PLEASEEEE help!!!! asap!!! thank u so much *will give brainliest*

Answer: \(y-1=\frac{1}{3} (x-6)\)

Point slope form is: \(y-y_{1} = m (x-x_{1} )\), where m=slope and y1 & x1 is the point given. So plug the slope 1/3 and point (6,1) into the spots to make it into point slope form.

For the first part, the number of license plates is: 65,780,800 and for the second part, the number of license plates with 3 consecutive digits is: 16,524,288.

If there are no restrictions on where the digits and letters are placed, the number of 8-place license plates consisting of 5 letters and 3 digits with no repetitions allowed can be found using the permutation formula:

nPr = n! / (n-r)!

where n is the number of available characters (26 letters and 10 digits) and r is the number of characters needed for the license plate (8).

Therefore, the number of license plates is:

(26 P 5) x (10 P 3) = (26!/21!) x (10!/7!) = 65,780,800

If the 3 digits must be consecutive, there are 8 possible positions for the block of digits (either the first 3, second 3, or last 3). Once the position of the block is chosen, the number of license plates can be found by counting the number of ways to arrange the letters and the block of digits. The number of ways to arrange the letters is 26 P 5, and the number of ways to arrange the block of digits is 10 (since there are only 10 possible sets of consecutive digits).

Therefore, the number of license plates with 3 consecutive digits is:

8 x (26 P 5) x 10 = 16,524,288

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The complete question goes thus:

If there are no restrictions on where the digits and letters are placed, how many 8

-place license plates consisting of 5

letters and 3

digits are possible if no repetitions of letters or digits are allowed? What if the 3

digits must be consecutive?

The 20 rational numbers between -3/7 and 2/3 are: -0.42857, -0.14286, 0.14286, 0.42857, 0.71429, -0.28571, -0.07143, 0.07143, 0.28571, 0.57143, -0.61538, -0.38462, -0.15385, 0.15385, 0.38462, 0.61538, -0.5, -0.25, 0.25, 0.5.

Rational numbers are numbers that can be expressed as a ratio of two integers, where the denominator is not equal to zero. In other words, any number that can be written in the form of a fraction a/b, where a and b are integers and b is not equal to zero, is a rational number.

To find 20 rational numbers between -3/7 and 2/3, we need to find the difference between 2/3 and -3/7, which is 65/21. Then we divide 65/21 by 21 to get the increment value of 3/7. Starting with -3/7, we add 3/7 to it 20 times to get the 20 rational numbers between -3/7 and 2/3.

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Answer:

scalene triangle for the first

equilateral triangle for the second

isosceles triangle for the 3rd

Answer:

The first is a scalene triangle. No two sides on a scalene trinagle will have the same length.

The second is an equilateral triangle. Equilateral literally means "equal sides".

The third is an isosceles triangle. These triangle will have perfect symmetry.

The magnitude of the vector of 15 units and the vector direction of 210° indicates that the horizontal and vertical components of the vector are;

(-12.99, -7.5)

The horizontal and vertical component of the vector can be found from the following formula;

Horizontal component = Magnitude × cos(direction angle)

Vertical component = Magnitude × sin(direction angle)

The magnitude of the vector = 15

The direction (angle) of the vector = 210°

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The expected value of a single spin would be number 8 because it the the most occuring number of the spinner.

To determine the expected value, the most occuring value should be identified which is equally the mode of the data set.

Of all the numbers in the spinner, the most occuring data is number 8 which appeared 4 times.

Therefore, the expected value as a spin should be number 8.

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The solution of the original equation with the given initial conditions is:

x(t) = ∫-e^(-t^2 - √2t)/2 dt * e^(√2t) + ∫e^(-t^2 + √2t)/2 dt * e^(-√2t)

The solution to the differential equation x'' - 2x = e^(-t^2) with the initial conditions x(0) = 0 and x'(0) = 0 can be found by using the method of variation of parameters. First, we need to find the general solution of the homogeneous equation x'' - 2x = 0. The characteristic equation is r^2 - 2 = 0, which has two real and distinct roots r1 = √2 and r2 = -√2. Therefore, the general solution of the homogeneous equation is xh(t) = c1e^(√2t) + c2e^(-√2t).

Now, we need to find a particular solution xp(t) of the non-homogeneous equation. We can use the method of variation of parameters and assume that xp(t) = u1(t)e^(√2t) + u2(t)e^(-√2t). Taking the first and second derivatives of xp(t) and substituting them into the original equation, we get:

u1'(t)e^(√2t) + u2'(t)e^(-√2t) = 0
u1''(t)e^(√2t) - 2u1'(t)e^(√2t) + u2''(t)e^(-√2t) + 2u2'(t)e^(-√2t) = e^(-t^2)

Solving this system of equations for u1'(t) and u2'(t), we get:

u1'(t) = -e^(-t^2 - √2t)/2
u2'(t) = e^(-t^2 + √2t)/2

Integrating both sides of these equations, we get:

u1(t) = ∫-e^(-t^2 - √2t)/2 dt
u2(t) = ∫e^(-t^2 + √2t)/2 dt

Substituting these expressions back into the formula for xp(t), we get:

xp(t) = ∫-e^(-t^2 - √2t)/2 dt * e^(√2t) + ∫e^(-t^2 + √2t)/2 dt * e^(-√2t)

Finally, the general solution of the original equation is x(t) = xh(t) + xp(t) = c1e^(√2t) + c2e^(-√2t) + ∫-e^(-t^2 - √2t)/2 dt * e^(√2t) + ∫e^(-t^2 + √2t)/2 dt * e^(-√2t).

Using the initial conditions x(0) = 0 and x'(0) = 0, we can find the values of c1 and c2:

0 = c1 + c2
0 = √2c1 - √2c2

Solving this system of equations, we get c1 = c2 = 0. Therefore, the solution of the original equation with the given initial conditions is:

x(t) = ∫-e^(-t^2 - √2t)/2 dt * e^(√2t) + ∫e^(-t^2 + √2t)/2 dt * e^(-√2t)

This is the solution to the differential equation x'' - 2x = e^(-t^2) with the initial conditions x(0) = 0 and x'(0) = 0 as a definite integral.

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Answer:

Step-by-step explanation:

A set of factors of a number (N) is a set of integers {f1, f2, f3, ...} whose product is the number:

  f1 × f2 × f3 × ... = N

Often the term "factors" is used to mean "prime factors," the set of prime numbers whose product is N.

When N = 25, the prime factors are {5, 5}. That is ...

  5×5 = 25

__

A "divisor" of a number is a sub-multiple of N. That is, the quotient N/k is an integer for some divisor k of N. Usually, we're interested in integer divisors. All prime factors will be integer divisors of N. The term "factor" is often used when the term "divisor" is meant.

Divisors of N = 25 include 1, 5, and 25. There will be an odd number of integer divisors (as here) if the number N is a perfect square.

__

For an integer k, the value k×N is called a multiple of N, often, the k-th multiple of N.

Multiples of 25 include 25, 50, 75, 100, 125, and any other decimal number ending in 25, 50, 75, or 00.

__

Another example

When N=30, the prime factors are {2, 3, 5}. The divisors are {1, 2, 3, 5, 6, 10, 15, 30}. (Note there are an even number of divisors.) Multiples of 30 include 30, 60, 90, 120, ....

\(19+1,5(10m-8)=127\\\\19+15m-12=127\\\\15m+7=127 \ \ /-7\\\\15m=120 \ \ /:15\\\\\huge\boxed{m=8}\)

Answer:

\(m=8\)

Step-by-step explanation:

\(19 + 1.5(10m-8) = 127\)

Multiply both sides by 10

\(190+15(10m-8)=1270\)

Subtract 190 from both sides

\(15(10m-8)=1080\)

Divide both sides by 15

\(10m-8=72\)

Add 8 to both sides

\(10m=80\)

Simplify

\(m=8\)

To decrease the width of the confidence interval, the researcher can take the following steps:

1. Decrease the confidence level: The confidence interval width is inversely proportional to the confidence level. By decreasing the confidence level, the researcher can have a narrower interval. However, it is important to note that decreasing the confidence level also increases the chance of the interval not capturing the true population mean.

2. Increase the sample size: The sample size affects the precision of the estimate. Increasing the sample size reduces the standard error, which leads to a narrower confidence interval. This is because a larger sample provides more information about the population.

Therefore, the researcher can decrease the width of the confidence interval by either decreasing the confidence level or increasing the sample size. Both approaches will result in a narrower interval, providing a more precise estimate of the mean fat content of hen's eggs.

The researcher can decrease the width of the confidence interval by either decreasing the confidence level or increasing the sample size. Both approaches will result in a more precise estimate of the mean fat content of hen's eggs.

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Rounded to two decimal places, the percentage increase per year in the winner's check over this period is approximately 15,332.33%.

To calculate the percentage increase per year in the winner's check over the period from 1895 to 2016, we can use the following formula:

Percentage increase = [(Final value - Initial value) / Initial value] * 100

Initial value: $150

Final value: $2,300,000

Number of years: 2016 - 1895 = 121 years

Percentage increase = [(2,300,000 - 150) / 150] * 100

= (2,299,850 / 150) * 100

= 15,332.33%

Rounded to two decimal places, the percentage increase per year in the winner's check over this period is approximately 15,332.33%.

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Answer:

The answer should be 12.5%

Answer:

c ik this because I took the same question as you

The product of the given algebraic expression is \(\frac{4}{(x-1)}\).

The given expression is \(\frac{8x+8}{x^2-2x+1} .\frac{x-1}{2x+2}\).

Multiplication is an operation that represents the basic idea of repeated addition of the same number. The numbers that are multiplied are called the factors and the result that is obtained after the multiplication of two or more numbers is known as the product of those numbers.

Here, \(\frac{4(2x+2)}{x^2-1x-1x+1} \times \frac{x-1}{2x+2}\)

= \(\frac{4(2x+2)}{x(x-1)-1(x-1)} \times \frac{x-1}{2x+2}\)

= \(\frac{4(2x+2)}{(x-1)(x-1)} \times \frac{x-1}{2x+2}\)

Cancel the same factors, that is

= \(\frac{4}{(x-1)} \times \frac{1}{1}\)

= \(\frac{4}{(x-1)}\)

Therefore, the product of the given algebraic expression is \(\frac{4}{(x-1)}\).

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Equations can have zero, one or more solutions

The solution to the equation is x = -0.5

The equation is given as:

\(6.8x+9.3 = -9.4+3.4(2-5x)\)

Distribute the expression

\(6.8x+9.3 = -9.4+3.4\times 2-3.4\times 5x\)

\(6.8x+9.3 = -9.4+6.8-17x\)

Apply addition properties

\(6.8x+9.3 = -2.6-17x\)

Combine terms

\(6.8x+17x =-9.3 -2.6\)

Apply addition properties

\(23.8x =-11.9\)

Apply division properties

\(x =-0.5\)

Hence, the solution to the equation is -0.5

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Answer:

Step-by-step explanation:

(a) To find the probability that it is raining when you arrive in Oneirabad on a random day, we need to use the law of total probability.

Let A be the event that it is raining, and B be the event that it is the Wet season.

P(A) = P(A|B)P(B) + P(A|B')P(B')

Given that the Wet season lasts for 1/3 of the year, we have P(B) = 1/3. The probability that it is raining during the Wet season is 3/4, so P(A|B) = 3/4.

The Dry season lasts for 2/3 of the year, so P(B') = 2/3. The probability that it is raining during the Dry season is 1/6, so P(A|B') = 1/6.

Now we can calculate the probability that it is raining when you arrive:

P(A) = (3/4)(1/3) + (1/6)(2/3)

= 1/4 + 1/9

= 9/36 + 4/36

= 13/36

Therefore, the probability that it is raining when you arrive in Oneirabad on a random day is 13/36.

(b) Given that it is raining when you arrive, we can use Bayes' theorem to calculate the probability that your visit is during the Wet season.

Let C be the event that your visit is during the Wet season.

P(C|A) = (P(A|C)P(C)) / P(A)

We already know that P(A) = 13/36. The probability that it is raining during the Wet season is 3/4, so P(A|C) = 3/4. The Wet season lasts for 1/3 of the year, so P(C) = 1/3.

Now we can calculate the probability that your visit is during the Wet season:

P(C|A) = (3/4)(1/3) / (13/36)

= 1/4 / (13/36)

= 9/52

Therefore, given that it is raining when you arrive, the probability that your visit is during the Wet season is 9/52.

(c) Given that it is raining when you arrive, the probability that it will be raining when you return to Oneirabad in a year's time depends on the season. If you arrived during the Wet season, the probability of rain will be different from if you arrived during the Dry season.

Let D be the event that it is raining when you return.

If you arrived during the Wet season, the probability of rain when you return is the same as the probability of rain during the Wet season, which is 3/4.

If you arrived during the Dry season, the probability of rain when you return is the same as the probability of rain during the Dry season, which is 1/6.

Since the season you arrived in is independent of the weather when you return, we need to consider the probabilities based on the season you arrived.

Let C' be the event that your visit is during the Dry season.

P(D) = P(D|C)P(C) + P(D|C')P(C')

Since P(C) = 1/3 and P(C') = 2/3, we can calculate:

P(D) = (3/4)(1/3) + (1/6)(2/3)

= 1/4 + 1/9

= 9/36 + 4/36

= 13/36

Therefore, the probability that it will be raining when you return to Oneirabad in a year's time, given that it is raining when you arrive, is 13/36.

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The boxplot required to answer the questions is attached below :

Answer:

The range for the Kentucky temperature :

(58.3 - 52.2) = 6.1

The range for the Illinois temperature :

(55 - 49.9) = 5.1

The IQR for the Kentucky temperature :

(57.3 - 54.5) = 2.8

The IQR for the Illinois temperature :

(53.3 - 50.9) = 2.4

Step-by-step explanation:

Range = maximum - minimum

Maximum and minimum values are given by the values at the end and start of the whisker.

The range for the Kentucky temperature :

(58.3 - 52.2) = 6.1

The range for the Illinois temperature :

(55 - 49.9) = 5.1

IQR = Q3 - Q1

Q3 = Value at the end of the box

Q1 = value of start of box

The IQR for the Kentucky temperature :

(57.3 - 54.5) = 2.8

The IQR for the Illinois temperature :

(53.3 - 50.9) = 2.4

Answer:

The range for the Kentucky temperatures is ✔ 6.1

The range for the Illinois temperatures is ✔ 5.1

The IQR for the Kentucky temperatures is ✔ 2.8

The IQR for the Illinois temperatures is ✔ 2.4

The box plots show ✔ more variation for Kentucky

Step-by-step explanation:

I just did it and got it correct.  

Answer:

2 sides are \(\frac{33}{13}\) or 2.5

2 sides are \(\frac{47}{13}\) or 3.6

Step-by-step explanation:

There must be 2   2x+3s  and 2   3x+2s

The perimeter is the sum of the sides (it has 4 sides)

2(2x+3) + 2(3x+2) = 4(4x-5)

distributive property

2*2x + 2*3

4x+6

other one

2*3x + 2*2

6x+4

perimeter

4*4x + 4*-5

16x-20

together...

(4x+6) + (6x+4) = 16x-20

add like terms

10x+10 = 16x-20

add 20 to both sides

10x+30 = 16x

subtract 10x from both sides

30 = 26x

divide by 26 on both sides

\(\frac{30}{26} = x\)

simplify

\(\frac{15}{13}\)

Substitute the x value

2 of the sides are 2x+3

so

\(2(\frac{15}{13} ) + 3\)

\(\frac{30}{13} + \frac{3}{1}\)

\(\frac{33}{13}\)

or 2.5

other sides

\(3(\frac{15}{13} ) + 2\)

\(\frac{45}{13} + \frac{2}{1}\)

\(\frac{47}{13}\) or 3.6

That's all I can say for now! Hope this helps and a thank won't hurt! :)

\(\huge \boxed{Answer}\)

there are two containers in four parts has to be filled in each, therefore, the fraction greater than one will be =

total number of part stobe filler in container

_________________________________________

number of parts in each container

= 8/4

the whole number greater than one will be = 2

hence, the whole number and a fraction greater than one to name the part filled is 2 and 8/4

Answer: The cost for the two pairs of shoes is $13.33

Step-by-step explanation:

Since we have given that

Cost of pair of flip flops = $3.46

Cost of pair of slip on shoes = $9.87

We need to find the cost of two pairs of shoes.

Cost of two pairs of shoes = Cost of pair of flip flops + Cost of pair of slip on shoes

\(cost=3.46+9.87=13.33\)

the cost for the two pairs of shoes is $13.33.

The coordinates are simple. Hope you understand this and you can draw it out.

The size of the quarterly deposits in Shelby's RRSP account was approximately $147.40.

Let's denote the size of the quarterly deposits as \(D\). The total number of deposits made over 9 years is \(9 \times 4 = 36\) since there are 4 quarters in a year. The interest rate per period is \(r = \frac{3.50}{100 \times 12} = 0.0029167\) (3.50% annual rate compounded monthly).

Using the formula for the future value of an ordinary annuity, we can calculate the accumulated value of the RRSP fund:

\[55,000 = D \times \left(\frac{{(1 + r)^{36} - 1}}{r}\right)\]

Simplifying the equation and solving for \(D\), we find:

\[D = \frac{55,000 \times r}{(1 + r)^{36} - 1}\]

Substituting the values into the formula, we get:

\[D = \frac{55,000 \times 0.0029167}{(1 + 0.0029167)^{36} - 1} \approx 147.40\]

Therefore, the size of the quarterly deposits, rounded to the nearest cent, is approximately $147.40.

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Answer:

40%

Step-by-step explanation:

Answer: 50, 60, Less.

Step-by-step explanation: The theoretical probability of the coin landing heads up is 50%

Based on​ Liz's results, the experimental probability of the coin landing heads up is 60

The theoretical probability is less than the experimental probability in this experiment.

Hope this helped have a nice day!

Answer:

196 miles

Step-by-step explanation:

distance (D) is calculated as

D = S × T ( S is average speed and T is time in hours )

here T = 3 and a half hours = 3.5 hours and S = 56 , then

D = 56 × 3.5 = 196 miles

Answer:

\(→ \frac{8 {a}^{( - 5) } {b}^{(4)} }{12 {a}^{( - 6)} {b}^{( - 2)} } = \frac{2 {a}^{(6 - 5)}{b}^{(4 + 2)} }{3} = \boxed{ \frac{2}{3}a {b}^{6} }✓ \\ \)

If the second container has a height of 30 m, the second container can hold 300 m³ of liquid.

Since the two containers have exactly the same shape, their volumes are proportional to the cubes of their corresponding dimensions. Let's denote the volume of the second container as V₂ and its height as h₂. Then we have:

(V₂ / V₁) = (h₂ / h₁)³

where V₁ and h₁ are the volume and height of the first container, respectively. Substituting the given values, we get:

(V₂ / 48) = (30 / 12)³

(V₂ / 48) = 2.5³

V₂ = 48 × 2.5³

V₂ = 300 m³

Therefore, the second container can hold 300 m³ of liquid.

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Answer:

A=17

Step-by-step explanation:

Answer:

If there is no 48 divided by 8 then it is 8 x _ = 48.

Step-by-step explanation:

Think about it. There are 6 times more children than adults *6 x 8 = 48.

Answer:

8 x______=48

Step-by-step explanation:

In the problem, they are telling you there are 48 children and 8 adults at the library. Then, they tell you how many times more adults than children are at the library. You aren't adding, and since they are saying times more, that is a sign of multiplication. Hope this helps!

Theorems are statements or propositions in mathematics that have been proven to be true based on logical reasoning or mathematical proof.

There are some general properties and theorems related to angle bisectors that can be considered:

Angle Bisector Theorem: The angle bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent sides. In other words, if AD is the angle bisector of ∠BAC, then BD/DC = AB/AC.

Concurrency of Angle Bisectors: In any triangle, the angle bisectors of the three angles are concurrent, meaning they intersect at a single point called the incenter. This point is equidistant from the sides of the triangle.

Equal Angles: The angle bisector divides an angle into two equal angles. In other words, if AD is the angle bisector of ∠BAC, then ∠BAD = ∠DAC.

Internal and External Angle Bisectors: An angle bisector can be internal or external to the triangle, depending on its position. Internal angle bisectors intersect within the triangle, while external angle bisectors extend outside the triangle.

These are some general statements and properties associated with angle bisectors.

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Answer:

Domain: {-3, -1, 1}

Range: {-1, 2, 4, 5}

Step-by-step explanation:

The domain is all the x-values, and the range is all the y-values. When listing values, repeated values are only written down once.

Answer:

12 pi cubic cm

Step-by-step explanation: