Answer:
5th or 6th grade
Step-by-step explanation:
The reason why we learn geography in 5th grade is to prepare us for 6th grade geography. The reason why we learn geography in 6th grade is because Geography helps us understand basic physical systems that affect everyday life. In other words, geography is a nice skill to have when you're learning about the water cycle or rock formations, or the moving of the tectonic plates (and other natural disasters). It's also a very important skill to have when you want to start traveling.
Which of the following best describes the set of complex numbers?
OA. The set of all numbers of the form a+bi, where a and bare any
real numbers and i equals -1
B. The set of all numbers of the form abi, where a and bare any real
numbers and i equals 1
C. The set of all numbers of the form a+bi, where a and b are any
real numbers and i equals √-1
OD. The set of all numbers of the form abi, where a and b are any real
numbers and / equals -1
Answer:
C
Step-by-step explanation:
The correct statement is option C.
What is complex number?A real number and an imaginary number are effectively combined to create a complex number. The complex number is written as a+ib, where a and ib are real and imaginary numbers, respectively. Additionally, i = √-1 and both a and b are real numbers.
Since we know that
Complex number is of the form a+ib
Where,
a is real number belongs to real axis
And b is also a real number belongs to imaginary axis.
And the value of i = √-1
Thus,
The set of all numbers of the form a+bi, where a and b are any
real numbers and i equals √-1 is the correct statement.
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A student wanted to find the sum of all the even numbers from 1 to 100. He said: The sum of all the even numbers from 1 to 100 is twice the sum of all the odd numbers from 1 to 100. The sum of all the odd numbers from 1 to 100 is 1002. Explain why each of these statements is incorrect. HELP ME ASAP
Statement 1
Twice the sum of the odd numbers would be:
[tex]2(1+3+5+\cdots+99)=2+6+10+\cdots+198[/tex], which is not equal to the sum of all the even numbers from 1 to 100.
Statement 2
The sum of all the odd numbers from 1 to 100 can be thought of as an arithmetic sequence containing 50 terms, with first term 1 and common difference 2. This means the sum of the series would be:
[tex]\frac{50}{2}[2(1)+(50-1)(2)]=2500[/tex]
which is not equal to 1002.
The statement 1 and the statement 2 are incorrect.
The sum of the arithmetic series of first term a₁ and common difference d will be s= n/2{2a₁+(n-1)d
Statement 1:
Given in statement 1, Twice the sum of the odd numbers would be:
2(1+3+5+7.......+99)=2+6+10+14+.....198
which is not equal to the sum of all the even numbers from 1 to 100.
Statement 2:
The sum of all the odd numbers from 1 to 100 can be calculated as follows where this is an arithmetic sequence of 50 terms, where the first term is 1 and the common difference is 2. This means the sum of the arithmetic series would be:
s= n/2{2a₁+(n-1)d}
putting the above formula
a₁=1
n=50
d=2
s= 50/2{2(1)+(50-1)2} = 25{2+98} =2500 ≠1002
which is not equal to 1002.
Therefore statement 1 and 2 are incorrect.
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The power 3 Superscript negative 3 equals StartFraction 1 Over 27 EndFraction . Which expression is equivalent to 3 Superscript negative 3?
Applying the negative exponent, the equivalent expression [tex]3^{-3}[/tex] is 1/27.
What are exponents?The exponents of a number are defined as the representation of a number that shows how many times a number is multiplied by itself.
When we have a negative exponent, we use a fraction, with the term with the exponent going to the denominator.
Hence, the equivalent expression is:
[tex]3^{-3} = \dfrac{1}{3^3} = \dfrac{1}{27}[/tex]
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three-quarters of a pile of bricks were used for a certain project. when two thirds of the reminder had been used, 50 bricks were left. how many bricks were there in the original pile?
Answer:
sorry I don't know the answer
Seismology In 1812, an earthquake of magnitude 7.9 shook New Madrid,
Missouri. Compare the amount of energy released by that earthquake to the
amount of energy released by each earthquake below.
magnitude 9.5 in Valdivia, Chile, in 1960
Solve the equation.
6(x−1)6/7=12
Answer:
Linear Equations In One Variable =
[6(x-1)6] / 7 = 12
[(6x - 6)6] = 84
[36x - 36] = 84
36x = 84 + 36
36x = 120
x = 120/36
x = 10/3
equation solved (Answer : 10/3)
PLEASEEEE PLEASEEEEEEEEE HELPPPPPPPPPP
how do i solve this equation?
Answer:
360
Step-by-step explanation:
using the definition
n [tex]P_{r}[/tex] = [tex]\frac{n!}{(n-r)!}[/tex]
where n! = n(n - 1)(n - 2)..... × 3 × 2 × 1
then
6[tex]P_{4}[/tex]
= [tex]\frac{6!}{(6-4)!}[/tex]
= [tex]\frac{6!}{2!}[/tex]
= [tex]\frac{6(5)(4)(3(2)(1)}{2(1)}[/tex] ← cancel 2(1) on numerator / denominator
= 6 × 5 × 4 × 3
= 360
What is the range of the function g(x) = –3sec(2x + 4) – 1?
(–∞, –2] ∪ [0, ∞)
(–∞, –4] ∪ [0, ∞)
(–∞, –4] ∪ [2, ∞)
(–∞, –5] ∪ [1, ∞)
The range of the function g(x) will be (–∞, –4] ∪ [2, ∞). Option C is correct.
What is the difference between domain and range?The domain denotes all potential x values, while the range denotes all possible y values.
When we plot the graph of the given function we will get the maximum and the minimum value till we get the function plot. The difference in the value of those coordinates is the range of the function.
The range of the function g(x) will be (–∞, –4] ∪ [2, ∞).
Hence option C is correct.
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if angle X is an acute angle with tan X = 8/7, what is the value of sec X?
Answer: [tex]\frac{\sqrt{113}}{7}[/tex]
Step-by-step explanation:
As X is an acute angle, all 6 trigonometric functions with an argument of X are positive.
Using the identity [tex]1+\tan^{2} X=\sec^{2} X[/tex],
[tex]1+\left(\frac{8}{7} \right)^{2}=\sec^{2} X\\\\\sec^{2} X=\frac{113}{49}\\\\\therefore \sec X=\boxed{\frac{\sqrt{113}}{7}}[/tex]
How to solve this one
Please step by step
Answer: 1/2
Step-by-step explanation:
Dividing the numerator and denominator by x, we get the limit is equal to:
[tex]\lim_{x \to -\infty} \frac{-\sqrt{1+\frac{5}{x}+\frac{1}{x^2}}+3}{4+\frac{7}{x}}\\\\=\frac{\lim_{x \to -\infty} \left(-\sqrt{1+\frac{5}{x}+\frac{1}{x^2}}+3 \right)}{\lim_{x \to -\infty} \left(4+\frac{7}{x} \right)}\\\\=\frac{2}{4}=\boxed{1/2}[/tex]
Somebody please assist me here
The base case of [tex]n=1[/tex] is trivially true, since
[tex]\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)[/tex]
but I think the case of [tex]n=2[/tex] may be a bit more convincing in this role. We have by the inclusion/exclusion principle
[tex]\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)[/tex]
with equality if [tex]E_1\cap E_2=\emptyset[/tex].
Now assume the case of [tex]n=k[/tex] is true, that
[tex]\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)[/tex]
We want to use this to prove the claim for [tex]n=k+1[/tex], that
[tex]\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)[/tex]
The I/EP tells us
[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)[/tex]
and by the same argument as in the [tex]n=2[/tex] case, this leads to
[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})[/tex]
By the induction hypothesis, we have an upper bound for the probability of the union of the [tex]E_1[/tex] through [tex]E_k[/tex]. The result follows.
[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)[/tex]
What is the best estimate of the perimeter of the figure on the grid if each square has side lengths of 1 mm?
Answer:
4mm
Step-by-step explanation:
you have to add both sides after substituting each of the 4 sides by 1mm
which gives you the total of 4
A firm operated at 80% of capacity for the past year, during which fixed costs were $197,000, variable costs were 70% of sales, and sales were $900,000. Operating profit was
a.$73,000
b.$630,000
c.$58,400
d.$270,000
The correct answer is option A which is the operating profit will be $73000.
What is an expression?Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.
Given that:-
A firm operated at 80% of capacity for the past year, during which fixed costs were $197,000, variable costs were 70% of sales, and sales were $900,000.We will consider the following notations and will make the expression for operating profit.
P = profit
S = sales = $900000
F = Fixed cost = $197000
V = 07S = variable cost
So the expression will be given as:-
P = S - F - V
P = S - F - 0.7S
P = 9000000 - 1797000 - ( 0.7 x 9000000)
P = 703000 - 630000
P = $73000
Therefore the correct answer is option A which is the operating profit will be $73000.
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what is the length of the squares side?
Answer:
side ≈ 4.24= [tex]\frac{6}{\sqrt{2} }[/tex]
Step-by-step explanation:
sides a = b
[tex]6^{2} =a^{2} +a^{2}[/tex]
[tex]6^{2} =2a^{2}[/tex]
[tex]a^{2}= \frac{6^{2} }{2} =\frac{36}{2} =18[/tex]
[tex]a=\sqrt{18} =4.24=\sqrt{(9)(2)} =\frac{6}{\sqrt{2} }[/tex]
Hope this helps
A line of best fit was drawn for 16 data points. What is the maximum number
of these data points that may not actually be on the line?
OA. 14
OB. 13
O C. 15
OD. 16
The maximum number of data points that may not actually be on the line is 16. so, the correct option is D.
How to find the line of best fit?We know that a line of best fit is basically a straight line drawn for a given data that may or may not pass through data points.
It is given that the line is drawn for 16 data points.
Hence, the line may or may not pass through all these 16 points.
Therefore, the maximum number of data points that may not actually be on the line is 16.
so, the correct option is D.
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According to American Airlines, flight 71098 from New York to Los Angeles is on time 88.9% of the time. Assume that we randomly select 150 flights, use the normal approximation to the binomial to do the following:
a) approximately the probability that exactly 124 flights are on time.
b) approximate the probability that between 113 and 130 flights ,inclusive, are on time.
Using the normal approximation to the binomial, it is found that the probabilities are given as follows:
a) 0.0055 = 0.55%.
b) 0.2296 = 22.96%.
Normal Probability DistributionThe z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The z-score measures how many standard deviations the measure is above or below the mean. Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].The parameters of the binomial distribution are given as follows:
n = 150, p = 0.889.
Hence the mean and the standard deviation of the approximation are:
[tex]\mu = E(X) = np = 150 x 0.889 = 133.35[/tex].[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150(0.889)(0.111)} = 3.8473[/tex]Item a:
Using continuity correction, the probability is P(123.5 < X < 124.5), which is the p-value of Z when X = 124.5 subtracted by the p-value of Z when X = 123.5, hence:
X = 124.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{124.5 - 133.35}{3.8473}[/tex]
Z = -2.3
Z = -2.3 has a p-value of 0.0107.
X = 123.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{123.5 - 133.35}{3.8473}[/tex]
Z = -2.56
Z = -2.56 has a p-value of 0.0052.
Hence the probability is 0.0107 - 0.0052 = 0.0055 = 0.55%.
Item b:
The probability is P(112.5 < X < 130.5), which is the p-value of Z when X = 130.5 subtracted by the p-value of Z when X = 112.5, hence:
X = 130.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130.5 - 133.35}{3.8473}[/tex]
Z = -0.74
Z = -0.74 has a p-value of 0.2296.
X = 112.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{112.5 - 133.35}{3.8473}[/tex]
Z = -5.42
Z = -5.42 has a p-value of 0.
Hence the probability is 0.2296 - 0 = 0.2296 = 22.96%.
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Can anyone solve this problem ?
Answer:
defectionist forme solb ifv
which equation has the solution x=6
Answer:
need more info
Step-by-step explanation:
what I can tell you is that when you change all the x's in the equation to 6s, both sides of the equation will equal eachother.
Match the scatter plot with its description
weak and positive
moderate and negative
strong and negative
strong and positive
weak and negative
moderate and positive
At the local Stop and Shop, Haagan Dazs ice cream is on sale for $3.67 for 14oz.
Breyer’s half gallons are on sale for $4.69. Use unit pricing to compare and determine
which is a better deal.
Answer:
Breyer
Step-by-step explanation:
1gallon=128oz half gallon=64oz
We can make both of their unit to 224oz: 14x16=224oz 64x3.5=224oz
Haagen Dazs:3.67x16=58.72/224oz
Breyer: 4.69x3.5=16.415/224oz
Breyer’s way cheaper
In 2010, the amount in Raja's account was 40 Lakhs. He debited 8000 in 2011, and in 2012, he debited 2,07,000. In 2013, he credited 1,16,000, and in 2014 he credited 12,000. How much more he has to credit to be worth what it was at at the start of 2010
How much more he has to credit to be worth what it was at at the start of 2010 is: 3,657,000.
Additional credit amountFirsts step
Total amount debited=8,000+207,000
Total amount debited=215,000
Total amount credited=116,000+12,000
Total amount credited=128,000
Second step
Additional amount to credit:
40 Lakhs is 4,000,000
Hence:
Additional amount to credit=4,000,000- (215,000+128,000)
Additional amount to credit=4,000,000-343,000
Additional amount to credit=
Therefore how much more he has to credit to be worth what it was at at the start of 2010 is: 3,657,000.
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Can someone help me answer fast
Answer:
2x+3+9
because plyer scored 9 points and throw 2point shotand same number 3
In the 30-60-90 triangle below, side s has a length of.
length of
30
90⁰
S
10
60"
A. 5.2: 5.2
B. 5,10,3
and side q has a
Answer: C, 5, 5√3
Step-by-step explanation:
The short leg of a 30-60-90 triangle equal half the hypotenuse, 10/2 = 5. The long leg equals √3 times the short leg, 5*√3 = 5√3.
Please help quick!!
Answer:
the answer will be similar, because if you divide 2, 3, 2.25 by 3 you will get the answer.
GOD PLEASE HELP!
Which of these is NOT a key factor that contributed to the financial crisis that started in 2007?
bond rating agencies giving CDOs very low ratings, discouraging investors to invest
highly leveraged investment accounts that magnified losses when bond prices decreased
excessive use of subprime mortgages to borrowers with poor credit
banks being incentivized to sell risky mortgages as mortgage-backed securities
Answer:
i think, bond rating agencies giving CDOs very low ratings, discouraging investors to invest
Step-by-step explanation:
Brittany removed half of the total number of marbles from her bag, and gave one fourth of the remaining marbles to her friend. What fraction of the total number of marbles is left in the bag? A) 1/4, B) 3/8, C) 1/2, D) 7/8
The fraction of the total number of marbles is left in the bag after removing and giving part to her friend is 3/8
Fractionlet
Number of marbles in her bag = xNumber of marbles removed from her bag = 1/2xRemaining marbles = x - 1/2x
= (2x-x) / 2
= x/2
= 1/2x
Number of marbles given to her friend = 1/4 of 1/2x
= 1/4 × 1/2x
= 1/8x
Number of marbles left in the bag = Remaining marbles - Number of marbles given to her friend
= 1/2x - 1/8x
= (4x-x) / 8
= 3x/8
= 3/8x
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Which absolute value functions will be narrower than the parent function, f(x) = |x|? Check all that apply.
f(x) = |x|
f(x) = |x – 2|
f(x) = |x| + 3
f(x) = 2.9|x|
f(x) = 1.2|x + 8|
f(x) = 0.7|x| – 3.2
The correct answer is option 4 which is f(x) = 2.9|x| is narrower than the parent function f(x) = |x|
What is a function?A function is defined as the expression that set up the relationship between the dependent variable and independent variable.
Given parent function is f(x)=|x|
Now we have to find which function from given choices will be narrower than the parent function.
Notice that adding or subtracting some number from the parent function only shifts the graph up, down, and left of the right side.
But that will not make the function narrower or broader.
So f(x) = |x – 2| and f(x) = |x| + 3, can't be the answer.
Multiplying by some positive real number which is more than 1, makes the function narrower.
Only f(x) = 2.9|x| from remaining choices fits that case.
Hence correct answer is option 4 which is f(x) = 2.9|x| is narrower than the parent function f(x) = |x|
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Answer:
D,E
Step-by-step explanation:
easy:
Clara buy: 300 apples, 74 potato, 15 eggs e 2 phone.
how many things did he buy in all?
Hard:
There are 92 boys in a school, the girl 100.How many more girl are there than boys?
Answer:
easy = 391 items
hard = there are 8 more girls than there are boys
If anyone can help me to solve this.
Explanation:
Equation: x + 1 = 9
Solving Steps:
x + 1 = 9 (subtract both sides by 1)
x + 1 - 1 = 9 - 1 (simplify the following)
x = 8
Then check for solution:
x + 1 = 9
[insert x = 8]
8 + 1 = 9
9 = 9
As both sides are equal, the statement is true.
[tex]\large\underline{ \cal{SOLUTION:}}[/tex]
[tex] \large \bold{x + 1 = 9}[/tex]
[tex] \large \bold{x = 9 - 1}[/tex]
[tex] \large \bold{x = \red{8}}[/tex]
[tex] \: [/tex]
[tex]\large\underline{ \cal{CHECKING:}}[/tex]
Put x = 8[tex] \large{ \bold{ \underline{ \red{8} }+ 1 = 9}}[/tex]
___________________________________[tex] \large \bold{-N} \frak{unx-}[/tex]
good morning i need help very fast
Step-by-step explanation:
all the details You can find in the attachment.
