Kristie has taken five tests in science class. The average of all five of Kristie's test scores is 94. The average of her last three test scores is 92. What is the average of her first two test scores?

Answer:

8000/4*100 = $200'000

Step-by-step explanation:

Answer:

Angle made by secant line equals[tex]56.7251^{o}[/tex]

Step-by-step explanation:

Solpe of a line joining points [tex](x_{1},y_{1}),(x_{2},y_{2})[/tex] is given by

[tex]tan(\theta)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

where [tex]y_{i}=f(x_{i})[/tex]

Applying values we get

[tex]tan(\theta)=\frac{7.1-3.9}{3.6-1.5}\\\\\theta =tan^{-1}\frac{32}{21}\\\\\theta=56.7251^{o}[/tex]

By gcd, I think you mean gcf ( Greatest Common Factor).

To find the gcf find all the factors of each number:

Factors of 20: 1, 2, 4,5 ,10 , 20

Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56

The largest common factor is 4.

LCM = Least Common Multiple

This is the smallest number that both numbers divide into evenly

Find the prime factors of each number:

Prime factors of 20: 2 * 2 * 5

Prime factors of 56: 2 * 2 * 2 * 7

To find the LCM, multiply all prime factors the most number of times they occur.

In the prime factor of 56, 2 appears 3 times and 7 appears once.

In the prime factor of 20 5 appears once.

LCM = 2 * 2 *2 * 7 * 5 = 280

Answer:

f(-4)=60 and g(-6)=17

Step-by-step explanation:

Plug in -4 for x-values

Square -4

Multiply 3 by 16 and -3 by -4 then solve

Simplify

Plug in -6 for x

Multiply -3 by -6

Subtract 1

Simplify

For this case we have the following functions:

[tex]f (x) = 3x ^ 2-3x\\g (x) = 3x-1[/tex]

We must find the value of the function [tex]f (x)[/tex] when [tex]x = -4[/tex], then:

[tex]f (-4) = 3 (-4) ^ 2-3 (-4)\\f (-4) = 3 * 16 12\\f (-4) = 48 12\\f (-4) = 60[/tex]

We must find the value of the function g (x) when [tex]x = -6[/tex], then:

[tex]g (-6) = 3 (-6) -1\\g (-6) = - 18-1\\g (-6) = - 19[/tex]

Answer:

[tex]f (-4) = 60\\g (-6) = - 19[/tex]

Answer:

a) 0; b) 4[tex]x^{2}[/tex]

Step-by-step explanation:

a) To compute f o g (0), first evaluate g(x) for x=0 and then evaluate f for x=g(0).

[tex]f \circ g (0)=f(2 \cdot 0)=f(0)=0^2[/tex]

b) To compute a mathematical expression for f o g do the same but instead of 0 use x,

[tex]f \circ g (x) = f( 2 \cdot x)= (2 \cdot x )^2[/tex]

In the question, f(x) = x², g(x) = 2x. We need to determine the value of function f composed with function g at 0 (f o g(0)), and the general expression for f o g(x). f o g(0) = 0 and (f o g)(x) = 4x².

To solve this problem, we first need to understand that 'f o g' denotes the composition of function f and function g, defined as (f o g)(x) = f(g(x)). In this case, function f(x) = x^2 and function g(x) = 2x.

(a) To evaluate f o g at 0, we substitute x = 0 into g(x), giving us g(0) = 2*0 = 0. Substituting g(0) into f(x), we get f(g(0)) = f(0) = 0. So, f o g(0) = 0.

(b) For a general form of f o g, we substitute g(x) = 2x into f(x), resulting in (f o g)(x) = f(2x) = (2x)^2 = 4x^2.

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Each calculation has been evaluated, taken all significant figures into consideration, and results have been presented in scientific notation. Special attention was given to rules related to multiplying numbers in scientific notation.

The given calculations require us to use scientific notation and proper treatment of significant figures. We are using the fundamentals of arithmetic with scientific notation, which is based on the rules of exponents. Each calculation is treated as follows:

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Answer:

The height is 28.57 cm.

The surface area is 9,628 cm^2.

Step-by-step explanation:

I assume the cooler is shaped like a rectangular prism with length and width of the base given, and with an unknown height.

volume = length * width * height

First, we convert the volume from liters to cubic centimeters.

60 liters * 1000 mL/L * 1 cm^3/mL = 60,000 cm^3

Now we substitute every dimension we have in the formula and solve for height, h.

60,000 cm^3 = 60 cm * 35 cm * h

60,000 cm^3 = 2,100 cm^2 * h

h = (60,000 cm^3)/(2,100 cm^2)

h = 28.57 cm

The height is 28.57 cm.

Now we calculate the internal surface area.

total surface area = area of the bases + area of the 4 sides

SA = 2 * 60 cm * 35 cm + (60 cm + 35 cm + 60 cm + 35 cm) * 28.57 cm

SA = 9,628 cm^2

The surface area is 9,628 cm^2.

Answer:

So the vectors are linearly independent.

Step-by-step explanation:

So if they are linearly independent then the following scalars in will have the condition a=b=c=0:

a(2,3,1)+b(2,-5,-3)+c(-3,8,-5)=(0,0,0).

We have three equations:

2a+2b-3c=0

3a-5b+8c=0

1a-3b-5c=0

Multiply last equation by -2:

2a+2b-3c=0

3a-5b+8c=0

-2a+6b+10c=0

Add equation 1 and 3:

0a+8b+7c=0

3a-5b+8c=0

-2a+6b+10c=0

Divide equation 3 by 2:

0a+8b+7c=0

3a-5b+8c=0

-a+3b+2c=0

Multiply equation 3 by 3:

0a+8b+7c=0

3a-5b+8c=0

-3a+9b+6c=0

Add equation 2 and 3:

0a+8b+7c=0

3a-5b+8c=0

0a+4b+13c=0

Multiply equation 3 by -2:

0a+8b+7c=0

3a-5b+8c=0

0a-8b-26c=0

Add equation 1 and 3:

0a+0b-19c=0

3a-5b+8c=0

0a-8b-26c=0

The first equation tells us -19c=0 which implies c=0.

If c=0 we have from the second and third equation:

3a-5b=0

0a-8b=0

0a-8b=0

0-8b=0

-8b=0 implies b=0

We have b=0 and c=0.

So what is a?

3a-5b=0 where b=0

3a-5(0)=0

3a-0=0

3a=0 implies a=0

So we have a=b=c=0.

So the vectors are linearly independent.

Final answer:

To find out if the vectors (2, 3, l), (2, -5, -3), and (-3, 8, -5) are linearly dependent or independent, set up a linear system with the vectors and look for non-trivial solutions.

Explanation:

To determine whether the vectors (2, 3, l), (2, -5, -3), and (-3, 8, -5) are linearly dependent or linearly independent, we set up the equation a(2, 3, l) + b(2, -5, -3) + c(-3, 8, -5) = (0, 0, 0), where a, b, and c are scalars.

If only the trivial solution exists, where a = b = c = 0, then the vectors are linearly independent. If a non-trivial solution exists, then the vectors are linearly dependent.

Let's solve the system of linear equations generated from the above equation:

Using the methods for solving systems of linear equations, such as Gaussian elimination, we can determine whether a unique solution exists.

If the determinant of the coefficients matrix is non-zero, the system has a unique solution, indicating linear independence. Otherwise, a non-unique solution indicates linear dependence, and we can express

Answer: 0.3758

Step-by-step explanation:

Given : The  probability that a milk container is underweight throughout of packaging line: [tex]p = 0.20[/tex]

The number of containers : n= 10

The formula binomial distribution formula :-

[tex]^nC_rp^{n-r}(1-p)^r[/tex]

The probability that at least nine milk containers in a box are properly filled is given by :-

[tex]P(X\geq9)=P(9)+P(10)\\\\=^{10}C_9(0.2)^{10-9}(1-0.20)^9+^{10}C_{10}(0.2)^{10-10}(1-0.2)^{10}\\=10(0.2)(0.8)^9+(1)(0.8)^{10}\\=0.3758096384\approx0.3758[/tex]

Answer:

t= 3.322 years

Step-by-step explanation:

investment made= $5000 (Principal)

amount obtained after a specific time= $6400

rate %= 7.5% per year compounded quarterly which means

r= 7.5/(100*4)= 0.01875

time = 4t ( compounded quarterly)

we know that Amount obtained is given by

[tex]A= P(1+r)^{4t}[/tex]

[tex]6400= 5000(1+0.01875)^{4t}[/tex]

[tex](1.01875)^{4t}=1.28[/tex]

taking log on both sides and solving we get

t= 3.322 years

hence my answer t= 3.322 years

Answer:

40514.837 kg

Step-by-step explanation:

The weight of the main axle of the 1893 ferris wheel built by George Washington Gale Ferris Jr. in Chicago, USA was 89,320 pounds (lb).

1 kg=2.20462 pounds (lb)

[tex]\Rightarrow 1 lb=\frac{1}{2.20426}[/tex]

[tex]\Rightarrow 1 lb=0.453592 kg[/tex]

[tex]\Rightarrow 89320 lb=89320\times 0.453592[/tex]

[tex]\therefore 89320 lb=40514.837 kg[/tex]

Answer:[tex]\frac{1}{1000}[/tex]

Step-by-step explanation:

For the lottery game three numbers must match in exact order

From 0 to 9 total 10 numbers are there

Therefore selecting exactly same numbers as of winner is

=[tex]^{10}C_1\times ^{10}C_1\times ^{10}C_1 [/tex]

Since numbers are repeatable therefore each time we have a choice of choosing 1 number out o 10

=[tex]10\times 10\times 10[/tex]

Probability of winning=[tex]\frac{1}{1000}[/tex]

In a lottery game where the player must match a repeatable sequence of three numbers ranging from 0-9, the probability of a single ticket having the winning sequence is 1/1000, or 0.1%.

The subject of this problem is probability in Mathematics, specifically in a lottery context. Each digit in the sequence can be any number from 0 to 9. Because these numbers are repeatable, this means there are 10 possible numbers for each of the three digits in the sequence. Therefore, to find the total number of possible sequences, you multiply the ten options for the first digit by the ten for the second and the ten for the third, which comes to 10 * 10 * 10 = 1000 total possible sequences.

Since you are only looking for one specific sequence being the winning number, that means there’s only 1 favorable outcome out of 1000. Therefore, the probability of your ticket having the winning number sequence is 1/1000, or 0.001 in decimal form or 0.1% in percentage form.

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Final answer:

To find the probability that a computer memory chip will have a lifetime of less than 8 years, we can use the properties of the Gamma distribution.

Explanation:

To find the probability that a computer memory chip will have a lifetime of less than 8 years, we can use the properties of the Gamma distribution. The average lifetime of the chip is given as 4 years, which corresponds to the mean. The variance is given as 16/3 years squared, which is equal to the mean squared.

Using these values, we can determine the shape and rate parameters of the Gamma distribution. The shape (α) is equal to the mean squared divided by the variance, which in this case is 16/3. The rate (β) is equal to the mean divided by the variance, which in this case is 4/(16/3).

To find the probability that the chip will have a lifetime of less than 8 years, we can calculate the cumulative distribution function (CDF) of the Gamma distribution with the shape and rate parameters we obtained.

Answer:

There are going to be 31 matches played in the soccer league.

Step-by-step explanation:

The soccer league has 6 teams, so if every team plays against the others twice, there are going to be played 30 matches:

-Team 1: v Team 2 (2), v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 2: v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 3: v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 4: v Team 5 (2), v Team 6 (2)

-Team 5: v Team 6 (2)

-Team 6: -

If there is a final championship game after the 30 regular season matches, there are going to be 31 matches played in the league.

Answer:

Expected Grade=2 i.e., C

Variance=1.2

Step-by-step explanation:

[tex]Expected\ value=E\left [ x \right ]=\sum _{i=1}^{k} x_{i}p_{i}[/tex]

The x values are

A = 4

B = 3

C = 2

D = 1

F = 0

Probability of each of the events

P(4)=0.1

P(3)=0.2

P(2)=0.4

P(1)=0.2

P(0)=0.1

[tex]E\left [ x \right ]=4\times 0.1+3\times 0.2+2\times 0.4+1\times 0.2+0\times 0.1\\\therefore E\left [ x \right ]=2[/tex]

Variance

[tex]Var\left ( x\right)=E\left [ x^2 \right ]-E\left [ x \right ]^2[/tex]

[tex]E\left [ x^2 \right ]=4^2 \times 0.1+3^2 \times 0.2+2^2 \times 0.4+1^2 \times 0.2+0^2 \times 0.1\\\Rightarrow E\left [ x^2 \right ]=5.2\\E\left [ x \right ]^2=2^2=4\\\therefore Var\left ( x\right)=5.2-4=1.2\\[/tex]

a. Substitute the given solutions and their derivatives into the ODE.

[tex]y_1=x\implies {y_1}'=1\implies{y_1}''=0[/tex]

[tex]x^2y''-xy'+y=-x+x=0[/tex]

[tex]y_2=x\ln x\implies{y_1}'=\ln x+1\implies{y_1}''=\dfrac1x[/tex]

[tex]x^2y''-xy'+y=x-x(\ln x+1)+x\ln x=0[/tex]

Both solutions satisfy the ODE.

b. The Wronskian determinant is

[tex]\begin{vmatrix}x&x\ln x\\1&\ln x+1\end{vmatrix}=x(\ln x+1)-x\ln x=x\neq0[/tex]

so the solutions are indeed independent.

c. The ODE has general solution [tex]y(t)=C_1x+C_2x\ln x[/tex]. Then with the given initial conditions, the constants satisfy

[tex]y(1)=7\implies 7=C_1[/tex]

[tex]y'(1)=2\implies2=C_1+C_2\implies C_2=-5[/tex]

So the ODE has the particular solution,

[tex]\boxed{y(t)=7x-5x\ln x}[/tex]

Final answer:

The functions y1 = x and y2 = x ln x are verified as solutions to the differential equation x^2y'' - xy' + y = 0. They are confirmed to be linearly independent through a non-zero Wronskian. Lastly, the particular solution is found to be y = 7x - 5x ln x using given initial conditions.

Explanation:

To verify that y1 = x and y2 = x ln x are solutions to the differential equation x2y'' - xy' + y = 0, we need to substitute each function into the equation and show that the left-hand side reduces to zero.

For y1 = x, its derivatives are y1' = 1 and y1'' = 0. Substituting these into the equation gives x2(0) - x(1) + x = 0, which simplifies to 0, confirming that y1 is a solution.

For y2 = x ln x, its first derivative is y2' = ln x + 1, and the second derivative is y2'' = 1/x. Substituting these into the equation gives x2(1/x) - x(ln x + 1) + x ln x = 0, which also simplifies to 0, confirming that y2 is a solution.

To demonstrate that y1 and y2 are linearly independent, we must calculate the Wronskian, W(y1,y2), and show that it is non-zero. The Wronskian is:
W(y1,y2) = y1y2' - y1'y2 = x(ln x + 1) - (x ln x) = x.
Since the Wronskian is not zero for all x
e 0, y1 and y2 are linearly independent.

For the particular solution of the differential equation with initial conditions y(1) = 7, y'(1) = 2, we express y as a linear combination of y1 and y2:
y = c1y1 + c2y2 = c1x + c2x ln x.
Applying the initial conditions, we get two equations:
1) y(1) = c1(1) + c2(1 ln 1) = 7
2) y'(1) = c1 + c2(ln 1 + 1) = 2
Simplifying these equations gives us c1 = 7 and c2 = -5, therefore the particular solution is y = 7x - 5x ln x.

Answer:

[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]

Step-by-step explanation:

We are given the following quadratic equation by completing the square:

[tex]z^2 - 3z - 5 = 0[/tex]

Rewriting the equation in the form [tex]x^2+2ax+a^2[/tex] to get:

[tex]z^2 - 3z - 5+(-\frac{3}{2} )^2-(-\frac{3}{2} )^2[/tex]

[tex]z^2-3z+(-\frac{3}{2} )^2=(z-\frac{3}{2} )^2[/tex]

Completing the square to get:

[tex] ( z - \frac{ 3 } { 2 } )^ 2 - 5 - ( - \frac { 3 } { 2 } ) ^ 2[/tex]

[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]

Answer: [tex]z_1=4.19\\\\z_2=-1.19[/tex]

Step-by-step explanation:

Add 5 to both sides of the equation:

[tex]z^2 - 3z - 5 +5= 0+5\\\\z^2 - 3z = 5[/tex]

Divide the coefficient of [tex]z[/tex] by two and square it:

[tex](\frac{b}{2})^2= (\frac{3}{2})^2[/tex]

Add it to both sides of the equation:

[tex]z^{2} -3z+ (\frac{3}{2})^2=5+ (\frac{3}{2})^2[/tex]

Then, simplifying:

[tex](z- \frac{3}{2})^2=\frac{29}{4}[/tex]

Apply square root to both sides and solve for "z":

[tex]\sqrt{(z- \frac{3}{2})^2}=\±\sqrt{\frac{29}{4} }\\\\z=\±\sqrt{\frac{29}{4}}+ \frac{3}{2}\\\\z_1=4.19\\\\z_2=-1.19[/tex]

Answer:

(d) reliability

Step-by-step explanation:

mostly we see that  validity and reliability is the key aspects of all research they help in differentiation between good and bad research so both are very necessary aspects of any research so the another name for validity in quantity research is reliability.

so the reliability will be the correct answer

so option (d) will be correct option

Answer:

.008000

Step-by-step explanation:

The first digit is either 0,1,2,3,4

P( right guess) = 1/5

The second digit is either 0,1,2,3,4

P( right guess) = 1/5

The third digit is either 0,1,2,3,4

P( right guess) = 1/5

Since they are independent

P( right,right,right) = 1/5*1/5*1/5 = 1/125 =.008

To six decimal places = .008000

To solve the problem, let's consider each piece of information step by step:
1. The password is 3 digits long.
2. Each digit can be any number from 0 to 4.
Since there are 5 choices for each digit (0, 1, 2, 3, or 4), we calculate the total number of distinct combinations possible for a 3-digit password where each digit has 5 possibilities.
For each place of the three digits, we have 5 choices, which gives us a total combination count using the Multiplication Principle:
- First digit: 5 choices (0-4)
- Second digit: 5 choices (0-4)
- Third digit: 5 choices (0-4)
To find the total number of different password combinations, we multiply the number of choices for each digit:
Total combinations = 5 (choices for the first digit) × 5 (choices for the second digit) × 5 (choices for the third digit) = \( 5^3 = 125 \) possible password combinations.
Each of these combinations is equally likely if the hacker guesses at random. Hence, the probability that the hacker guesses the correct password on the first try is 1 out of the total number of combinations.
Therefore, the probability is:
\( P(\text{correct on first try}) = \frac{1}{125} \)
Let's convert this probability to a decimal and then round it to six decimal places:
\( P(\text{correct on first try}) = \frac{1}{125} = 0.008 \)
When rounded to six decimal places, the probability is:
\( P(\text{correct on first try}) \approx 0.008000 \)
So, the probability that the hacker guesses the password correctly on the first try is approximately 0.008000.

Answer:

$1.710

Step-by-step explanation:

Mike deposited $850 into the bank in July.

In August his balance will be: $850×1.15 = $977.5

In September his balance will be: $977.5×1.15 = $1124.125

In October his balance will be: $1124.125×1.15 = $1292.74375

In November his balance will be: $1292.74375×1.15 = $1.486,6553125

In December his balance will be: $1.486,6553125×1.15 = $1.709,653609375

Therefore, the amount of money he will have after december will be $1.710

Answer:

The total number of samples that contain at least 3 red balls is 115.

Step-by-step explanation:

Total number of balls = 11

Total number of red balls = 6

Total number of white balls = 5

A sample of 4 balls is to be selected that contain at least 3 red. It means either 3 out of 4 balls are red or 4 out of 4 ball are red.

[tex]\text{Total ways}=\text{Three balls are red}+\text{Four balls are red}[/tex]

[tex]\text{Total ways}=^6C_3\times ^5C_1+^6C_4\times ^5C_0[/tex]

Combination formula:

[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Using this formula we get

[tex]\text{Total ways}=\frac{6!}{3!(6-3)!}\times \frac{5!}{1!(5-1)!}+\frac{6!}{4!(6-4)!}\times \frac{5!}{0!(5-0)!}[/tex]

[tex]\text{Total ways}=20\times 5+15\times 1[/tex]

[tex]\text{Total ways}=115[/tex]

Therefore the total number of samples that contain at least 3 red balls is 115.

Using the combination formula, it is found that 115 samples contain at least 3 red balls.

The balls are chosen without replacement, which is why the combination formula is used.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem, the outcomes with at least 3 red balls are:

Hence:

[tex]T = C_{6,3}C_{5,1} + C_{6,4} = \frac{6!}{3!3!}\frac{5!}{1!4!} + \frac{6!}{4!2!} = 20(5) + 15 = 100 + 15 = 115[/tex]

115 samples contain at least 3 red balls.

A similar problem is given at https://brainly.com/question/24437717

Answer:

The probability that the racket is wood or defective is 0.57.

Step-by-step explanation:

Let W represents wood racket, G represents the graphite racket and D represents the defective racket,

Given,

n(W) = 100,

n(G) = 100,

⇒ Total rackets = 100 + 100 = 200

n(W∩D) = 15,

n(G∩D) = 14,

⇒ n(D) = n(W∩D) + n(G∩D) = 15 + 14 = 29,

We know that,

n(W∪D) = n(W) + n(D) - n(W∩D)

= 100 + 29 - 15

= 100 + 14

= 114,

Hence, the probability that the racket is wood or defective,

[tex]P(W\cup D) = \frac{114}{200}[/tex]

[tex]=0.57[/tex]

Answer:

Suppose,

(C - A) ∩ (B - A) ≠ ∅

Let x is an element of (C - A) ∩ (B - A),

That is, x ∈ (C - A) ∩ (B - A),

⇒ x ∈ C - A and x ∈ B - A

⇒ x ∈ C, x ∉ A and x ∈ B, x ∉ A

⇒ x ∈ B ∩ C and x ∉ A

⇒ B ∩ C ⊄ A

But we have given,

B ∩ C ⊂ A

Therefore, our assumption is wrong,

And, there is no common elements in (C - A) and (B-A),

That is,  (C - A) ∩ (B - A) = ∅

Hence proved...

Answer:

The answer is

[tex]W=\frac{S-2LH}{2H+2L}[/tex]

Step-by-step explanation:

The formula for the area of a solid rectangle is

[tex]S = 2HW+2LW+2LH[/tex]

Solve it for W

[tex]2HW+2LW=S-2LH\\\\W(2H+2L)=S-2LH\\\\W=\frac{S-2LH}{2H+2L}[/tex]

Final answer:

To calculate the population in the year 2010 based on exponential growth from 1990, use a growth rate factor and the known population figures from the given years.

Explanation:

Population Growth Calculation:

Determine the growth rate factor from 1990 to 1996: 25 million / 21.7 million = 1.152

Apply the growth rate to find the population in 2010: 21.7 million * (1.152)¹⁴ (14 years from 1996 to 2010) = 48.9 million

Answer: a) 48

b) 43

Step-by-step explanation:

Given : The number of students Professor Date has in his Calculus class = 31

The number of students Professor Date has in his Discrete Mathematics class = 17

(a) If we assume that there are no students who take both classes, then the total number of students Professor Date Has = 31+17=48

(b) If we assume that there are five students who take both classes, then the total number of students Professor Date Has = 31+17-5=43

Answer:

Yes;Each side of triangle PQR is the same length as the corresponding side of triangle STU

Step-by-step explanation:

You can observe  the sides of both triangles to see if this property holds

Lets check the length of AB

A(0,3)   and B(0,-1)

[tex]AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\\\\\AB=\sqrt{(0-0)^2+(-1-3)^2} \\\\\\\\AB=\sqrt{0^2+-4^2} \\\\\\AB=\sqrt{16} =4units[/tex]

Now check the length of the corresponding side DE

D(1,2)  and E(1,-2)

[tex]DE=\sqrt{(1-1)^2+(-2-2)^2} \\\\\\DE=\sqrt{0^2+-4^2} \\\\\\DE=\sqrt{16} =4units[/tex]

The side AB has the same length as side DE.This is also true for the remaining corresponding sides.

Answer:

The cost difference between lowest and highest bid $ 251.20

Step-by-step explanation:

Given,

The average quote for the traffic control subcontractor = $ 1570,

Also, the lowest bid is 4 % under average,

That is, lowest bid = average quote - 4% average quote

= 1570 - 4% of 1570

[tex]=1570-\frac{4\times 1570}{100}[/tex]

[tex]=1570-\frac{6280}{100}[/tex]

[tex]=1570-62.80[/tex]

[tex]=\$1507.2[/tex]

While, the highest bid is 12% above average,

That is, the highest bid = average quote + 12% average quote

= 1570 + 12% of 1570

[tex]=1570+\frac{12\times 1570}{100}[/tex]

[tex]=1570+\frac{18840}{100}[/tex]

[tex]=1570+188.4[/tex]

[tex]=\$1758.4[/tex]

Hence,  the cost difference between lowest and highest bid = $ 1758.4 - $ 1507.2 = $ 251.20

Frequency--

It is the number of times an outcome occurs while performing an experiment some " n " number of times.

Relative frequency--

It is the ratio of the frequency of an outcomes to the total number of times an experiment is been performed.

Here there are TOTAL : 120 responses and three outcomes A , B and C.

The frequency table is given as follows:

Outcome         A           B            C

Frequency       64         23           33

and the Relative frequency table is given as follows:

Outcome                          A              B               C

Relative frequency         64/120     23/120      33/120

i.e. the Relative frequency table is given by:

Outcome                         A              B               C

Relative frequency         0.53         0.19          0.28    

The frequency distribution for the given sample is: A: 64, B: 23, C: 33. The relative frequency distribution is: A: 0.53, B: 0.19, C: 0.28.

To find the frequency distribution, we simply count the number of occurrences of each response. For the given sample of 120 responses, we have:

A: 64 responses

B: 23 responses

C: 33 responses

To find the relative frequency distribution, we divide the frequency of each response by the total number of responses (120). The relative frequencies, rounded to two decimal places, are:

A: 0.53

B: 0.19

C: 0.28

Answer:

[tex]a_{n}=15 + (-1)^n * 5[/tex]

Step-by-step explanation:

First, we notice that the when n is odd, [tex]a_{n}[/tex] = 10. And when n is even, [tex]a_{n}[/tex] = 20.

The average of 10 and 20 is [tex](10+20)/2 = 15[/tex]. So, the distance between 15 and 10 is the same that between 15 and 20.

That distance is 5.  

From 15, we need to subtract 5 to get 10 when n is odd and we need to add 5 to get 20 when n is even.

The easiest way to express that oscilation is using [tex](-1)^n[/tex], because it is (-1) when n is odd and 1 when is even. And when multiplied by 5, it will add or subtract 5 as we wanted.