Answer:
option d)378
Step-by-step explanation:
Given that a researcher at a major hospital wishes to estimate the proportion of the adult population of the United States that has high blood pressure.
Margin of error should be at most 6% = 0.06
Let us assume p =0.5 as when p =0.5 we get maximum std deviation so this method will give the minimum value for n the sample size easily.
We have std error = [tex]\sqrt{\frac{pq}{n} } =\frac{0.5}{\sqrt{n} }[/tex]
For 98%confident interval Z critical score = 2.33
Hence we have margin of error = [tex]2.33(\frac{0.5}{\sqrt{n} } <0.06\\n>377[/tex]
Hence answer is option d)378
The size of the sample needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 6% is; 376
What is the size of the sample?We are told that Margin of error should be at most 6% = 0.06
Formula for margin of error is;
M = z√(p(1 - p)/n)
we are given the confidence level to be 98% and the z-score at this confidence level is 2.326
Since no standard deviation then we assume it is maximum and as such assume p =0.5 which will give us the minimum sample required.
Thus;
0.06 = 2.326√(0.5(1 - 0.5)/n)
(0.06/2.326)² = (0.5²/n)
solving for n gives approximately n = 376
Thus, the size of the sample required is 376
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How many dfferent strings can be made from the letters in MISSISSIPPI, using all letters? 15 013837
Answer: 34650
Step-by-step explanation:
The number of permutations of n objects, where one object is repeated [tex]n_1[/tex] times , another is repeated [tex]n_2[/tex] times and so on is :
[tex]\dfrac{n!}{n_1!n_2!....n_k!}[/tex]
Given : The number of letters in string MISSISSIPPI = 11
Here I is repeated 4 times, S is repeated 4 times and P is repeated 2 times.
Then , the number of different strings can be made from the letters in MISSISSIPPI, using all letters is given by :-
[tex]\dfrac{11!}{4!4!2!}=34650[/tex]
Therefore , there are 34650 different strings can be made from the letters in MISSISSIPPI.
To find the number of different strings that can be formed from the letters in MISSISSIPPI, we use the formula for permutations of a multi set. The total is 34,650 unique permutations.
The question asks how many different strings can be made from the letters in MISSISSIPPI, using all letters.
This is a problem of calculating permutations of a multi set.
The word MISSISSIPPI has 11 letters with the following counts of each letter: M occurs 1 time, I occurs 4 times, S occurs 4 times, and P occurs 2 times.
To find the total number of unique permutations, we use the formula for the permutations of a multiset:
The formula is:
Permutations = n! / (n1! * n2! * ... * nk!),
where n is the total number of items to arrange, and n1, n2, ..., nk are the counts of each distinct item.
For MISSISSIPPI:
Total permutations = 11! / (1! * 4! * 4! * 2!) = 34650
Therefore, there are 34,650 different strings that can be made from the letters in MISSISSIPPI using all the letters.
Of the 122 students who took a mathematics exam, 76 correctly answered the first question, 60 correctly answered the second question, and 38 correctly answered both questions. How many students answered the first question correctly, but not the second? students answered the first question correctly, but not the second
Answer: 38
Step-by-step explanation:
If P and Q are two different set the their difference is given by P-Q i.e. the number of elements in P bit not Q .i.e. [tex]P-Q=n(P)-n(P\cap Q)[/tex]
Let A be the number of students who correctly answered the first question and B be the number of students who correctly answered the second question .
Given : [tex]n(A)=76[/tex]
[tex]n(B)=60[/tex]
[tex]n(A\cap B)=38[/tex]
Then the number of students who answered the first question correctly, but not the second is given by :-
[tex]A-B=n(A)-n(A\cap B)\\\\=76-38=38[/tex]
Hence, the number of students who answered the first question correctly, but not the second is 38.
Prove by induction that 3n(n 1) is divisible by 6 for all positive integers.
We are asked to prove by the method of mathematical induction that:
3n(n+1) is divisible by 6 for all positive integers.
for n=1 we have:[tex]3n(n+1)=3\times 1(1+1)\\\\i.e.\\\\3n(n+1)=3\times 2\\\\i.e.\\\\3n(n+1)=6[/tex]
which is divisible by 6.
Hence, the result is true for n=1
Let the result is true for n=ki.e. 3k(k+1) is divisible by 6.
Now we prove that the result is true for n=k+1Let n=k+1
then
[tex]3n(n+1)=3(k+1)\times (k+1+1)\\\\i.e.\\\\3n(n+1)=3(k+1)(k+2)\\\\i.e.\\\\3n(n+1)=(3k+3)(k+2)\\\\i.e.\\\\3n(n+1)=3k(k+2)+3(k+2)\\\\i.e.\\\\3n(n+1)=3k^2+6k+3k+6\\\\i.e.\\\\3n(n+1)=3k^2+3k+6k+6\\\\i.e.\\\\3n(n+1)=3k(k+1)+6(k+1)[/tex]
Since, the first term:
[tex]3k(k+1)[/tex] is divisible by 6.
( As the result is true for n=k)
and the second term [tex]6(k+1)[/tex] is also divisible by 6.
Hence, the sum:
[tex]3k(k+1)+6(k+1)[/tex] is divisible by 6.
Hence, the result is true for n=k+1
Hence, we may say that the result is true for all n where n belongs to positive integers.
To prove by induction that the expression 3n(n+1) is divisible by 6, we start with the base case of n=1, which is divisible by 6, and then show that if it holds for an integer k, it also holds for k+1. By factoring and using the induction hypothesis, we demonstrate the expression's divisibility by 6 for all positive integers.
Proof by Induction of Divisibility by 6
To prove by induction that 3n(n+1) is divisible by 6 for all positive integers, we follow two steps: the base case and the inductive step.
Base Case
Let's check for n=1:
Inductive Step
Assume the statement holds for a positive integer k, so 3k(k+1) is divisible by 6.
Now, we must show that 3(k+1)((k+1)+1) = 3(k+1)(k+2) is also divisible by 6.
Factoring out the common term we get:
Notice that (k+1) is an integer, hence 3(k+1) is divisible by 3. If k is even, then k+1 is odd, so 3(k+1) is still divisible by 3 but not necessarily by 6. However, if k is odd, k+1 is even and 3(k+1) is divisible by 6. Since the divisible by 3 part is always true, and the divisible by an additional factor of 2 part is true every other time, the sum 6m + 3(k+1) is divisible by 6 regardless of whether k is odd or even. This is because adding a multiple of 6 to either another multiple of 6 or a multiple of 3 always results in a multiple of 6.
Therefore, 3n(n + 1) is divisible by 6 for all positive integers n.
A bag of 100 tulip bulbs purchased from a nursery contains 20 red tulip bulbs, 20 yellow tulip bulbs, and 60 purple tulip bulbs. (a) What is the probability that a randomly selected tulip bulb is red? (b) What is the probability that a randomly selected tulip bulb is purple? (c) Interpret these two probabilities.
Answer: a) 0.2 b) 0.6
c) The event of selecting red tulip is not likely to occur.
The event of selecting purple tulip is likely to occur.
Step-by-step explanation:
Given : Total number of tulips = 100
The number of red tulips = 20
The number of purple tulips =60
The probability that a randomly selected tulip bulb is red :-
[tex]\dfrac{\text{Number of red tulips}}{\text{Total tulips}}\\\\=\dfrac{20}{100}=0.2[/tex]
Since 0.2 is less than 0.5.
It means that the event of selecting red tulip is not likely to occur.
The probability that a randomly selected tulip bulb is purple :-
[tex]\dfrac{\text{Number of purple tulips}}{\text{Total tulips}}\\\\=\dfrac{60}{100}=0.6[/tex]
Since 0.6 is more than 0.5.
It means that the event of selecting purple tulip is likely to occur.
Final answer:
The probability of selecting a red tulip bulb is 20%, and the probability of selecting a purple tulip bulb is 60%. These probabilities reflect the likelihood of picking a bulb of a particular color at random from the bag.
Explanation:
The question involves calculating the probability of selecting a red or purple tulip bulb from a bag.
Probability of Selecting a Red Tulip Bulb
The probability, P(Red), is calculated by dividing the number of red bulbs by the total number of bulbs:
P(Red) = Number of Red Bulbs / Total Number of Bulbs = 20 / 100 = 0.2
Probability of Selecting a Purple Tulip Bulb
Similarly, the probability, P(Purple), is:
P(Purple) = Number of Purple Bulbs / Total Number of Bulbs = 60 / 100 = 0.6
Interpretation of Probabilities
These probabilities indicate that there is a 20% chance of selecting a red bulb and a 60% chance of selecting a purple bulb from the bag. The higher the probability, the more likely it is to select a bulb of that color at random.
2. Let a, b, cE Z such that ged(a, c)d for some integer d. Prove that if a | bc then a | bd. [3
Answer with explanation:
It is given that, a, b and c belong to the set of integers.
→ gcd(a,c)=d
→GCD=Greatest Common Divisor
→The greatest number which divides both a and c is d.
It means d divides a, and d divides c.
a=d k, for some integer k.-------(1)
c= d m, for some integer m.-------(2)
Now, it is given that, a divides bc.
So,→ 'a' will divide "bdm".--------[using 2, as c=d m]
It shows that, a divides bd, that is a| bd.
Hence proved
Prove Corollary 6.2. If L : V ? W is a linear transformation of a vector space V into a vector space W and dim V=dim W, then the following statements are true: (a) If L is one-to-one, then it is onto. (b) If L is onto, then it is one-to-one.
Answer with explanation:
Given L: V\rightarrow W is a linear transformation of a vector space V into a vector space W.
Let Dim V= DimW=n
a.If L is one-one
Then nullity=0 .It means dimension of null space is zero.
By rank- nullity theorem we have
Rank+nullity= Dim V=n
Rank+0=n
Rank=n
Hence, the linear transformation is onto. Because dimension of range is equal to dimension of codomain.
b.If linear transformation is onto.
It means dimension of range space is equal to dimension of codomain
Rank=n
By rank nullity theorem we have
Rank + nullity=dimV
n+nullity=n
Nullity=n-n=0
Dimension of null space is zero.Hence, the linear transformation is one-one.
To prove Corollary 6.2, we show that a one-to-one linear transformation between vector spaces of equal dimensions is onto, and conversely, an onto transformation is one-to-one. This conclusion is based on the properties of linear transformations and the significance of preserving vector space dimensions.
Explanation:To prove Corollary 6.2 regarding a linear transformation L: V → W, where both vector spaces V and W have the same dimension, we must show two parts:
If L is one-to-one, then it is onto. Assuming L is one-to-one, for every vector v in V, there is a unique image in W, guaranteeing that all vectors in W can be reached since dim V = dim W. Thus, L covers all of W, making it onto.
If L is onto, then it is one-to-one. When assuming L is onto, every vector in W is the image of some vector in V. Because dim V = dim W, there can't be more vectors in V than in W, preventing multiple vectors in V from mapping to the same vector in W, thus L is one-to-one.
These parts rely heavily on the concept that linear transformations maintain vector operations, such as the addition and scalar multiplication, and the relationship between dimensions of the domain and codomain for linear transformations.
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8) The monthly worldwide average number of airplane crashes of commercial airlines is 3.5. What is the probability that there will be (a) at least 2 such accidents in the next month; (b) at most 1 accident in the next month? Explain your reasoning!
Answer: (a) 0.8641
(b) 0.1359
Step-by-step explanation:
Given : The monthly worldwide average number of airplane crashes of commercial airlines [tex]\lambda= 3.5[/tex]
We use the Poisson distribution for the given situation.
The Poisson distribution formula for probability is given by :-
[tex]P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]
a) The probability that there will be at least 2 such accidents in the next month is given by :-
[tex]P(X\geq2)=1-(P(X=1)+P(X=0))\\\\=1-(\dfrac{e^{-3.5}(3.5)^0}{0!}+\dfrac{e^{-3.5}(3.5)^1}{1!})\\\\=1-(0.1358882254)=0.8641117746\approx0.8641[/tex]
b) The probability that there will be at most 1 accident in the next month is given by :-
[tex]P(X\leq1)=(P(X=1)+P(X=0))\\\\=\dfrac{e^{-3.5}(3.5)^0}{0!}+\dfrac{e^{-3.5}(3.5)^1}{1!}\\\\=0.1358882254\approx0.1359[/tex]
A random sample of 225 items from a population results in 60% possessing a given characteristic. Using this information, the researcher constructs a 99% confidence interval to estimate the population proportion. The resulting confidence interval is _______.
Answer: [tex](0.52,\ 0.68)[/tex]
Step-by-step explanation:
The formula for a [tex]\alpha[/tex]- level confidence interval for the population proportion:-
[tex]p\pm z_{\alpha/2}\times\sqrt{\dfrac{p(1-p)}{n}}[/tex]
Given : n = 225 ; p = 0.60 ; [tex]\alpha= 1-0.99=0.01[/tex]
By using the given information , the confidence interval for the population proportion:-
[tex]0.6\pm z_{0.005}\times\sqrt{\dfrac{0.6(1-0.6)}{225}}\\\\=0.6\pm(2.576)(0.0327)\\\\=0.6\pm0.08\\\\=0.6-0.08,\ 0.6+0.08\\\\=(0.52,\ 0.68)[/tex]
Hence, the resulting confidence interval is [tex](0.52,\ 0.68)[/tex] .
To construct a 99% confidence interval for the population proportion, we can use the sample proportion, critical value, and standard error. The resulting confidence interval is (0.516, 0.684).
Explanation:To calculate a confidence interval for a population proportion, we can use the formula:
CI = sample proportion ± (critical value)(standard error)
Given that the sample proportion is 60% (0.6) and the confidence level is 99%, we need to find the critical value corresponding to a 99% confidence level. Using a normal distribution, the critical value is approximately 2.576. The standard error can be calculated using the formula:
SE = √((sample proportion)(1 - sample proportion) / sample size)
Let's assume the sample size is 225. Plugging these values into the formula, we can calculate the standard error:
SE = √((0.6)(1 - 0.6) / 225) ≈ √(0.24 / 225) ≈ √0.001067 ≈ 0.0326
Now we can calculate the confidence interval:
CI = 0.6 ± (2.576)(0.0326) ≈ 0.6 ± 0.084
Therefore, the 99% confidence interval for the population proportion is approximately (0.516, 0.684).
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Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their paths intersect. (If an answer does not exist, enter DNE.)
Answer:
DNE
Step-by-step explanation:
Given that two particles travel along the space curves
[tex]r_1(t) = (t, t^2, t^3)\\ r_2(t) = (1 + 2t, 1 + 6t, 1 + 14t )[/tex]
To find the points of intersection:
At points of intersection both coordinates should be equal.
i.e. r1 =r2
Equate corresponding coordinates
[tex]t=1+2t\\t^2=1+6t\\t^3=1+14t[/tex]
I equation gives t =-1
Substitute in II equation to get [tex]t^2 = -5[/tex]
i.e. t cannot be real
Hence no point of intersection
DNE
Type the correct answer in each box. Use numerals instead of words.
Consider the systems of equations below.
Answer:
System A has 2 real solutions.
System B has 0 real solutions.
System C has 1 real solutions.
Step-by-step explanation:
If the graph of system of equation intersect each other at n points then the system of equation has n real solutions.
System A:
[tex]x^2+y^2=17[/tex] .... (1)
[tex]y=-\frac{1}{2}x[/tex] .... (2)
Plot the graph of these equations.
From graph (1) it is clear that the graph of equation (1) and (2), intersect each other at two points, (-3.688,1.844) and (3.688,-1.844).
Therefore, System A has two real solutions.
System B:
[tex]y=x^2-7x+10[/tex] .... (3)
[tex]y=-6x+5[/tex] .... (4)
Plot the graph of these equations.
From graph (2) it is clear that the graph of equation (3) and (4), never intersect each other.
Therefore, System B has 0 real solutions.
System C:
[tex]y=-2x^2+9[/tex] .... (5)
[tex]8x-y=-17[/tex] .... (6)
Plot the graph of these equations.
From graph (3) it is clear that the graph of equation (5) and (6), intersect each other at one points, (-2,1).
Therefore, System C has 1 real solutions.
Bill Mason is considering two job offers. Job 1 pays a salary of $41,300 with $5,525 of nontaxable employee benefits. Job 2 pays a salary of $40,400 and $7,125 of nontaxable benefits. Use a 25 percent tax rate.
Calculate the monetary value for both the jobs.
Final answer:
To calculate the monetary value of Bill Mason's job offers, you subtract the taxes from the salary and add the nontaxable benefits. Job 1 results in a total monetary value of $36,500, while Job 2 is higher, with a total monetary value of $37,425. Therefore, Job 2 offers a higher total monetary value.
Explanation:
Bill Mason is considering two job offers, each with a different combination of salary and nontaxable benefits. To determine the monetary value for both jobs, considering a 25 percent tax rate, we must calculate the Net Annual Income for each job separately. Net Annual Income is the salary after taxes have been subtracted. Nonetheless, nontaxable benefits do not affect the Net Annual Income as they do not get taxed.
Calculations for Job 1:
Gross Salary: $41,300
Taxable Income (Tax Rate 25%): 0.25 x $41,300 = $10,325
Net Salary after Tax: $41,300 - $10,325 = $30,975
Nontaxable Benefits: $5,525
Total Monetary Value: Net Salary + Benefits = $30,975 + $5,525 = $36,500
Calculations for Job 2:
Gross Salary: $40,400
Taxable Income (Tax Rate 25%): 0.25 x $40,400 = $10,100
Net Salary after Tax: $40,400 - $10,100 = $30,300
Nontaxable Benefits: $7,125
Total Monetary Value: Net Salary + Benefits = $30,300 + $7,125 = $37,425
The final answer for the total monetary value of Job 1 is $36,500 and for Job 2 is $37,425. As per the given calculations, Job 2 offers a higher total monetary value compared to Job 1 when taking into account the salary after tax plus nontaxable benefits. This explanation should provide clarity on how to assess job offers based on their financial merits.
What would the seasons be like if the axis of Earth's rotation was tilted 0 degrees to the ecliptic, instead of the 23.5 degrees we find it today? What about if it was tilted 0 degrees?
Answer:
The seasons would become constant. It would be equinox throughout the year.
Step-by-step explanation:
The earth would be in a state of constant equinox i.e., the length of day and night would be same in a particular place.
The season of a place would be what it is when it is normally titled at equinox.
The animal and plant life which depend on the seasons would be affected.
Snow would only occur at parts where it normally snows at equinoxes.
if 4 oz of solution is required for 2 Gallons of water. How much of the solution should I use if the amount of water required is 22 ounces?
1 Gallon = 128 Ounces
Answer:
You will requiere 0.34375 oz of the solution for 22 ounces of water.
Step-by-step explanation:
We should start by matching the units: in the example we have 2 gallons of water, while the question refers to 22 ounces of water.
So, by the equivalence we are given, we now know that 2 gallons of water are equal to 256 ounces.
From this point the question can be solved using an arithmetic method known as cross-multiply (sometimes refered as the Rule of Three). This consist of wrinting an equation of the form:
[tex]\frac{a}{b} = \frac{c}{x}\\[/tex]
Where b is the amount of solution required for a ounces of water, and x (unknown variable) is the amount of solution for c ounces of water.
So we have the next equation:
[tex]\frac{256}{4} =\frac{22}{x}[/tex]
We apply the correspondent arithmetic rules so we can calculate the x as follows:
[tex]x=\frac{(4)(22)}{256}[/tex]
[tex]x=\frac{88}{256}[/tex]
[tex]x=0.34375[/tex]
This way we can reply that 0.34375 oz of solution are required for 22 oz of water.
Mr. and Mrs. Rose have six old railroad ties that they would like to use to border two different triangular flower beds. They have two ties that are 4 feet long, two ties that are 6 feet long, one tie that is 9 feet long, and one tie that is 5 feet long. Can these ties be used to border two flower beds without having to cut them? If so, what are the total possible dimensions for each set of flower beds?
Answer:
2 ties of 4 feet and 1 tie of 5 feet
2 ties of 6 feet and 1 tie of 9 feet
Step-by-step explanation:
Given data
2 ties = 4 feet
2 ties = 6 feet
1 tie = 9 feet
1 tie = 5 feet
to find out
what are the total possible dimensions for each set of flower beds
solution
there are many combination but
the best possible combination are for 1st triangular
when we use 2 ties of 4 feet and 1 tie of 5 feet
and
for 2nd triangular
when we use 2 ties of 6 feet and 1 tie of 9 feet
only these are best combination
Susan is 3 times as old as Paul and Paul is 3/2 times as old as Mary. What fraction of Susan's age is Mary?
a) 3/8 b) 2/9 c) 15/4 d) 6 e) 13/2
Answer:
2/9
Step-by-step explanation:
We are given Susan (S) is 3 times as old as Paul (P) so S=3P.
And Paul (P) is 3/2 times as old as Mary (M) so P=(3/2)M.
What fraction of Susan's age is Mary?
So if S=3P and P=(3/2)M, then by substitution we have
[tex]S=3(\frac{3}{2})M[/tex]
[tex]S=\frac{9}{2}M[/tex]
Multiply both sides by the reciprocal of 9/2 resulting in:
[tex]\frac{2}{9}S=M[/tex]
This says Mary is two-ninths the age of Susan.
Answer:
b) 2/9.
Step-by-step explanation:
Let Mary's age be x , then:
Paul's age is (3/2)x.
Susan's age = 3(3/2)x = 9/2 x.
So Mary is 2/9 the age of Susan.
(5). (10 points) There are 5 hotels in Stony Brook. If 3 people check into hotels on September 12, what is the probability that they each check into a different hotel? (What assumptions are you making?) Make sure to define any notation you use to describe elements of the sample space.
Answer:
0.48
Step-by-step explanation:
Probability that the first person chooses a hotel
⁵C₁
[tex]^5C_1=\frac{5!}{(5-1)!1!}\\=\frac{120}{24}=5[/tex]
Probability that the second person chooses a different hotel
⁴C₁
[tex]^4C_1=\frac{4!}{(4-1)!1!}\\=\frac{24}{6}=4[/tex]
because the choice of hotels has reduced by 1 as one hotel is occupied by the first person
Probability that the second person chooses a different hotel
³C₁
[tex]^3C_1=\frac{3!}{(3-1)!1!}\\=\frac{6}{2}=3[/tex]
because the choice of hotels has reduced by 2 as two different hotels are occupied by the first person and second person
∴ The favorable outcomes are =⁵C₁×⁴C₁׳C₁=5×4×3=60
The total number of outcomes=5³=125
∴Probability that they each check into a different hotel=60/125=0.48
If a population is recorded at 1,200 in the year 2000 and the rate of increase is a steady 50 people each year, what will be the population in 2018?
Answer:
2100
Step-by-step explanation:
50*18=900
900+1,200=2100
To find the population in 2018, we calculate the total increase from 2000 to 2018 by multiplying the yearly increase of 50 people by 18 years, resulting in an additional 900 people. Adding this to the initial population of 1,200 people gives us a total population of 2,100 people in 2018.
If a population is recorded at 1,200 in the year 2000 and increases at a steady rate of 50 people each year, we can calculate the population in 2018 using a linear growth model. First, we need to determine the number of years between 2000 and 2018, which is 18 years. Next, we multiply the annual increase (50 people) by the number of years (18) to find the total increase over this period.
The calculation would be as follows:
Total Increase = Annual Increase times Number of YearsTotal Increase = 50 people/year times 18 yearsTotal Increase = 900 peopleWe then add this total increase to the initial population to get the population in 2018:
Population in 2018 = Initial Population + Total IncreasePopulation in 2018 = 1,200 people + 900 peoplePopulation in 2018 = 2,100 peopleThe population in 2018 would be 2,100 people.
For the lines x=3t, y=1-2t, z=2-3t and x (3, 1, 4) +s(-9, 6, 9) (a) Show that the lines are parallel. (b) Calculate the distance between the paralle lines.
Answer:
Step-by-step explanation:
Given lines in parametric form
line [tex]L_1[/tex]
[tex]\frac{x}{3}=\frac{y-1}{-2} =\frac{z-2}{-3}[/tex]
direction vector of [tex]L_1 v_1=<3,-2,-3 >[/tex]
Line [tex]L_2[/tex]
direction vector of [tex]L_2 v_2=<-9,6,9 >[/tex]
therefore
[tex]v_2=-3v_1[/tex]
thus lines are parallel.
(ii)distance between two lines is
[tex]L_2[/tex] is given by
[tex]\frac{x-3}{-9}=\frac{y-1}{6} =\frac{z-4}{9}[/tex]=s
[tex]\frac{x-3}{-3}=\frac{y-1}{2} =\frac{z-4}{3}[/tex]=3s
[tex]\left | \frac{\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1 \\ a_1&b_1&c_1 \\ a_2&b_2&c_2\end{vmatrix}}{\sqrt{\left ( a_1b_2-a_2b_1 \right )^2+\left ( b_1c_2-b_2c_1 \right )^2+\left ( c_1a_2-c_2a_1 \right )^2}}\right |[/tex]
where [tex]a_1[/tex]=3
[tex]b_1[/tex]=-2
[tex]c_1[/tex]=-3
[tex]a_2[/tex]=-9
[tex]b_2[/tex]=6
[tex]c_2[/tex]=9
distance(d)=0 units since value of the matrix
[tex]\begin{vmatrix}x_2-x_1 &y_2-y_1 &z_2-z_1\\ a_1&b_1&c_1 \\a_2&b_2 &c_2 \end{vmatrix}[/tex]
is zero
Tickets for a play cost 2 pounds for a child, and 4 pounds for an adult. one adult brought 4 children with him and the remaining adults each brought 2 children with them. The total ticket sales were 60 pounds. how many adults and children were present in that play?
Solve using augmented matrix.
Answer:
Number of adults = 7
Number of children = 16
Step-by-step explanation:
Tickets for a play cost 2 pounds for a child, and 4 pounds for an adult.
Let x number of adults and y number of children.
1 child ticket cost = 2 pound
y children ticket cost = 2y pound
1 adult ticket cost = 4 pound
x adults ticket cost = 4x pound
Total number of ticket sales were 60 pounds
Therefore, 4x + 2y = 60 ------------- (1)
One adult brought 4 children with him and the remaining adults each brought 2 children with them.
Remaining number of adult whose brought 2 children = x-1
Number children = 2(x-1)
Total number of children = 2(x-1)+4
Therefore, y=2x+2 ---------------------(2)
System of equation,
2x + y = 30
-2x + y = 2
Using augmented matrix to solve system of equation.
[tex]\begin{bmatrix}2&1&\ |30\\-2&1&|2\end{bmatrix}\\\\R_2\rightarrow R_2+R_1\\\\\begin{bmatrix}2&1& |30\\0&2&|32\end{bmatrix}\\\\R_2\rightarrow\dfrac{1}{2}R_2\\[/tex]
[tex]\begin{bmatrix}2&1&\ |30\\0&1&|16\end{bmatrix}\\\\R_1\rightarrow R_1-R_2\\\\\begin{bmatrix}2&0&\ |14\\0&1&|16\end{bmatrix}\\\\\\[/tex]
[tex]R_1\rightarrow \dfrac{1}{2}R_1\\\\\begin{bmatrix}1&0&|7\\0&1&|16\end{bmatrix}\\\\[/tex]
Now, we find the value of variable.
[tex]x=7\text{ and }y=16[/tex]
Hence, Number of adults are 7 and Number of children are 16.
A container contains 12 diesel engines. The company chooses 5 engines at random, and will not ship the container if any of the engines chosen are defective. Find the probability that a container will be shipped even though it contains 2 defectives if the sample size is 5.
Answer:[tex]\frac{^{10}C_5}{^{12}C_5}[/tex]
Step-by-step explanation:
Given a total of 12 diesel engines
out of which 2 are defective
Company have to choose 5 good engines to ship.
Therefore no of ways in good engines are selected is [tex]^{10}C_5[/tex]
no of ways in which any 5 engines are selected from a total of 12 engines is
[tex]^{12}C_5[/tex]
Therefore the required probability =[tex]\frac{^{10}C_5}{^{12}C_5}[/tex]
=[tex]\frac{7}{22}[/tex]=0.318
Alfred collects black and brown socks but loses a lot of them. If 60% of all his socks are brown, 20% of his brown socks are in the wash and 120 brown socks are in his sock drawer, how many socks does Alfred have?
Answer:
Alfred has 250 socks in his collection.
Step-by-step explanation:
Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.
Based on the information given to us we can see that out of All his Brown socks he has 20% in the wash and the rest are in his drawer. Meaning 80% of the Brown socks are in his drawer. So we first need to find how many Brown socks are in the wash. We can solve this using the Rule of Three property as shown in the picture below.
120 drawer ⇒ 80%
x wash ⇒ 20%
[tex]\frac{120*20}{80} = 30[/tex]
Now that we have the amount of Brown socks in the washer we can add that to the amount in the drawer to find the total amount of Brown socks.
[tex]Br = 120+30\\Br = 150[/tex]
So we now know that there are a total of 150 Brown socks. Since the question states that the Brown socks are 60% of the total we can use the Rule of Three to find the total.
150 Brown ⇒ 60%
T Total ⇒ 100%
[tex]\frac{150*100}{60} = T[/tex]
[tex]250 socks = T[/tex]
Finally, we can see that Alfred has 250 socks in his collection.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Alfred owns a total of 250 socks. This was derived by first finding out that Alfred has 150 brown socks, which represent 60% of his total socks' collection. Thus, the total number of socks Alfred owns is 250.
Explanation:The question deals with the calculation of a total number of socks owned by Alfred.
If we know that 120 brown socks represent 80% of all of Alfred's brown socks (because 20% of them are in the wash), we can calculate the total number of brown socks. To do this, divide 120 by 0.8, which equals 150. Hence, Alfred has 150 brown socks.
We know from the problem that the brown socks account for 60% of all his socks. Hence the total number of socks (brown and black) is calculated by dividing the total number of brown socks (150) by 0.6. After performing this percentage calculation, we find that Alfred owns 250 socks.
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Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular solution requested (Label the Gen. Sol. and Particular Sol.). Write both solutions in EXPLICIT FORM (solved for y). x dy/dx = x^3 + 2y subject to: y(2) = 6
Answer:
General Solution is [tex]y=x^{3}+cx^{2}[/tex] and the particular solution is [tex]y=x^{3}-\frac{1}{2}x^{2}[/tex]
Step-by-step explanation:
[tex]x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}[/tex]
This is a linear diffrential equation of type
[tex]\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)[/tex]..................(i)
here [tex]p(x)=\frac{-2}{x}[/tex]
[tex]q(x)=x^{2}[/tex]
The solution of equation i is given by
[tex]y\times e^{\int p(x)dx}=\int e^{\int p(x)dx}\times q(x)dx[/tex]
we have [tex]e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}[/tex]
Thus the solution becomes
[tex]\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+c[/tex][tex]y=x^{3}+cx^{2[/tex]
This is the general solution now to find the particular solution we put value of x=2 for which y=6
we have [tex]6=8+4c[/tex]
Thus solving for c we get c = -1/2
Thus particular solution becomes
[tex]y=x^{3}-\frac{1}{2}x^{2}[/tex]
1. Solve the equation x = (2x+ 3)1/2
2. Consider f1(x) = ln(x + 1) + ln (x-1) and f2(x) = ln(x^2-1).
a) State domains and ranges of f1 and f2.
b) Sketch the curves y = f1(x) and y = f2(x).
Answer:
1. [tex]x=3[/tex]; 2. Domain: [tex]x>1[/tex] Range: all real numbers
Step-by-step explanation:
Let's find the solutions.
1. Solve the equation [tex]x=\sqrt{2x+3}[/tex] so:
[tex](x)^2=(\sqrt{2x+3})^2[/tex]
[tex]x^2=2x+3[/tex]
[tex]x^2-2x-3=0[/tex]
[tex]x1=\frac{-b+\sqrt{b^{2}-4ac}}{2a}[/tex]
[tex]x1=\frac{2+\sqrt{(-2)^{2}-(4*1*(-3))}}{2*1}[/tex]
[tex]x1=3[/tex]
[tex]x2=\frac{-b-\sqrt{b^{2}-4ac}}{2a}[/tex]
[tex]x2=\frac{2-\sqrt{(-2)^{2}-(4*1*(-3))}}{2*1}[/tex]
[tex]x2=-1[/tex]
Although we have two answers, remember that from the original equation the result of [tex]\sqrt{2x+3} > 0[/tex] is never negative. So -1 do not solve the equation.
In conlcusion, the equation is solved by x=3.
2A. Domains and ranges of f1(x) and f2(x)
[tex]f1(x)=ln(x+1)+ln(x-1)[/tex]
Using logarithmic property [tex]ln(a)+ln(b)=ln(a*b)[/tex] we have:
[tex]f1(x)=ln(x^2-1)[/tex] because:
[tex]ln(x)[/tex] is defined by [tex]x>0[/tex] then:
[tex]x^2-1>0[/tex]
[tex]x>\sqrt{1}[/tex] so the domain of f1(x) is [tex]x>1[/tex]
Now for the range:
[tex]f1(x)=ln(x+1)+ln(x-1)[/tex]
[tex]y=ln(x^2-1)[/tex]
[tex]e^y=x^2-1[/tex]
[tex]\sqrt{e^y+1}=x^2-1[/tex] notice that [tex]e^y+1[/tex] is always positive, so the range of f1(x) is all real numbers.
Be aware that although point number two of the problem mentioned two equations, f1(x)=f2(x) by logarithmic properties, so their domains and ranges are the same.
2B. Graph of f1(x) is attached. Because f1(x)=f2(x) both functions plot equal.
Beginning one month after birth of their son, Noah, the Nelsons deposited $100 each month in an annuity for his college fund. The annuity earned interest at an average rate of 6.8% compounded monthly until his 18th birthday. What was the amount of Noah's college fund on his 18th birthday? Referring to question 4, how much interest did Noah's college fund earn in total on his 18th birthday?
Answer:
$100 * (1 + 6.8%/12)^216 + $100*(1+6.8%/12)^215 + ... + $100*(1+6.8%/12)^1
Now note that
x + x^2 + x^3 + ... + x^N = x ( 1 + x + ... + x^(N-1) )
= x ( (x^N -1)/(x-1) )
Here, x = 1+6.8%1 = 1.00566666 and N = 216, so
$100 * ( 1.00566666 ( 1.00566666^216 -1) / 0.00566666 )
= $ 42398.33
The total interest earned is $42,398 - $21,600 = $20,798
Step-by-step explanation:
The generic formula used in this compound interest calculator is V = P(1+r/n)^(nt)
V = the future value of the investment
P = the principal investment amount
r = the annual interest rate
n = the number of times that interest is compounded per year
t = the number of years the money is invested for
A simple random sample of size nequals40 is drawn from a population. The sample mean is found to be 104.3, and the sample standard deviation is found to be 18.2. Is the population mean greater than 100 at the alphaequals0.01 level of significance?
Answer:
We conclude that population mean is equal to 100 at the α=0.01 level of significance.
Step-by-step explanation:
Given information:
Sample size, n=40
Sample mean=104.3
sample standard deviation, s=18.2
We need to check whether the population mean greater than 100 at the α=0.01 level of significance.
Null hypothesis:
[tex]H_0:\mu=100[/tex]
Alternative hypothesis:
[tex]H_1:\mu>100[/tex]
Let as assume that the data follow the normal distribution. It is a right tailed test.
The formula for z score is
[tex]z=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]z=\frac{104.3-100}{\frac{18.2}{\sqrt{40}}}[/tex]
[tex]z=1.494263[/tex]
[tex]z\approx 1.49[/tex]
Using the standard normal table the p-value at z=1.49 and 0.01 level of significance is 0.068112.
(0.068112 > 0.01) p-value is greater than α, so we accept the null hypothesis.
Therefore, we conclude that population mean is equal to 100 at the α=0.01 level of significance.
Find 10^(5^101) (mod 21).
note: 10^(5^101) is not 10^(501)
We have [tex]\lambda(21)=6[/tex], where [tex]\lambda[/tex] is the Carmichael function. So we have
[tex]10^{5^{101}}\equiv10^{5^{101}\pmod6}\pmod{21}[/tex]
The powers of 5 modulo 6 follow a periodic pattern
[tex]5^1\equiv5\pmod6[/tex]
[tex]5^2\equiv25\equiv1\pmod6[/tex]
[tex]5^3\equiv1\cdot5\equiv5\pmod6[/tex]
[tex]5^4\equiv5^2\equiv1\pmod6[/tex]
and so on, with odd powers of 5 equivalent to 5 modulo 6. So
[tex]10^{5^{101}}\equiv10^{5^{101}\pmod6}\equiv10^5\pmod{21}[/tex]
The rest is easy to deal with. We have
[tex]10^2\equiv16\pmod{21}[/tex]
[tex]10^3\equiv160\equiv13\pmod{21}[/tex]
[tex]10^4\equiv130\equiv4\pmod{21}[/tex]
[tex]10^5\equiv40\equiv19\pmod{21}[/tex]
and so the answer is 19.
In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. The expected frequency for the Business College is
Answer:
Step-by-step explanation:
Given that in the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College.
Table is prepared as follows
Bus coll. Lib Arts coll Educ. coll Total
Observed 90 120 90 300
Expected p.c. 35 35 30 100
Expected number 105 105 90 300
Percent*300/100
Hence expected frequency for business college = 105
This is for mathematics!p
Answer:
$933.12
Step-by-step explanation:
This is a composite figure: an upper rectangle and a lower one. We need to find the volumes of each one individually, add the volumes together, then multiply the total volume by .02
The upper rectangle has a volume of:
V = 72×12×24
V = 20,736 cubic inches
The lower rectangle has a volume of:
V = 72×36×10
V = 25,920 cubic inches
The sum of the two volumes is
V = 46,656 inches cubed
Multiply that by .02:
46,656(.02) = $933.12
Write the equation in the slope-intercept form 7x +4y - 20 0 Find the y-intercept of the corresponding line. (x, y)
Step-by-step explanation:
write the equation in slope intercept form
7x + 4y - 20=0
find the slope of corresponding line.
????
then
find the y intercept of corresponding line
(x,y)=???.
A real estate company wants to build a parking lot along the side of one of its buildings using 800 feet of fence. If the side along the building needs no fence, what are the dimensions of the largest possible parking lot?
Answer:
80,00[tex]ft^{2}[/tex]
Step-by-step explanation:
According to my research, the formula for the Area of a rectangle is the following,
[tex]A = L*W[/tex]
Where
A is the AreaL is the lengthW is the widthSince the building wall is acting as one side length of the rectangle. We are left with 1 length and 2 width sides. To maximize the Area of the parking lot we will need to equally divide the 800 ft of fencing between the Length and Width.
800 / 2 = 400ft
So We have 400 ft for the length and 400 ft for the width. Since the width has 2 sides we need to divide 60 by 2.
400/2 = 200 ft
Now we can calculate the maximum Area using the values above.
[tex]A = 400ft*200ft[/tex]
[tex]A = 80,000ft^{2}[/tex]
So the Maximum area we are able to create with 800 ft of fencing is 80,00[tex]ft^{2}[/tex]
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
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The largest possible parking lot, given 800 feet of fencing with one side not requiring a fence, would be a rectangular lot with dimensions of 400 feet by 200 feet.
Explanation:We're dealing with a rectangular parking lot here where one of its sides is bordered by a building, so it doesn't need a fence. We have 800 feet of fence available for the three remaining sides. Let's denote the length of the rectangular parking lot by 'x' and the width by 'y'.
Because we only need to fence three sides, we can establish the following equation based on the total amount of fence available:
x + 2y = 800
To calculate the maximum possible area of a rectangle, we need to use the formula for the area of a rectangle, which is length times width (Area = x * y). However, we want to express the area in terms of a single variable. To do this, we can rearrange our fence equation to solve for y:
y = (800 - x) / 2
Now replace y in the area equation:
Area = x * (800 - x) / 2
For the area to be maximum, the derivative of the area with respect to x must be equal to zero. Differentiating and solving for x, we get the dimensions as x = 400 and y = 200.
So the largest possible parking lot would have dimensions of 400 feet by 200 feet.
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