cm²
the perimeter is 52 cm
Step-by-step explanation:
The rhombus area is given by:
A = D * d / 2
that is, the larger diagonal D by the smaller diagonal d, between two, we know that D = 24 cm and d = 10 cm, replacing:
A = 24 * 10/2
A = 120
The area is 120 cm²
To calculate the perimeter use the Pythagorean theorem
h² = a² + b²
Since if you look at the rhombus it is formed by four right triangle we will take 1 of them with the following measures 5 cm (10/2) and height 12 cm (24/2) and replace:
h² = 5² + 12²
h² = 169
h = 13
now, the perimeter would be the sum of all its sides, which in this case are equal and measure 13 cm, therefore:
p = 4 * 13
p = 52
which means that the perimeter is 52 cm
Giraffes have excellent vision. Refer to the African Zoology (Oct. 2013) study of a giraffe’s eyesight. Recall that the researchers measured the eye mass for a sample of 27 giraffes native to Zimbabwe, Africa, and found x = 53.4 grams and s = 8.6 grams. Suppose the objective is to sample enough giraffes in order to obtain an estimate of the mean eye mass to within 3 grams of its true value with a 99% confidence interval.A. Identify the cofidence coefficient for this study. B. Identify the desired sampling error for this study. C. Fnd the sample size required to obtain the desired estimate of the true mean.
Answer:
(a) The confidence coefficient is 2.779.
(b) The desired sampling error for this study is the margin of error, ±3.
(c) The sample size required to obtain the desired estimate of the true mean is 55.
Step-by-step explanation:
The information provided is:
[tex]\bar x=53.4\ \text{grams}\\s=8.6\ \text{grams}\\n=27\\\text{Confidence Level} = 99\%[/tex]
(a)
As the sample size is n = 27 < 30 and the population standard deviation is not provided a t-interval would be used to estimate the true mean.
So, the confidence coefficient for the 99% confidence interval for true mean is:
[tex]t_{\alpha/2, (n-1)}=t_{0.01/2, (27-1)}=t_{0.005, 26}=2.779[/tex]
*Use the t-table.
Thus, the confidence coefficient is 2.779.
(b)
The error caused due to incorrect sample selection, i.e. a sample that is not a true representative of the population is known as the sampling error.
In this case the sampling error is known as the margin of error of the confidence interval.
The margin of error is:
MOE = 3 grams
Thus, the desired sampling error for this study is the margin of error, ±3.
(c)
Assume that the ample size required to obtain the desired estimate of the true mean is quite large, such that the sampling distribution of sample mean follows a normal distribution according to the central limit theorem.
Then the margin of error for a z-interval for mean is:
[tex]\text{MOE}=z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]
MOE = 3 grams
The critical value of z for 99% confidence level is:
[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58[/tex]
*Use a z-table.
Compute the sample size as follows:
[tex]\text{MOE}=z_{\alpha/2}\times \frac{s}{\sqrt{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\times s}{MOE}]^{2}[/tex]
[tex]=[\frac{2.58\times 8.6}{3}]^{2}\\\\=54.700816\\\\\approx 55[/tex]
Thus, the sample size required to obtain the desired estimate of the true mean is 55.
Consider the following system of linear equations: Instructions: Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. At the end of the procedure, clearly state the solution to the system outside of a matrix. 1. If the solution is unique, express the solution in real numbers. 3. If there are infinitely many solutions, express the solution in parameter(s). 3. If there is no solution, say so, and explain why.All of the following are possible ranks of a 4x3 matrix except:0123 4 How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-
Answer:
Consider the following system of linear equations: 2 + 3y + 2z = 5 - 2x + y - z= -2 2x + 3z = 11 Instructions: • Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. • Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. • At the end of the procedure, clearly state the solution to the system outside of a matrix. • If the solution is unique, express the solution in real numbers. • If there are infinitely many solutions, express the solution in parameter(s). . If there is no solution, say so, and explain why.
All of the following are possible ranks of a 4x3 matrix EXCEPT O 1 2 3 4
How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-r
Step-by-step explanation:
x + 3y +2z = 5
-2x + y - z = -2
2x + 3z = 11
Here,
[tex]A = \left[\begin{array}{ccc}1&3&2\\-2&1&-1\\2&0&3\end{array}\right][/tex]
[tex]B =\left[\begin{array}{ccc}5\\-2\\11\end{array}\right][/tex]
[tex]X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex]
i.e AX=B
We can write as augmented matrix
[tex]\left[\begin{array}{ccc|c}1&3&2&5\\-2&1&-1&-2\\2&0&3&11\end{array}\right][/tex]
[tex]\frac{R_2\rightarrow R_2+2R_1}{R_3\rightarrow R_3-2R_1} \left[\begin{array}{ccc|c}1&3&2&5\\0&7&3&8\\0&-6&-1&1\end{array}\right][/tex]
[tex]\frac{R_3\rightarrow R_3+\frac{6R_2}{7} }{R_1\rightarrow R_1-\frac{3R_2}{7} } \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&7&3&8\\0&0&11/7&55/7\end{array}\right][/tex]
[tex]\frac{R_2\rightarrow\frac{R_2}{7}}{R_3\rightarrow\frac{7}{11}R_3} \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&1&3/7&8/7\\0&0&11/7&55/7\end{array}\right][/tex]
[tex]\frac{R_1\rightarrow R_1 -\frac{5}{7}R_3}{R_2\rightarrow R_2 -\frac{3}{7}R_3} \left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&-1\\0&0&1&5\end{array}\right][/tex]
Since Rank (A|B) = Rank (A) = 3 = number of variables
⇒ systems has unique solution and x = -2 , y = -1 , z = 5Given a regular octagon, in how many ways can we color one diagonal red and another diagonal blue so that the two colored diagonals cross (in the interior)? Consider rotations and reflections distinct.
Answer:
Rotation = 1.
Reflection = 5 or 4.
Step-by-step explanation:
A regular octagon is the octagon that has eight, 8 sides and all these eight sides are equal.
So, if we are going to consider rotation and reflection distinct in order to color one diagonal red and another diagonal blue so that the two colored diagonals cross (in the interior) we can do that by following the steps Below;
=> Make sure that the first diagonal with the red color is drawn first.
=> to another make fixing to another end.
=> Determine the possibilities.
The numbers of times can be gotten by using Burnside Lemma which gives Rotation = 1 And Reflection = 5 or 4.
which graph represents this equation 5y=x+5
Answer: A
Step-by-step explanation: 5y=x+5 simplified to slope-intercept form is y = 1/5x + 1
The equation 5y = x + 5 is a linear equaton, and the graph A represents the equation 5y = x + 5
How to determine the graph?The equation is given as:
5y = x + 5
Make x = 0.
So, we have:
5y = 0 + 5
This gives
5y = 5
So, we have:
y = 1
The point is represented as (0,1)
Make y = 0.
So, we have:
5 * 0 = x + 5
This gives
0 = x + 5
So, we have:
x = -5
The point is represented as (-5,0)
This means that the graph of 5y = x + 5 passes through the points (0,1) and (-5,0)
From the list of options, only graph (A) passes through the points (0,1) and (-5,0)
Hence, the graph A represents the equation 5y = x + 5
Read more about linear equations at:
https://brainly.com/question/14323743
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In the Journal of Shell and Spatial Structures (December 1963), environmental researcher Vivek Ajmani studied the performance of truss and frame structures subjected to uncertain loads. The load was assumed to have a normal distribution with a mean of 20,000 pounds. Also, the probability that the load is between 10,000 and 30,000 pounds is 0.95. Based on this information, find the standard deviation of the load distribution. Put your answer in three decimal places.
Answer:
The standard deviation of the load distribution is of 5102.041 pounds.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 20000[/tex]
Also, the probability that the load is between 10,000 and 30,000 pounds is 0.95.
10,000 pounds and 30,000 pounds are equidistant from the mean. Due to this, and the probability of 0.95 of having a value in this range, 10000 is the (100-95)/2 = 2.5th percentile and 30000 is the (100+95)/2 = 97.5th percentile. Applying one of them, we find the standard deviation.
30,000 is the 97.5th percentile:
This means that when X = 30000, Z has a pvalue of 0.975. So when X = 30000, Z = 1.96. Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.96 = \frac{30000 - 20000}{\sigma}[/tex]
[tex]1.96\sigma = 10000[/tex]
[tex]\sigma = \frac{10000}{1.96}[/tex]
[tex]\sigma = 5102.041[/tex]
The standard deviation of the load distribution is of 5102.041 pounds.
Write an equation that represents the following:
The ages of two sisters are consecutive integers . The sum of the older sister's age and three times the younger sister's age is 61.
Answer:
(o+y) + 3y = 61
to solve divide by 5 because there is 4 younger sister age + 1 for the older sister. 61-1=60. we subtract because the sister is older by 1 year
60/5=12 so the younger sister is 12
(13+12) + 36 = 61
Consecutive means in order like 3,4,5 or 11,12,13
Hope this helps
Step-by-step explanation:
Find the general solution of the simple homogeneous "system" below, which consists of a single linear equation. Give your answer as a linear combination of vectors. Let x2 and x3 be free variables. 3x1 - 6x2 9x3
Answer:
[tex]= \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]
Step-by-step explanation:
Given: 3x1 - 6x2 + 9x3 = 0
x2 and x3 are free variables
We have:
3x1 = 6x2 - 9x3
divide all sides by 3, we have:
x1 = 2x2 - 3x3
Finding the general solution, we have:
[tex] \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{ccc}2x_2 - 3x_3\\x_2\\x_3\end{array}\right] [/tex]
[tex] = \left[\begin{array}{ccc}2x_2\\x_2\\0\end{array}\right] + \left[\begin{array}{ccc}-3x_3\\0\\x_3\end{array}\right][/tex]
[tex]= x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]
The general solution is
[tex]= \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]
Find the relationship between zeros and x-intercepts of a2 + 5a + 6.
Answer:
The x-intercepts, or zeros, which are the values of x(or a in this problem) for which the function is 0, are x = a = -2 and x = a = -3.
Step-by-step explanation:
Suppose we have a function y = f(x). The zeros, which are the values of x for which y = 0, are also called the x-intercepts of the function.
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
In this question:
I will write the function as a function of x, just exchanging a for x.
[tex]f(x) = x^{2} + 5x + 6[/tex]
[tex]\bigtriangleup = 5^{2} - 4*1*6 = 1[/tex]
[tex]x_{1} = \frac{-5 + \sqrt{1}}{2*1} = -2[/tex]
[tex]x_{2} = \frac{-5 - \sqrt{1}}{2*1} = -3[/tex]
The x-intercepts, or zeros, which are the values of x(or a in this problem) for which the function is 0, are x = a = -2 and x = a = -3.
% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.
Complete Question:
41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.
Answer:
a) P(exactly 5) = 0.209
b) P(at least six) = 0.183
c) P(less than four) = 0.358
Step-by-step explanation:
Sample size, n = 10
Proportion of adults that have very little confidence in newspapers, p = 41% p = 0.41
q = 1 - 0.41 = 0.59
This is a binomial distribution question:
[tex]P(X=r) = nCr p^{r} q^{n-r}[/tex]
a) P(exactly 5)
[tex]P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 252 * 0.01159 * 0.072\\P(X=5) = 0.209[/tex]
b) P(at least six)
[tex]P(X \geq 6) = P(6) + P(7) + P(8) + P(9) + P(10)[/tex]
[tex]P(X\geq6) = (10C6 * 0.41^6*0.59^4) + (10C7*0.41^7*0.59^3) + (10C8*0.41^8*0.59^2) + (10C9 *0.41^9*0.59^1) + (10C10 *0.41^{10})\\P(X\geq6) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001\\P(X\geq6) = 0.183[/tex]
c) P(less than four)
[tex]P(X < 4) = 1 - [x \geq 4][/tex]
[tex]P(X<4)= 1 - [P(4) + P(5) + P(x \geq 6)][/tex]
[tex]P(X <4)= 1 - [(10C4*0.41^4*0.59^6) + 0.209 + 0.183]\\P(X <4)= 0.358[/tex]
A wire 18cm long is cut into 2 pieces. the longer piece is 2 cm longer than the shorter piece. Find the lenght of the shorter piece
Answer:
8cm
Step-by-step explanation:
If the pieces are 8cm and 10cm they will still add up to 18cm but the longer piece is 2cm longer so both conditions are satisfied.
A real estate agent has 17 properties that she shows. She feels that there is a 40% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling less than 5 properties in one week. Round your answer to four decimal places.
Answer:
0.126 = 12.6% probability of selling less than 5 properties in one week.
Step-by-step explanation:
For each property, there are only two possible outcomes. The chance of selling any one property is independent of selling another property. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
A real estate agent has 17 properties that she shows.
This means that [tex]n = 17[/tex]
She feels that there is a 40% chance of selling any one property during a week.
This means that [tex]p = 0.4[/tex]
Compute the probability of selling less than 5 properties in one week.
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{17,0}.(0.4)^{0}.(0.6)^{17} = 0.0002[/tex]
[tex]P(X = 1) = C_{17,1}.(0.4)^{1}.(0.6)^{16} = 0.0019[/tex]
[tex]P(X = 2) = C_{17,2}.(0.4)^{2}.(0.6)^{15} = 0.0102[/tex]
[tex]P(X = 3) = C_{17,3}.(0.4)^{3}.(0.6)^{14} = 0.0341[/tex]
[tex]P(X = 4) = C_{17,4}.(0.4)^{4}.(0.6)^{13} = 0.0796[/tex]
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0019 + 0.0102 + 0.0341 + 0.0796 = 0.126[/tex]
0.126 = 12.6% probability of selling less than 5 properties in one week.
Solve for e - 9e+4=−5e+14+13e
Answer:
e=1o ajjajajajajkananwnwnwhhshssh
Rangers tagged and released 300 salmon into a Maine lake. A
month later, fishermen on the lake were surveyed. They reported
catching 80 salmon, 12 of which had tags. Using this sample,
estimate the salmon population in the lake.
Answer:
2,000 salmon
Step-by-step explanation:
The point estimate of the fraction of salmon that corresponds to the tagged Salmon released by the rangers, based on the fishermen report, is:
[tex]p = \frac{12}{80}\\p=0.15[/tex]
This means that the 300 salmon released into the lake correspond to 15% of the total salmon population. The estimate for the salmon population is:
[tex]P= \frac{300}{0.15}\\P=2,000\ salmon[/tex]
The estimate is 2,000 salmon.
Answer:
The estimate of the salmon population will be 2,000
Step-by-step explanation:
Rangers tagged and released 300 salmon into a Maine lake.
The fishermen reported catching 80 salmon of which 12 had tags
Thus we have 12/80 as a fraction of the tagged and released 300
12/80 = 0.15 which is about 15% of the population of tagged and released 300.
To then estimate the salmon population, we have 300/0.15 = 2,000
3) For a certain good we have LaTeX: q=f\left(p\right)=200e^{-0.4p}q = f ( p ) = 200 e − 0.4 p.
a) Find the elasticity of demand at price p = $50.
b) At p = $50, is the demand elastic, inelastic, or does it have unit elasticity? Explain what this means for this product.
c) Find the elasticity of demand at price p = $20.
d) At p = $20, is the demand elastic, inelastic, or does it have unit elasticity? Explain what this means for this product.
Answer:
(a)20
(b)Elastic
(c)8
(d) Elastic
Step-by-step explanation:
Elasticity of demand(E) indicates the impact of a price change on a product's sales.
The general formula for an exponential demand curve is given as:
[tex]y=ae^{-bp}[/tex]
Given the demand curve formula
[tex]q=f\left(p\right)=200e^{-0.4p}[/tex]
The formula for Elasticity of demand, E
[tex]E = -\dfrac{p}{q}\dfrac{\text{d}q}{\text{d}p}[/tex]
(a)When Price, p = $50
p=50
[tex]q=200e^{-0.4*50}=200e^{-20}[/tex]
[tex]\dfrac{\text{d}q}{\text{d}p}=-0.4*200e^{-0.4p}=-80e^{-0.4p}[/tex]
Therefore:
[tex]E = -\dfrac{50}{200e^{-20}}*-80e^{-0.4*50}\\=\dfrac{1}{4e^{-20}}*80e^{-20}\\\\E=20[/tex]
(b)At p = $50, Since elasticity is greater than 1, the demand is elastic.
An elasticity value of 20 means that a 1% increase in price causes a 20% decrease in demand.
(c)At p=$20
p=20
[tex]q=200e^{-0.4*20}=200e^{-8}[/tex]
[tex]\dfrac{\text{d}q}{\text{d}p}=-0.4*200e^{-0.4p}=-80e^{-0.4p}[/tex]
Therefore:
[tex]E = -\dfrac{20}{200e^{-8}}*-80e^{-0.4*20}\\=\dfrac{1}{10e^{-20}}*80e^{-20}\\\\E=8[/tex]
(d)At p = $20, the demand is elastic.
An elasticity value of 8 means that a 1% increase in price causes a 8% decrease in demand.
When I visit a local cafe, there's an $80\%$ probability that I'll order a sandwich, and a $70\%$ probability that I'll order a soup. If there's a $5\%$ probability that I don't order the sandwich or soup, then what's the probability that I order both the sandwich and soup
Answer:
There is 55% probability that I order both the sandwich and soup
Step-by-step explanation:
P(sandwich) = 0.8
P(Soup) = 0.7
P(neither sandwich nor soup) = 0.05
P(sandwich or soup) = 1 - P(neither sandwich nor soup)
P(sandwich or soup) = 1 - 0.05 = 0.95
P(Sandwich & Soup) = x
P(Sandwich only) = 0.8 - x
P(Soup only) = 0.7 - x
P(sandwich or soup) = P(Sandwich only) + P(Soup only) + P(Sandwich & Soup)
Note that P(neither sandwich nor soup) has already been used to get the P(sandwich or soup) and should not be included in the above formula. Don't make that mistake!
0.95 = 0.8 - x + 0.7 - x + x
0.95 = 1.50 - x
x = 1.50 - 0.95
x = 0.55
There is 55% probability that I order both the sandwich and soup
In a data distribution, the first quartile, the median and the means are 30.8, 48.5 and 42.0 respectively. If the coefficient skewness is −0.38
a) What is the approximate value of the third quartile (Q3 ), correct to 2 decimal places.
b)What is the approximate value of the variance, correct to the nearest whole number
Answer:
a) The third quartile Q₃ = 56.45
b) The variance = 2633.31
Step-by-step explanation:
a) The coefficient of skewness formula is given as follows;
[tex]SK = \dfrac{Q_{3}+Q_{1}-2Q_{_{2}}}{Q_{3}-Q_{1}}[/tex]
Plugging in the values, we have;
[tex]-0.38 = \dfrac{Q_{3}+30.8-2 \times 48.5_{_{}}}{Q_{3}-30.8}[/tex]
Solving gives Q₃ = 56.45
b) To determine the variance, we use the skewness formula as follows;
[tex]SK_{p} = \dfrac{Mean-\left (3\times Median - 2\times Mean \right )}{\sigma } = \dfrac{3\times\left ( Mean - Median \right )}{\sigma }[/tex]
Plugging in the values, we get;
[tex]-0.38= \dfrac{42-\left (3\times 48.5- 2\times 42\right )}{\sigma } = \dfrac{-19.5}{\sigma}[/tex]
[tex]\therefore \sigma =\dfrac{-19.5}{-0.38} = 51.32[/tex]
The variance = σ² = 51.32² = 2633.31.
What is the quotient of (x^3+3x^2+5x+3) divided by (x+1)
Answer:
x^2+2x+3 I sent the answer
The city manager made a scatter plot of the number of retail stores in a
city over many years. The trend line for the scatter plot has the equation y =
11x + 12, where y is the total number of stores and x is the number of years
after 1990. Predict how many retail stores there were in 1997.
Answer:
There were 89 retail stores in 1997.
Step-by-step explanation:
You know:
y = 11*x + 12
where y is the total number of stores and x is the number of years after 1990.
Being this a linear function (Linear functions are those functions that have the form y = mx + b as it happens in this case), to know the total number of stores and, then you must know the value of x, that is, the number of years after 1990, and replace that value in the given function.
To find out how many retail stores there were in 1997, then since 1990 7 years have passed, calculated by subtracting: 1997 - 1990 = 7
So, 7 is the number of years after 1990, then x=7
Replacing in the function y=11*x + 12 you get:
y=11*7+12
Solving:
y=77+12
y=89
There were 89 retail stores in 1997.
coin is tossed and an eight-sided die numbered 1 through 8 is rolled. Find the probability of tossing a tail and then rolling a number greater than 5. The probability of tossing a tail and then rolling a number greater than 5 is nothing. (Round to three decimal places as needed.)
Answer:
The probability of tossing a tail and then rolling a number greater than 5 is 0.188
Step-by-step explanation:
Independent events:
If two events, A and B, are independent, we have that:
[tex]P(A \cap B) = P(A)*P(B)[/tex]
In this question:
The coin and the die are independent. So
Event A: Tossing a tail.
Event B: Rolling a number greater than 5.
Probability of tossing a tail:
Coin can be heads or tails(2 outcomes), so the probability of a tail is [tex]P(A) = \frac{1}{2} = 0.5[/tex]
Probability of rolling a number greater than 5:
8 numbers(1 through 8), 3 of which(6,7,8) are greater than 5. So the probability of rolling a number greater than 5 is [tex]P(B) = \frac{3}{8} = 0.375[/tex]
Probability of A and B:
[tex]P(A \cap B) = P(A)*P(B) = 0.5*0.375 = 0.188[/tex]
The probability of tossing a tail and then rolling a number greater than 5 is 0.188
The table shows the weekly income of 20 randomly selected full time students. If the
Answer: is that the full question?
Step-by-step explanation: