There were approximately 586,800 elephants in Africa, rounded up to the nearest elephant.
In 1980, there were 1.2 million elephants in Africa. With a decrease of 6.8% per year, we need to calculate the population in 1987, which is 7 years later.
To find the population in 1987, we use the formula:
Final population = Initial population * (1 - decrease rate) ^ number of years
Final population = 1,200,000 * (1 - 0.068) ^ 7
Final population ≈ 1,200,000 * 0.489
Final population ≈ 586,800
In 1987, there were approximately 586,800 elephants in Africa, rounded up to the nearest elephant.
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Write all integers that belong to the intervals:[-1, 1]
Okay, here we have this:
Considering that the intervals are closed, we have the following integers in this interval: -1,0,1.
solve the following system of equations using lu factorization with partial pivoting: 2x1−6x2−x3=−38 −3x1−x2 7x3=−34 −8x1 x2−2x3=−40
x = [2; 5; 1] is the solution to the given system of equations 2x1−6x2−x3=−38 −3x1−x2 7x3=−34 −8x1 x2−2x3=−40 using LU factorization with partial pivoting.
To solve the system of equations using LU factorization with partial pivoting, we first write the system in matrix form as:
A*x = b
where A is the coefficient matrix:
A = [2 -6 -1; -3 -1 7; -8 1 -2]
x is the vector of unknowns:
x = [x1; x2; x3]
and b is the vector of constants:
b = [-38; -34; -40]
We then perform LU factorization with partial pivoting on A to get:
P*A = L*U
where P is the permutation matrix that represents the row exchanges done during the factorization, L is the lower triangular matrix with ones on the diagonal, and U is the upper triangular matrix. We can solve the system using the following steps:
Step 1: Perform partial pivoting to interchange the first and third rows of A:
A = [−8 1 −2; −3 −1 7; 2 −6 −1]
P = [0 0 1; 0 1 0; 1 0 0]
Step 2: Compute the LU factorization of A:
L = [1 0 0; −0.25 1 0; −0.25 0.8571 1]
U = [−8 1 −2; 0 −0.75 6.5; 0 0 −3.1429]
Step 3: Solve Ly = Pb for y:
Pb = [−40; −34; −38]
y = [−38; 0.25; 1.2857]
Step 4: Solve Ux = y for x:
x = [2; 5; 1]
Therefore, the solution to the system of equations is:
x1 = 2
x2 = 5
x3 = 1
Note: The LU factorization with partial pivoting is a numerical method for solving systems of linear equations. It is not always necessary to use this method for small systems like the one in this example, but it can be useful for larger systems or for matrices that are difficult to solve using other methods.
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How many parallel side and perpendicular side have a triangle
triangles do not have parallel lines, and usually the only triangle that has two perpendicular sides is the right angle triangle.
what is the number of terms for 1000z + 800y
Answer:
2 terms
Step-by-step explanation:
hope this helps! :D
have a MIRACULOUS day!! <3
Help me find x please!!!
Answer:
Step-by-step explanation:
2) 133 + a = 180 {Linear pair}
a = 180 - 133
a = 47
a +b = 180 {linear pair}
47 + b = 180
b = 180 - 47
b = 133
x = b + 30 {Exterior angle property}
x = 133 +30
x = 163
3) m + 35+ 35 = 180 {Angle sum property of triangle}
m + 70 = 180
m = 180 - 70
m = 110
z = 35 {alternate interior angles are congruent}
x = m + z
x = 110 + 35
x = 145
The volume of a right rectangular prism can be found by using the product 6(4)(18). Which expression could also be used to determine the volume of this prism?
A. 6 + 4 + 18
B. 2(6 + 4 + 18)
C. 32(18)
D. 4(6)(18)
Answer:
c
Step-by-step explanation:
Answer:
guys its D
Step-by-step explanation:
Can someone please help me on this question ASAP!! thank you. ' what are the coordinates of A & B? '
Answer:
First photo:
A: (2, 6)B: (8, 4)Second photo:
A: (0, 4)B: (5, 0)Third photo:
A: (8.5, 4)B: (2.5, 1.5)Fourth photo:
A: (4, 2)B: (-2, 3)Fifth photo:
A: (3, -5)B: (-4, -1)I hope this helps!
Read the paragraph.
Not long ago, I decided I needed more exercise, so I began riding my bicycle
almost every day. I like to ride on old country roads because there is a lot to
see.
Last Saturday morning, I saw a horse riding my bicycle down the street.
What a magnificent animal it was!
Which revision corrects the misplaced modifier in the paragraph?
Last Saturday morning, while riding my bicycle down the street, I saw a horse.
OI saw a horse riding my bicycle down the street last Saturday morning.
Last Saturday morning, just down the street, I saw a horse riding my bicycle
I saw a horse riding my bicycle last Saturday morning down the street.
5 points
The revision that corrects the misplaced modifier in the paragraph is option A. Last Saturday morning, while riding my bicycle down the street, I saw a horse.
What is a modifier?In English grammar, a modifier is a word or group of words that provides more information about another word in a sentence. It can either modify a noun, a verb, or another modifier.
Modifiers that describe or provide more information about a noun are called adjectives. For example, in the sentence "The yellow sun is shining," "yellow" is an adjective modifying the noun "sun." Modifiers that describe or provide more information about a verb are called adverbs. For example, in the sentence "She sings beautifully," "beautifully" is an adverb modifying the verb "sings."
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Help me for brainlist please
Answer:
C 80in^2
Step-by-step explanation:
Multiply 12*3*2 for the area of both sides of the H to get 72
For the middle multiply 2*4 to get 8
72+8=80
3/4 cup juice per 2/3 cup sugar, find cups juice per cup sugar
Answer:
9/8
Step-by-step explanation:
divide 3/4 by 2/3
use the KCF method
so your gonna get 3/4 divided 2/3 then change the division to multiplcation and then flip the 2/3 to 3/2
now multiply across: 3/4 × 3/2=9/8
The Busy Bee farm is famous for their wildflower honey. On average, their beehives produce
6 cups of honey every day. If the beekeepers harvest honey every 4 weeks, how many
gallons of honey do they collect?
Answer:
10.5.
Step-by-step explanation:
If the bee farm produces 6 cups of honey every day, and there are 7 days in a week, we can create an equation. you can first do 7×4 which is 28, so with this given you can create another equation; 28×6=168. if there are 16 cups in one gallon, you can divide 168 by 16, which is 10.5 gallons.
Answer:
10.5 gallons in 4 weeks.
Step-by-step explanation:
1 Gallon = 16 Cups
If the farmer gathers 6 cups a day then in a week the farmer gathers 42 cups.
You can figure this out by multiplying the amount of cups the farmer gathers in a day (6) with the amount of days in a week (7).
Then you multiply 42 by the amount of weeks that the farmer gathers honey for to get 168 cups of honey.
Then you convert cups of honey to gallons of honey by dividing 168 by 16.
Your final answer should be 10.5
2. 1 to the power of 2 ( + 10)
Determine whether each relation is a function
Please help I dont wanna fail math this year
Thanks
(I am in Algebra 1 btw)
Find the exact length of the third side.
Answer:
b=6.708
Step-by-step explanation:
Use the pythagorean theorem to solve. \(a^{2} +b^{2} =c^{2}\)
\(6^{2} +b^{2} =9^{2}\)
\(36+b^{2} =81\)
\(b^2=45\)
\(\sqrt{b^{2} } =\sqrt{45}\)
\(b=3\sqrt{5}\)
\(b=6.708\)
If 12 oranges cost $4.80, what would be the cost for 5 oranges
Answer: 24
Step-by-step explanation:
Because if you do 4.80 times 5 you’ll get 24 because each orange cost $4.80
Answer:
the cost of 5 oranges would be $2.00
Step-by-step explanation:
To solve this we need to find the unit rate and when we do this we can solve this problem. To find that we need to divide 4.80 by 12 once we do that we just have to multiply that number by 5 to get our answer, when you divide you'll get 0.4 and then multiply that by 5 and you'll get $2.00.
A lecture class has 30 full-time students and 20 part-time students. A random sample of 10 students is drawn. UseRcommands to answer the following questions: (a) What is the probability that exactly 6 of the students will be full-time? (b) What is the probability that at most 8 of the students will be full-time? (c) Use the binomial approximation that exactly 6 of the students will be full-time, wherep=30/50=0.6. How does this compare to your answer in part a? Is it appropriate to use the binomial approximation in this scenario?
It's important to remember that using the binomial approximation assumes a large population size, which may not be the case here.
To answer your questions, we will use R commands to calculate the probabilities.
(a) To find the probability that exactly 6 of the students will be full-time, we can use the "dhyper" function in R. The parameters are x = 6, m = 30 (number of full-time students), n = 20 (number of part-time students), and k = 10 (sample size).
R command:
```R
dhyper(x = 6, m = 30, n = 20, k = 10)
```
(b) To find the probability that at most 8 of the students will be full-time, we can use the "phyper" function in R, which calculates the cumulative probability. We need to calculate P(X ≤ 8).
R command:
```R
phyper(q = 8, m = 30, n = 20, k = 10)
```
(c) To use the binomial approximation, we can use the "dbinom" function in R with p = 0.6 (probability of full-time student) and n = 10 (sample size). We want to calculate P(X = 6).
R command:
```R
dbinom(x = 6, size = 10, prob = 0.6)
```
Compare this result with the answer in part (a). If the difference is relatively small, it's appropriate to use the binomial approximation in this scenario.
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What are the coordinates of the point?
A) 6, – 4)
B) (-6, – 4)
C) (6,4)
D) to (-4, 6)
Answer:
A) (6,-4)
Step-by-step explanation:
find a equation line perpendicular to y-8=-2x that passes through the point (8,-5)
Answer:
Step-by-step explanation:
y = -2x + 8
perp.: 1/2
y + 5 = 1/2(x - 8)
y + 5 = 1/2x - 4
y = 1/2x - 9
Find 100% when 12 is 10%
Answer:
I think it's 120??
Step-by-step explanation:
10 x 10 = 100
12 x 10 = 100
Answer:
if 10%=12
then
1%=12/10
or,1%=1.2
then, for 100%
100%=1.2*100
hence, 100%=120...
Consider the following NLP: min s.t. 2x12+2x1x2+x22−10x1−10x2
x12+x22≤5
3x1+x2≤6
x1,x2≥0 (a) Aside from regularity and the given constraints, what are the first order necessary conditions for this problem? (Be as specific as possible.) (b) Find a solution by assuming the first Lagrangian multiplier constraint is active and the second one is inactive. (c) Does this satisfy the first order necessary conditions? Explain.
The first-order necessary conditions for the given NLP problem involve the KKT conditions, and a specific solution satisfying these conditions needs further analysis.
(a) The first-order necessary conditions for constrained optimization problems are defined by the KKT conditions. These conditions require that the gradient of the objective function be orthogonal to the feasible region, the constraints be satisfied, and the Lagrange multipliers be non-negative.
(b) Assuming the first Lagrangian multiplier constraint is active means that it holds with equality, while the second one is inactive implies that it does not affect the solution. By incorporating these assumptions into the KKT conditions and solving the resulting equations along with the given constraints, a solution can be obtained.
(c) To determine if the solution satisfies the first-order necessary conditions, one needs to verify if the obtained values satisfy the KKT conditions. This involves checking if the gradient of the objective function is orthogonal to the feasible region, if the constraints are satisfied, and if the Lagrange multipliers are non-negative. Only by performing this analysis can it be determined if the solution satisfies the first-order necessary conditions.
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Olivia needs 54 licks for every 3 lollipops they eat.
Answer:
so what is the question ??
Identify the highlighted part of circle O shown below
Central angle
Secant
Inscribed angle
Chord
Answer:
Chord
Step-by-step explanation:
Notice that the highlighted part is the line segment that joins the points J and E on the circle, which is known as a chord.
A racquet tennis is on sale for 30% off. Find the original price if the sale price is $21.
11. Write an expression with all fractional
terms that applies four properties of
multiplication to simplify the expression to
one term, Include at least one term being
divided. Show your work and explain how
your expression simplifies.
The statement claims that the phrase may be reduced to the single fractional term 8/5.
What is fundamental expression?Expressions serve as the fundamental construction blocks of Statements since every BASIC word is composed of both expressions and keywords (such as GOTO, TO, and STEP). Therefore, in addition to basic mathematical operations and boolean expressions (such as 1 + 2,) expressions can also include functions, constants, and lvalues (scalar constants or arrays).
Let's use the expression: (2/3) x (4/5) ÷ (1/6)
Using the four properties of multiplication:
Associativity: (2/3) x (4/5) ÷ (1/6) = (2/3) x (4/5) ÷ (1/6)
Commutativity: (2/3) x (4/5) ÷ (1/6) = (4/5) x (2/3) ÷ (1/6)
Distributivity: (4/5) x (2/3) ÷ (1/6) = 4/5 x (2/3 ÷ 1/6) = 4/5 x (12/6) = 4/5 x 2
Identity: 4/5 x 2 = 4/5 x 2 = 8/5
As a result, the equation is reduced to the fraction 8/5.Each fractional term in the equation is streamlined using the characteristics of multiplication until only a fractional term is left.
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MATH GENIUSES HELLPP PLEASE A circle has a circumference 200π feet. What is the radius of the circle?
Answer:
31.83 feet
Step-by-step explanation:
circumeference = π x diameter
200 = π x diameter
diameter = 200/π = 63.662
radius is half of diameter so divide by 2
= 31.83
Pepe's science test scores for the semester are 94, 69, 74, 96, He wants to get all of his test score to be an average of at least eighty eight. What score must he get on his fifth test? (Enter a numeric value)
Answer:
Step-by-step explanation:
1. Find the mean
94 + 69+74+96 = 333
Divide 333 by 4 = 83.25
Subtract 83.35 from 88
Answer : 4.75
Now follow the next steps
What is the standard form of 5.58 • 10(5)
Answer:
558000
Step-by-step explanation:
Becouse you add five spaces from the desimal
What is the length of side a?
\(\huge \bf༆ Answer ༄\)
According to given figure, The Triangle BDC is a right angled Triangle (Angle D = 90°)
So, we can apply Pythagoras theorem to find the Side a (Hypotenuse)
Now, let's solve ~
\( \sf \: BD {}^{2} + CD {}^{2} = BC {}^{2} \)\( \sf4 {}^{2} + {3}^{2} = {a}^{2} \)\( \sf16 + 9 = {a}^{2} \)\( \sf \: a {}^{2} = 25\)\( \sf \: a = \sqrt{25} \)\( \sf \: a = 5 \: units\)Therefore, Length of side a is 5 units ~
If you go twice as fast, will your stopping distance increase by: A. Two times. B. Three times. C. Four times. D. Five times
If you go twice as fast, your stopping distance will increase by four times (option C).
This relationship is based on the laws of physics and the principles of motion.
When an object is in motion, its stopping distance is influenced by its initial speed, reaction time, and braking capabilities. The stopping distance consists of two components: the thinking distance (the distance traveled during the reaction time) and the braking distance (the distance needed to bring the object to a complete stop).
According to the laws of physics, the braking distance is directly proportional to the square of the initial speed. This means that if you double your speed, the braking distance will increase by a factor of four. In other words, going twice as fast will require four times the distance to come to a stop.
It is important to note that this relationship assumes other factors, such as road conditions and braking efficiency, remain constant. However, in real-world scenarios, these factors may vary and can affect the stopping distance to some extent.
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Find an equation for the hyperbola with foci (0,±5) and with asymptotes y=± 3/4 x.
The equation for the hyperbola with foci (0,±5) and asymptotes y=± 3/4 x is:
y^2 / 25 - x^2 / a^2 = 1
where a is the distance from the center to a vertex and is related to the slope of the asymptotes by a = 5 / (3/4) = 20/3.
Thus, the equation for the hyperbola is:
y^2 / 25 - x^2 / (400/9) = 1
or
9y^2 - 400x^2 = 900
The center of the hyperbola is at the origin, since the foci have y-coordinates of ±5 and the asymptotes have y-intercepts of 0.
To graph the hyperbola, we can plot the foci at (0,±5) and draw the asymptotes y=± 3/4 x. Then, we can sketch the branches of the hyperbola by drawing a rectangle with sides of length 2a and centered at the origin. The vertices of the hyperbola will lie on the corners of this rectangle. Finally, we can sketch the hyperbola by drawing the two branches that pass through the vertices and are tangent to the asymptotes.
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