Answer:
The statement is true for every n between 0 and 77 and it is false for [tex]n\geq 78[/tex]
Step-by-step explanation:
First, observe that, for n=0 and n=1 the statement is true:
For n=0: [tex]\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4[/tex]
For n=1: [tex]\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4[/tex]
From this point we will assume that [tex]n\geq 2[/tex]
As we can see, [tex]\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4[/tex] and [tex](4n)^4=256n^4[/tex]. Then,
[tex]\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4[/tex]
Now, we will use the formula for the sum of the first 4th powers:
[tex]\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}[/tex]
Therefore:
[tex]\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0[/tex]
and, because [tex]n \geq 0[/tex],
[tex]465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1[/tex]
Observe that, because [tex]n \geq 2[/tex] and is an integer,
[tex]n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5[/tex]
In concusion, the statement is true if and only if n is a non negative integer such that [tex]n\leq 77[/tex]
So, 78 is the smallest value of n that does not satisfy the inequality.
Note: If you compute [tex](4n)^4- \sum^{n}_{i=0} (2i)^4[/tex] for 77 and 78 you will obtain:
[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=53810064[/tex]
[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992[/tex]
We will use mathematical induction to prove this inequality.
Explanation:We will prove this inequality using mathematical induction. First, let's check the base case when n = 0. The left-hand side (LHS) of the inequality is 0 and the right-hand side (RHS) is (4*0)^4 = 0. So, the inequality holds for n = 0.
Next, assume that the inequality holds for some positive integer k, i.e.,
∑([2i]^4) ≤ (4k)^4 (where the sum is taken from i = 0 to k)
We will prove that it also holds for k + 1. Adding the (k+1)th term to both sides of the inequality:
∑([2i]^4) + ([2(k+1)]^4) ≤ (4k)^4 + ([2(k+1)]^4)
Now, simplifying the LHS and RHS:
(∑([2i]^4)) + ([2(k+1)]^4) ≤ (4k)^4 + ([2(k+1)]^4)
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The New Mexico Division of Fish and Wildlife keeps track of the silvery minnow population in the Rio Grande River. They tagged 54 silvery minnows and released them. A week later, they captured 62 silvery minnows, including 12 tagged silvery minnows. What is a good estimate of the silvery minnow population in the Rio Grande River?
Answer: There are 104 silvery minnow population in the Rio Grande River.
Step-by-step explanation:
Since we have given that
Number of tagged silvery minnows = 54
Number of captured silvery minnows = 62
It includes 12 tagged silvery minnows.
So, Number of silvery minnows without tags = 62 -12 =50
So, Good estimate of the silvery minnow population in the Rio Grande River would be
[tex]54+50\\\\=104[/tex]
Hence, there are 104 silvery minnow population in the Rio Grande River.
Find a vector equation for the line through the point P = (-3,3, -5) and parallel to the vector V= (-2,-4, -4). Assume 1-(0) =-3i+ 3j - 5k. It) = Enter your answer with the angle bracket notation using ''. For example, enter"" for the vector $ _ ܀
Answer:
The vector equation of the line is
l(t)= "-3,3,-5" + t "-2,-4,-4", where t is a parameter.
a. Every set has an element. b. The smallest perfect number is 28. c. Is 1.5 an irrational number? d. Please find the popular approximation of pi. e. Can you extract the root of -25?
Answer:
a) False. b) False. c) No, it's Rational. d) pi=355/113 e) 5i (for Complex Set of Numbers)
Step-by-step explanation:
a) Since there is the empty set. And an axiom assures us the existence of this Set. "There is a set such no element belongs to it"
∅ has no elements.
b) A perfect number is a positive integer equals to the sum of proper divisions
The smallest is 6. Since the proper divisors of 6={3,2,1}. 6=1+2+3
28 is a perfect number, but not the smallest. It is perfect since
28 proper divisors={14,7,4,2} 28=1+2+3+4+5+6+7
c) No, An Irrational number cannot be written as a fraction a/b where "a" and "b" are integers. 1.5 is a rational one, since 3/2 =1.5
d)
[tex]\pi[/tex]=22/7 -1/791= 355/113
e) Not for Real Numbers, since it is not defined for Real numbers. But for the set of Complex 5i
I had this class in college where the semester’s four exams weighed 10%, 15%, 25%, and 50%, respectively. The class average on each of the exams where 75%, 91%, 63%, 87%, respectively. Create two vectors in to represent the data. Calculate the dot product of your two vectors. What does the scalar value represent in terms of the class?
Answer:
[tex]v_{1}.v_{2} = 0.804[/tex]
In terms of the class, the dot product represents the weighed class average.
Step-by-step explanation:
The two vectors are:
-[tex]v_{1}:[/tex] The weight of each of the semester's exams.
[tex]v_{1} = (10%, 15%, 25%, 50%)[/tex]
In decimal:
[tex]v_{1} = (0.10, 0.15, 0.25, 0.50)[/tex]
-[tex]v_{2}:[/tex] The class average on each of the exams
In decimal:
[tex]v_{2} = (0.75, 0.91, 0.63, 0.87)[/tex]
-----------------------
Dot product:
Suppose there are two vectors, u and v
u = (a,b,c)
v = (d,e,f)
There dot product between the vectors u and v is:
u.v = (a,b,c).(d,e,f) = ad + be + cf
------------------
So
[tex]v_{1}.v_{2} = (0.10, 0.15, 0.25, 0.50).(0.75, 0.91, 0.63, 0.87) = 0.10*0.75 + 0.15*0.91 + 0.25*0.63 + 0.50*0.87 = 0.804[/tex]
[tex]v_{1}.v_{2} = 0.804[/tex]
In terms of the class, the dot product represents the weighed class average.
For each function below, determine whether or not the function is injective and whether| or not the function is surjective. Be sure to justify your answers. (a) f : N -> N given by f(n) =n+ 2
(b) f P({1,2, 3}) -» N given by f(A) = |A| (Note: P(S) denotes the power set of a set S.)
Answer:
a) injective but not surjective. b) neither injective nor surjective.
Step-by-step explanation:
A function is injective if there aren't repeated images. To check if a function is injective we are going to suppose that for two values in the domain the image is equal, then we need to find that the two values are equal.
a) f : N → N with f(n) = n+2.
suppose that for n and m natural numbers, f(n)=f(m). Then
n+2 = m+2
n+2-2 = m
n = m.
Then, f(n) is injective.
Now, a function is surjective if every term m in the codomain there exists a pre-image of that element, that is to say, there exists an n such that f(n) = m. That is, the range of the function is equal to the codomain.
In this case, f(n) is not surjective. For example, if we would have that
n+2 = 1
n = 1-2
n = -1
but -1 is not a natural number, then for m=1 we don't have a pre image in f.
b) f: P({1, 2, 3}) → N with f(A) = |A| (amount of elements in the subset A).
Now, for {1,2,3} we can have subsets of 0, 1, 2 or 3 elements. Then, the range of the function f is {1, 2, 3} (0 is not included because 0 is not natural). Then, the range is not all the natural numbers and therefore the function is not surjective.
Now, let's check if f is injective. Let {1} and {2} subsets of {1, 2, 3}. Then
f({1}) = |{1}| = 1.
f ({2}) = |{2}| = 1.
We have two different subsets with the same image, then f is not injective.
Ronna has volleyball practice every 4 days. Ronna also has violin lessons every 10 days. She has both activities today after school. When will she have both activities again on the same day?
Answer:
She have both activities again on 20th day from now.
Step-by-step explanation:
Given :Ronna has volleyball practice every 4 days
Ronna also has violin lessons every 10 days.
To Find : When will she have both activities again on the same day?
Solution:
We will find the LCM of 4 and 10
2 | 4 ,10
2 | 2 , 5
5 | 1 , 5
| 1 , 1
So, [tex]LCM = 2\times 2 \times 5[/tex]
[tex]LCM =20[/tex]
Hence she have both activities again on 20th day from now.
Prove or disprove that the intersection of any collection
ofclosed sets is closed.
Answer:
Intersection of collection of any closed set is a closed set.
Step-by-step explanation:
Let F be a collection of arbitrary closed sets and let [tex]B_i[/tex] be closed set belonging to F.
We define a closed set as the set that contains its limit point or in other words it can be described that the complement or not of a closed set is an open set.
Thus, we can write R as
[tex]R =\bigcap\limits_{B_i \in F }^{} B_i[/tex]
Now, applying De-Morgan's Theorem, we have
[tex]R^c = (\bigcap\limits_{B_i \in F }^{} B_i)^c[/tex]
[tex]R^c = \bigcup\limits_{B_i \in F }^{} B_i^c[/tex]
Since we knew[tex]B_i[/tex] are closed set, thus, [tex]B_i^c[/tex] is an open set.
We also know that union of all open set is an open set.
Thus, [tex]R^c[/tex] is an open set.
Thus, R is a closed set.
Hence, the theorem.
The intersection of any collection of closed sets is indeed closed. This can be proven using the properties of complements in topology and by employing a contradiction approach where the assumption that the intersection is not closed leads to a contradiction, hence proving it must be closed.
The question addresses whether the intersection of any collection of closed sets is closed. In topology, a closed set is one where its complement (the elements not in the set) is open. One way to approach the proof is considering De Morgan's Laws, which in topology state that the intersection of closed sets is the complement of the union of their complements, which are open sets. Since the union of open sets is open, it follows that the complement (the intersection of our original sets) is closed.
For example, consider two closed intervals on the real line, A and B. The intersection, A B, would be the set of all points that are in both A and B. The fact that both A and B contain their boundary points ensures that their intersection also contains boundary points, maintaining closedness.
Proof strategy involves contradiction: assume the intersection is not closed. This would imply that its complement is not open, violating the definition that complements of closed sets are open, thus leading to a contradiction and proving the proposition is true.
In 1970 the male incarceration rate in the U.S. was approximately 190 inmates per 100,000 population. In 2008 the rate was 960 inmates per 100,000 population. What is the percent increase in the male incarceration rate during this period?
Answer:
405.26%
Step-by-step explanation:
We have been given that in 1970 the male incarceration rate in the U.S. was approximately 190 inmates per 100,000 population. In 2008 the rate was 960 inmates per 100,000 population.
[tex]\text{Percent increase}=\frac{\text{Final amount}-\text{Initial amount}}{\text{Initial amount}}\times 100[/tex]
[tex]\text{Percent increase}=\frac{960-190}{190}\times 100[/tex]
[tex]\text{Percent increase}=\frac{770}{190}\times 100[/tex]
[tex]\text{Percent increase}=4.052631578947\times 100[/tex]
[tex]\text{Percent increase}=405.2631578947\%[/tex]
[tex]\text{Percent increase}\approx 405.26\%[/tex]
Therefore, the percent increase in the male incarceration rate during the given period is 405.26%.
What is the numeral preceding and succeeding each of the following.
A) 640 base 7
B) 100000 base 2
C) 555 base 6
D) 100 base 5
E) 10000 base 4
F) 405 base 6
Answer:
a) [tex]636_{7}[/tex] and [tex]641_{7}[/tex]
b) [tex]11111_{2}[/tex] and [tex]100001_{2}[/tex]
c) [tex]554_{6}[/tex] and [tex]1000_{6}[/tex]
d)[tex]44_{5}[/tex] and [tex]101_{5}[/tex]
e)[tex]3333_{4}[/tex] and [tex]10001_{4}[/tex]
f) [tex]404_{6}[/tex] and [tex]410_{6}[/tex]
Step-by-step explanation:
The logics followed in order to find them:
Base numbers work exactly the same as base 10 numbers (the ones we use on a daily basis). Take for example, in a decimal system, we have 10 digits available:
0,1,2,3,4,5,6,7,8,9.
when counting, when we get to the 9, we start over again, but writting a 1 to the left.
0,1,2,3,4,5,6,7,8,9,10,11,12....20,21,22,23....,90,91,92,93,94,95,96,97,98,99,100
when reaching 99, we have no more digits to use, so we start the count again and go from 99 to 100.
The same works with any other base number, take for example the base 7 numbet. When counting in base 7, we only have 7 digits available: (0,1,2,3,4,5,6) So when we re0_{7},ach the digit 6, we go immediately to 10, like this:
[tex]0_{7},1_{7},2_{7},3_{7},4_{7},5_{7},6_{7},10_{7},11_{7},12_{7},13_{7},14_{7},15_{7},16_{7},20_{7}...[/tex]
notice how it went from 6 to 10 and from 16 to 20. This is because in base 7, there is no such thing as digits from 7 to 9, so we don't have any other option to go directly to 20.
So, on part A) if we were working with decimal numbers, the previous value for 640 would be 639, but notice that in base 7, there is no such thing as the digit 9, so the greatest digit we can use there would be 6, therefore, the previous value for the [tex]640_{7}[/tex] number would be [tex]636_{7}[/tex]. The next number would be [tex]641_{7}[/tex] because the 1 does exist for a base 7 system.
The same logics is followed for the rest of the problems.
One less than the given number is known as preceding number whereas one more than the given number is the succeeding number.
The logics followed in order to find them:
Base numbers work exactly the same as base 10 numbers (the ones we use on a daily basis). Take for example, in a decimal system, we have 10 digits available:
0,1,2,3,4,5,6,7,8,9.
When counting, when we get to the 9, we start over again, but writing a 1 to the left.
0,1,2,3,4,5,6,7,8,9,10,11,12....20,21,22,23....,90,91,92,93,94,95,96,97,98,99,10
When reaching 99, we have no more digits to use, so we start the count again and go from 99 to 100.
The same works with any other base number, take for example the base 7 number. When counting in base 7, we only have 7 digits available: (0,1,2,3,4,5,6) So when 0_{7},each the digit 6, we go immediately to 10, like this:
It went from 6 to 10 and from 16 to 20. This is because in base 7, there is no digits from 7 to 9, so we have directly to 20.
So, if working with decimal numbers, the previous value for 640 would be 639, but notice that in base 7,
There is no such thing as the digit 9, so the greatest digit we can use there 6,
Therefore, the previous value[tex]636_7[/tex] for the number would be [tex]636_7[/tex] . The next number would be [tex]632_7[/tex] because the 1 does exist for a base 7 system.
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A rectangle field has a perimeter of 220 meters and an area of 2856 square meters. What are the dimensions?
Answer:
The dimensions are 68 meters by 42 meters.
Step-by-step explanation:
Let the length be x
Let the width be y
Perimeter = 220 meter
So, [tex]2x+2y=220[/tex]
or [tex]x+y=110[/tex] or [tex]x=110-y[/tex] ...........(1)
Area = 2856 square meter
So, [tex]xy=2856[/tex] ............(2)
Substituting the value of x from (1) in (2)
[tex](110-y)y=2856[/tex]
=> [tex]110y-y^{2}=2856[/tex]
=> [tex]y^{2}-110y+2856=0[/tex]
Solving this equation, we get roots as y = 68 and y = 42
So, if we put y = 68, we get x = 42
If we put y = 42, we get x = 68
As length is longer than width, we will take length = 68 meters and width = 42 meters.
Hence, the dimensions are 68 meters by 42 meters.
At Tech High there are three math teachers, Mary, Tom, and Alex. All students at Tech High take math, with 21% taking class with Mary, 15% with Tom, and 64% taking class with Alex. Every year 4% of Mary’s students fail, 6% of Tom’s students fail, and 13% of Alex’s student fail. What is the probability that a Tech High student fails math? If needed, round to FOUR decimal places.
Answer:
The probability that a Tech High student fails math is 0.1006
Step-by-step explanation:
21% students are taking class with Mary.
Every year 4% of Mary’s students fail,
15% students are taking class with Tom
Every year 6% of Tom’s students fail.
64% students are taking class with Alex
Every year 13% of Alex's students fail.
Now we are supposed to find the probability that a Tech High student fails math.
[tex]P(\text{Tech High student fails math}) = P(\text{taking class with Mary})\times P(\text{Mary students fail})+P(\text{taking class with Tom}) \times P(\text{Tom students fail})+P(\text{taking class with Alex}) \times P(\text{Alex student fail})[/tex]
[tex]P(\text{Tech High student fails math}) = 0.21 \times 0.04+0.15 \times 0.06+0.64 \times 0.13[/tex]
[tex]P(\text{Tech High student fails math}) =0.1006[/tex]
Hence the probability that a Tech High student fails math is 0.1006
If n is an odd, prime integer and 10n19, which is true about the mean of all possible values of n? 0 A. It is greater than the median and greater than the mode. O B. It is greater than the median. ○ C. It is equal to the median. D. It is less than the median.
Answer:
B. It is greater than the median.
Step-by-step explanation:
If n is an odd, prime integer and 10<n<19, then the true statements about the mean of all possible values of n are:
B. It is greater than the median.
Here, n can be 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18.
But as n is odd , it can be 11, 13, 15 and 17.
And also it is given that n is prime, so we will cancel out 15.
Now we are left with 11, 13 and 17.
Their mean is = [tex]\frac{11+13+17}{3}[/tex] = 13.66
Median is 13, that is smaller than 13.66.
So, we can see that the mean is greater than median.
Calculate:
3 pounds (lbs) =——grams (g)
Find the missing length to the nearest tenth of a meter of the
right triangle. One side is 1.4m and the other is 3.1. What is the
third side?
Answer:
length of the third side may be either 3.4 m or 2.8 m.
Step-by-step explanation:
Measure of two sides of a right triangle are 1.4 m and the other side is 3.1 m.
then we have to find the measure of third side.
If the third side of the triangle is its hypotenuse then
(Third side)² = (1.4)² + (3.1)²
Third side = [tex]\sqrt{(1.4)^{2}+(3.1)^{2}}[/tex]
= [tex]\sqrt{1.96+9.61}[/tex]
= [tex]\sqrt{11.57}[/tex]
= 3.4 m
If the third side is one of the perpendicular sides of the triangle and 3.1 m is hypotenuse, then
(Third side)² + (1.4)² = (3.1)²
(Third side)²= (3.1)² - (1.4)²
Third side = [tex]\sqrt{(3.1)^{2}-(1.4)^{2}}[/tex]
= [tex]\sqrt{9.61-1.96}[/tex]
= [tex]\sqrt{7.65}[/tex]
= 2.76 m
≈ 2.8 m
Therefore, length of the third side may be either 3.4 m or 2.8 m.
How do I estimate 542,817
-27,398
___________
To estimate the result of 542,817 - 27,398, we can round to the nearest ten thousand and perform the subtraction, yielding an estimation of 510,000.
Explanation:The student wants to estimate the result of 542,817 - 27,398. One way of doing this is by rounding to the nearest ten thousand. In this case, 542,817 becomes 540,000 and 27,398 becomes 30,000.
Now, perform the calculation: 540,000 - 30,000 = 510,000. Hence, the estimated result of 542,817 - 27,398 is roughly 510,000.
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fill in the missing number 8×____=4×8
Answer:
4
Step-by-step explanation:
8 *4 is the same as 4*8; commutative property
Answer:
[tex]4[/tex]
Step-by-step explanation:
8 x 4 is exactly the same as 4 x 8, both equaling 32.
No matter how insert 4 and 8, it will always be the same
[tex]x \times y = y \times x [/tex]
^^^
What angles are congruent to angle 4
Answer:
1, 5, 8
Step-by-step explanation:
1 is a vertical angle with angle 4, so is congruent.
5 is an alternate interior angle with angle 4, so is congruent.
8 is a corresponding angle with angle 4, so is congruent.
To determine the angles that are congruent to angle 4, we need to find angles with the same measure.
Explanation:In geometry, two angles are congruent if they have the same measure. To determine which angles are congruent to angle 4, we need to find other angles that have the same measure as angle 4.
Let's assume angle 4 measures x degrees. In this case, any angle that also measures x degrees would be congruent to angle 4.
if angle 4 measures 60 degrees, any other angle that measures 60 degrees.
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Suppose the transit system of a large city has a lost and found department that keeps detailed records, including the estimated value in dollars of each lost item. Each year the department publishes a report with a histogram showing the proportion of items lost in various price ranges.
The bar on the histogram with endpoints 100 to 200 has a height of 0.11. What does this mean?
(A) of the lost items, 11% cost $150.
(B) Between 100 and 200 items had an estimated value of 11% of the total cost of all items.
(C) Of the lost items, 11% cost between $100 and $200.
(D) It is impossible to answer the question because it does not say what variable is displayed on the horizontal axis.
(E) Of the lost items, 11% had an average cost of $150.
Answer:
(C) Of the lost items, 11% cost between $100 and $200
Step-by-step explanation:
On the Y-axis is represented the frequency of lost items found.
In this case in the form of a percentage.
On the X-axis is represented the range of prices
See picture attached.
So, this means that 11% of the found items have an estimated price between $100 and $200
Using direct substitution, verify that y(t) is a solution of the given differential equations 17-19. Then using the initial conditions, determine the constants C or c1 and c2.
17. y ′′ + 4y = 0, y(0) = 1, y ′ (0) = 0, y(t) = c1 cos 2t + c2 sin 2t
18. y ′′ − 5y ′ + 4y = 0, y(0) = 1, y ′ (0) = 0, y(t) = c1et + c2e4t
19. y ′′ + 4y ′ + 13y = 0, y(0) = 1, y ′ (0) = 0, y(t) = c1e-2t cos 3t + c2e-3tsin 3t
Answer:
17. C1 = 1 and C2 = 0
18. C1 = 4/3 and C2 = -1/3
Step-by-step explanation:
See it in the picture
A test tube contains 25 bacteria, 5 of which are can stay alive for atleast 30 days, 10 of which will die in their second day. 10 of which are already dead.
Given that a randomly chosen bacteria for experiment is alive. What is the probability it will still be alive after one week?
(a)1 ⁄ 3(b)2 ⁄3
(c) 1⁄5 (d)4 ⁄5
The probability that a randomly selected bacterium that's currently alive will still be alive after one week is 1/3. The calculation is done by dividing the number of bacteria that can stay alive for at least 30 days by the total number of bacteria that are currently alive.
Explanation:The question pertains to the concept of probability in mathematics and it's asking for the likelihood that a randomly chosen bacteria from a test tube would still be alive after one week. Given that 15 bacteria are alive (5 will stay alive for at least 30 days, and 10 will die on the second day), and you have chosen a bacterium that is currently alive. After one week (7 days), only the 5 that can stay alive for at least 30 days will still be alive.
As a result, the probability that the chosen bacterium will still be alive after one week can be calculated by dividing the number of bacteria that can stay alive for at least 30 days (5) by the total number of currently alive bacteria (15).
Probablitiy = number of favourable outcomes / Total number of outcomes
Therefore, the probability that it will still be alive after one week is 5 / 15 = 1/3.
Therefore, the answer is (a) 1/3.
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A research company desires to know the mean consumption of meat per week among males over age 25. They believe that the meat consumption has a mean of 3.8 pounds, and want to construct a 85% confidence interval with a maximum error of 0.06 pounds. Assuming a standard deviation of 1.3 pounds, what is the minimum number of males over age 25 they must include in their sample? Round your answer up to the next integer.
To construct an 85% confidence interval with a maximum error of 0.06 pounds (with a standard deviation of 1.3 pounds), a sample size of 275 males over age 25 would be required.
Explanation:The subject of this question revolves around the mathematical concept of
confidence intervals
by utilizing the formula to find the appropriate sample size. As per the question about meat consumption among males over age 25, the company wish to construct an
85% confidence interval
with a maximum error of 0.06 pounds, with the standard deviation being 1.3 pounds. To find the minimum sample size necessary for the research, we'll have to use the following formula:
n=z²*σ²/E²
, where z is the z-score representative of the desired confidence level, σ is the known standard deviation, and E is the maximum acceptable error. Looking up the Z table or Z score table, 85% confidence corresponds to a z score of approximately 1.44. Substituting these values into the formula gives us n=1.44² * 1.3² / 0.06². This results to roughly 274.94, but since we can't have a fraction of a person, we round up to the nearest whole number, giving us a minimum sample size of 275.
Learn more about Confidence Intervals here:https://brainly.com/question/34700241
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A classroom has ten students. Three students are freshman, two are sophomores, and five are juniors. Three students are randomly selected (without replacement) to participate in a survey. Consider the following events: A = Exactly 1 of the three selected is a freshman B = Exactly 2 of the three selected are juniors Find the following probability. If needed, round to FOUR decimal places. Pr(A∩B) = ___________
Answer:
0.25
Step-by-step explanation:
We have a total of ten student, and three students are randomly selected (without replacement) to participate in a survey. So, the total number of subsets of size 3 is given by 10C3=120.
On the other hand A=Exactly 1 of the three selected is a freshman. We have that three students are freshman in the classroom, we can form 3C1 different subsets of size 1 with the three freshman; besides B=Exactly 2 of the three selected are juniors, and five are juniors in the classroom. We can form 5C2 different subsets of size 2 with the five juniors. By the multiplication rule the number of different subsets of size 3 with exactly 1 freshman and 2 juniors is given by
(3C1)(5C2)=(3)(10)=30 and
Pr(A∩B)=30/120=0.25
Final answer:
To find the probability that exactly one freshman and exactly two juniors are selected, we calculate the combination of selecting one from three freshmen and two from five juniors, then divide by the total combinations of selecting three students from ten. The probability is 0.25.
Explanation:
We need to find the probability Pr(A∩B) where:
A is the event that exactly 1 of the three selected is a freshman.B is the event that exactly 2 of the three selected are juniors.For both events A and B to occur simultaneously, we must select one freshman and two juniors in our three student picks. The number of ways to choose one freshman out of three is C(3,1), and the number of ways to choose two juniors out of five is C(5,2). The total number of ways to choose any three students out of ten is C(10,3). Hence, the probability is:
Pr(A∩B) = (C(3,1) × C(5,2)) / C(10,3)
Calculating this gives:
Pr(A∩B) = (3 × 10) / 120 = 30 / 120
Pr(A∩B) = 0.25
A restaurant sold 6 hamburgers every day for a week. How many hamburgers were sold during the week
Answer:
42 hamburgers
Step-by-step explanation:
Your answer is 42.
6 x 7 = 42
Find m∠ABC. 20 POINTS AND BRAINLIEST FOR CORRECT ANSWER
A. 10°
B. 21°
C. 37°
D. 53°
The measure of angle ABC in the figure is 53 degrees.
Option D) 53° is the correct answer.
From the figure in the image;
Angle ABC and angle CBD are complementary angles, hence, their sum equals 180 degrees.
From the figure:
Angle ABC = ( 5x + 3 )
Angle CBD = ( 3x + 7 )
First, we solve for x:
Since the two angles are complementary:
Angle ABC + Angle CBD = 90
( 5x + 3 ) + ( 3x + 7 ) = 90
5x + 3x + 3 + 7 = 90
8x + 10 = 90
8x = 90 - 10
8x = 80
x = 80/8
x = 10
Now, we find the measure of angle ABC:
Angle ABC = ( 5x + 3 )
Plug in x = 10:
Angle ABC = 5(10) + 3
Angle ABC = 50 + 3
Angle ABC = 53°
Therefore, angle ABC measures 53 degrees.
The correct option is D) 53°
which expressions are equivalent to 2(4f+2g)?
choose 3 answers:
a) 8f+2g
b) 2f(4+2g)
c) 8f+4g
d) 4(2f+g)
e) 4f+4f+4g
$62.98=____% of 185.95
Divide:
62.98 / 185.95 = 0.338693
Multiply the decimal by 100:
0.338693 x 100 = 33.8693%
Round the answer as needed.
At a large department store, the average number of years of employment for a cashier is 5.7 with a standard deviation of 1.8 years. If the number of years of employment at this department store is normally distributed, what is the probability that a cashier selected at random has worked at the store for over 10 years?
Answer: 0.0019
Step-by-step explanation:
Let x be the random variable that represents the number of years of employment at this department store.
Given : The number of years of employment at this department store is normally distributed,
Population mean : [tex]\mu=5.7[/tex]
Standard deviation : [tex]\sigma=1.8[/tex]
Z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Now, the z-value corresponding to 10 : [tex]z=\dfrac{10-5.7}{1.8}\approx2.39[/tex]
P-value = [tex]P(x>10)=P(Z>2.89)=1-P(z\leq2.89)[/tex]
[tex]=1-0.9980737=0.0019263\approx0.0019\text{ (Rounded to nearest ten thousandth)}[/tex]
Hence, the probability that a cashier selected at random has worked at the store for over 10 years = 0.0019
T F Any set of vectors containing the zero vector is linearly dependent
Answer:
True
Step-by-step explanation:
Let B={[tex]v_1, v_2,\cdots,v_k,0[/tex]} a set of vectors containing the zero vector.
B is linear dependent if exist scalars [tex]a_0,a_1, a_2,\cdots,a_k[/tex] not all zero such that [tex]a_0*0+a_1*v_1+\cdots+a_k*v_k=0[/tex].
Observe that if [tex]a_1=a_2=\cdots=a_k=0[/tex] and [tex]a_0=\lambda\neq 0[/tex] then
[tex]\lambda*0+0*v_1+\cdots+0*v_k=0[/tex].
Then B is linear dependent.
Final answer:
True, any set of vectors that contains the zero vector is linearly dependent because a nontrivial combination of the vectors (including the zero vector scaled by any scalar) can produce the null vector.
Explanation:
True. Any set of vectors containing the zero vector is linearly dependent. This is because the null vector, which has all components equal to zero (0î + 0ç + 0k) and therefore no length or direction, can always be represented as a linear combination of other vectors in the set by simply scaling it by any scalar. In more technical terms, for a set of vectors that includes the zero vector, you can find a nontrivial combination (not all scalars being zero) such that when these scalars are applied to their respective vectors, their sum equals the zero vector. This is the very definition of linear dependency. For instance, if you consider any scalar 'a' not equal to zero, a∗0 + 0∗0 + ... + 0∗0 = 0, which demonstrates linear dependence because not all the scalars are zero (namely 'a' is not).
Linear independence would require that the only way to represent the zero vector as a linear combination of the set of vectors is to have all scalars that multiply each vector in the combination be zero. Since the set includes the zero vector itself, this condition is violated, and the set is dependent.
Find the area of the triangle with the vertices (2,1), (10,-1),
and(-1,8).
Answer:
The area of triangle is 25 square units.
Step-by-step explanation:
Given information: Vertices of the triangle are (2,1), (10,-1), and (-1,8).
Formula for area of a triangle:
[tex]A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|[/tex]
The given vertices are (2,1), (10,-1), and (-1,8).
Using the above formula the area of triangle is
[tex]A=\frac{1}{2}|2(-1-8)+10(8-1)+(-1)(1-(-1))|[/tex]
[tex]A=\frac{1}{2}|2(-9)+10(7)+(-1)(1+1)|[/tex]
[tex]A=\frac{1}{2}|-18+70-2|[/tex]
On further simplification we get
[tex]A=\frac{1}{2}|50|[/tex]
[tex]A=\frac{1}{2}(50)[/tex]
[tex]A=25[/tex]
Therefore the area of triangle is 25 square units.
Let p, q, and r represent the following statements"
p : Sam has pizza last night.
q : Chris finished her homework.
r : Pat watched the news this morning.
Give a formula (using appropriate symbols) for each of these statements:
a) Sam had pizza last night and Chris finished her homework.
b) Chris did not finish her homework and Pat watched the news this morning.
c) Sam did not have pizza last night or Chris did not finish her homework
d) Either Chris finished her homework or Pat watched the news this morning, but not both.
e) If Sam had pizza last night then Chris finished her homework.
f) Pat watched the news this morning only if Sam had pizza last night.
g) Chris finished her homework if Sam did not have pizza last night.
h) It is not the case that if Sam had pizza last night, then Pat watched the news this morning.
i) Sam did not have pizza last night and Chris finished her homework implies that Pat watched the news this morning. Express, in words, the statements represented by the following formulas.
j) q ⇒ r
k) p ⇒ (q ∧ r)
l) ¬p ⇒ (q ∨ r)
m) r ⇒ (p ∨ q)
Answer:
Step-by-step explanation:
[tex]$$a. Sam had pizza last night and Chris finished her homework.\\p\wedge q\\\\$b. Chris did not finish her homework and Pat watched the news this morning.$\\\neg q \wedge r\\\\$c. Sam did not have pizza last night or Chris did not finish her homework.$\\\neg p \vee \neg q\\\\[/tex]
[tex]$$d. Either Chris finished her homework or Pat watched the news this morning, but not both.$\\q\vee r\\\\$e. If Sam had pizza last night then Chris finished her homework.$\\p \rightarrow q\\\\$f. Pat watched the news this morning only if Sam had pizza last night.$\\p\leftrightarrow r\\\\[/tex]
[tex]$$g. Chris finished her homework if Sam did not have pizza last night.$\\\neg p \rightarrow q\\\\$h. It is not the case that if Sam had pizza last night, then Pat watched the news this morning.$\\\neg (p\rightarrow r)\\\\[/tex]
[tex]$$i. Sam did not have pizza last night and Chris finished her homework implies that Pat watched the news this morning.$\\(\neg p \wedge q) \Rightarrow r\\\\ $j. q\Rightarrow r$\\Chris finished her homework implies that Pat watched the news this morning.\\\\[/tex]
[tex]$$k. p \Rightarrow (q \wedge r)$\\Sam has pizza last night implies that Chris finished her homework and Pat watched the news this morning.$\\\\[/tex]
[tex]$$l. \neg p \Rightarrow (q \vee r)$\\Sam did not have pizza last night implies that Chris finished her homework or Pat watched the news this morning.$\\\\[/tex]
[tex]$$m. r \Rightarrow (p \vee q)$\\Pat watched the news this morning implies that Sam had pizza last night or Chris finished her homework$[/tex]
Final answer:
Logical formulas for the given statements about Sam, Chris, and Pat are provided using logical operators such as 'and', 'or', 'not', and 'implies'. The logical symbols ∧, ∨, ¬, and ⇒ are utilized to form the statements. Some formulas are also expressed in words to match their logical predictions with linguistic intuitions.
Explanation:
The logical formulas for the statements given about Sam, Chris, and Pat using p, q, and r are as follows:
a) Sam had pizza last night and Chris finished her homework: p ∧ q
b) Chris did not finish her homework and Pat watched the news this morning: ¬q ∧ r
c) Sam did not have pizza last night or Chris did not finish her homework: ¬p ∨ ¬q
d) Either Chris finished her homework or Pat watched the news this morning, but not both: q ⊕ r
e) If Sam had pizza last night then Chris finished her homework: p ⇒ q
f) Pat watched the news this morning only if Sam had pizza last night: r ⇒ p
g) Chris finished her homework if Sam did not have pizza last night: ¬p ⇒ q
h) It is not the case that if Sam had pizza last night, then Pat watched the news this morning: ¬(p ⇒ r)
i) Sam did not have pizza last night and Chris finished her homework implies that Pat watched the news this morning: (¬p ∧ q) ⇒ r
Now let's express some formulas in words:
j) q ⇒ r: If Chris finished her homework, then Pat watched the news this morning.
k) p ⇒ (q ∧ r): If Sam had pizza last night, then Chris finished her homework and Pat watched the news this morning.
l) ¬p ⇒ (q ∨ r): If Sam did not have pizza last night, then Chris finished her homework or Pat watched the news this morning.
m) r ⇒ (p ∨ q): If Pat watched the news this morning, then Sam had pizza last night or Chris finished her homework.