Will mark BRAINLIEST

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Explanation:

Find the prime factorization of each value. A factor tree may help.

10 = 2*5

30 = 2*3*5

45 = 3*3*5

We see that '5' is found in each of the factorizations. This is the only factor that shows up in all three values, and it only shows up one time. So that's why 5 is the GCF.

Answer:

y₂ = 3

Step-by-step explanation:

Calculate the slope m using the slope formula and equate to - \(\frac{3}{4}\)

m = \(\frac{y_{2}-y_{1} }{x_{2}-x_{1} }\)

with (x₁, y₁ ) = (7, - 6) and (x₂, y₂ ) = (- 5, y₂ )

m = \(\frac{y_{2}+6 }{-5-7}\) , that is

\(\frac{y_{2}+6 }{-12}\) = - \(\frac{3}{4}\) ( cross- multiply )

4(y₂ + 6) = 36 ( divide both sides by 4 )

y₂ + 6 = 9 ( subtract 6 from both sides )

y₂ = 3

Sin(pi - theta) has a value of 4/9. sin(pi - theta) is equal to sin for any value of theta (theta). sin(pi - theta) has a value of 4/9.

We may calculate the value of sin(pi - theta) using the trigonometric identity sin(pi - theta) = sin(pi)cos(theta) - cos(pi)sin(theta):

Pi minus theta equals sin (pi)

cos(pi)sin - cos(theta) (theta)

We may simplify the expression to: since sin(pi) = 0 and cos(pi) = -1

0*cos(theta) - (-1)*sin = sin(pi - theta) (theta)

Pi minus theta equals sin (theta)

The value of sin(theta) can now be substituted to obtain:

Theta = 4/9 sin(pi - theta)

Hence, sin(pi - theta) has a value of 4/9.

The value of theta determines the value of sin(pi - theta). In terms of the trigonometric functions of theta, there is a generic formula to get sin(pi - theta): Pi minus theta equals sin (pi)

cos(pi)sin - cos(theta) (theta)

We may simplify the expression to: since sin(pi) = 0 and cos(pi) = -1

0*cos(theta) - (-1)*sin = sin(pi - theta) (theta)

Pi minus theta equals sin (theta)

Hence, sin(pi - theta) is equal to sin for any value of theta (theta).

Hence, sin(pi - theta) has a value of 4/9.

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The probability that a randomly selected black bear weighs more than 375 pounds is approximately 0.0119, or 1.19%.

To find the probability that a randomly selected black bear weighs more than 375 pounds, we can utilize the properties of the Normal distribution.

Given:

Mean (μ) = 250 pounds

Standard deviation (σ) = 55 pounds

We can standardize the weight value of 375 pounds using the Z-score formula:

Z = (X - μ) / σ

Where X is the value we want to standardize, μ is the mean, and σ is the standard deviation.

Plugging in the values:

Z = (375 - 250) / 55

Z = 125 / 55

Z ≈ 2.27

Next, we need to find the probability of a randomly selected bear weighing more than 375 pounds. This corresponds to finding the area under the Normal curve to the right of the Z-score.

Using a standard Normal distribution table or a statistical calculator, we can find the area to the left of the Z-score of 2.27, which is approximately 0.9881.

Since we want the probability of the weight being more than 375 pounds, we subtract the area to the left from 1:

Probability = 1 - 0.9881

Probability ≈ 0.0119

Thus, the likelihood that a black bear chosen at random weighs more than 375 pounds is roughly 0.0119, or 1.19%.

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Answer:

\(Rate = \$0.23\)   and    \(Rate = \$0.094\)

Step-by-step explanation:

Given

Sew-on logos

Cost = $2.30

Units = 10

Stick-on logos

Cost = $9.40

Units = 100

Required

Divide cost by units, in both cases

When we divide cost by units, it gives the unit rate.

i.e.

\(Rate = \frac{Cost}{Units}\)

FOR SEW-ON LOGOS

Substitute $2.30 for Cost and 10 for Units

\(Rate = \frac{\$2.30}{10}\)

\(Rate = \$0.23\)

FOR STICK-ON LOGOS

Substitute $9.40 for Cost and 100 for Units

\(Rate = \frac{\$9.40}{100}\)

\(Rate = \$0.094\)

Hence, the solution is

\(Rate = \$0.23\)   and    \(Rate = \$0.094\)

the answers are: A) P(x) = -0.001x^4 + 6.8x - 60 B) R(150) = 1050, C(150) = 843.75, P(150) = 815.625.

To find P(x), we need to understand what it represents. P(x) is the profit function and is calculated by subtracting the cost function from the revenue function. Therefore, P(x) = R(x) - C(x).
Substituting the given values of R(x) and C(x), we get:
P(x) = 7x - (0.001x^4 + 1.2x + 60)
Simplifying further, we get:
P(x) = -0.001x^4 + 6.8x - 60

To find R(150), we simply substitute 150 for x in the given function of R(x), giving us:
R(150) = 7(150)
R(150) = 1050

To find C(150), we substitute 150 for x in the given function of C(x), giving us:
C(150) = 0.001(150)^4 + 1.2(150) + 60
C(150) = 843.75

To find P(150), we substitute 150 for x in the derived function of P(x), giving us:
P(150) = -0.001(150)^4 + 6.8(150) - 60
P(150) = 815.625

Therefore, the answers are:
A) P(x) = -0.001x^4 + 6.8x - 60
B) R(150) = 1050, C(150) = 843.75, P(150) = 815.625.

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The rise and run on the graph is, 5

And, The distance increase 5 meters in 1 second

We have to given that;

A graph is shown in figure.

Now, Take two points on line are (4, 20) and (8, 40)

Hence, The rate of change is,

⇒ m = (40 - 20) / (8 - 4)

⇒ m = 20/4

⇒ m = 5/1

Thus, the rise and run on the graph is, 5

And, The distance increase 5 meters in 1 second

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Answer:

Answer:

lol

Step-by-step explanation:

Using combination, number of ways in which two rotten apples be chosen is f) None of the above.

To determine the number of ways two rotten apples can be chosen from a sample of three apples, we can use the concept of combinations.

The number of ways to choose two rotten apples from a set of two rotten apples is given by the combination formula:

C(n, k) = n! / (k! * (n - k)!)

Where n is the total number of objects and k is the number of objects to be chosen.

In this case, n = 2 (two rotten apples) and k = 2 (two rotten apples to be chosen). Plugging these values into the formula:

C(2, 2) = 2! / (2! * (2 - 2)!) = 2! / (2! * 0!) = 1

Therefore, there is only 1 way to choose two rotten apples from the sample of three apples.

Among the given options, none of them match the correct answer of 1. So, the correct option is "f. None of the above."

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Answer:

3000cm3

Step-by-step explanation:

(a) Length =10cm, breadth =8cm, height =3cm

Volume of cuboid =lbh=10×8×3=240cm  

3

[V=240cm  3 ]

(b) Length =1.5m, breadth =25cm, height =15cm

[1m=100cm]

Volume of cuboid =lbh=150×25×15=56250

[V=56,250cm3 ]

(c) Length =15cm, breadth=2.5dm, height =8cm

[1dm]=10cm

Volume of cuboid =lbh=15×25×8

[V=3000cm3 ].

Step-by-step explanation:

The probability that you will walk at least 2 dogs this week is 0.74

The probability that you will walk at most 2 dogs this week is 0.41

The probability that you will walk 3 dogs this week is 0.23

The probability that the number of dogs you will walk this week is not less than 5 is 0.18

The expected number of dogs you will walk this week is 2.83

The expected value of x is 11.29

The variance of dogs you will walk this week is 3.2811

The standard deviation of dogs you will walk this week is 1.8113

X: Random variable denoting how many dogs you walk this week.

Table attached at the end of solution

a) Probability that you will walk at least 2 dogs this week

\(P(x \geqslant 2) & =1-P(x < 2) \\\)

= 1 - [P(x = 0) + P(x = 1)]

= 1 - [0.14 + 0.12]

= 1 - 0.26

\(P(x \geqslant 2) &\) = 0.74

b) What is the probability that you will walk at most 2 dogs this week

P(x < 2) = P(x = 0) + P(x = 1) + P(x = 2)

= 0.14 + 0.12 + 0.15

P(x < 2) = 0.41

c) Probability that you will walk 3 dogs this week

P(x = 3) = 0.23

d) Probability that the number of dogs you will walk this week is not less than 5

\(P(x \geqslant 5) &\) = P(x = 6) + P(x = 7) + P(x = 5)

= 0.08 + 0.01 + 0.09

= 0.09 + 0.09

= 0.18

e) Expected number of dogs you will want this week

F(x) = \(\sum_i p_i x_i\)

E(x) = 0 × 0.14 + 1 × 0.12 + 2 × 0.15 + 3 × 0.23 + 4 × 0.18 + 5 × 0.09 + 6 × 0.08 + 7 × 0.01

E(x) = 2.83

f) Expected value of \($x^2$\)

\(E\left(x^2\right) & =\sum_1 p_i x_i^2 \\\)

= 0 × 0.14 + 1 × 0.12 + 4 × 0.15 + 9 × 0.23 + 16 × 0.18 + 25 × 0.09 + 36 × 0.08 + 49 × 0.01

\(E\left(x^2\right) &\) = 11.29

g) Variance of dogs you will walk this week

\(V(x) & =E\left(x^2\right)-[E(x)]^2 \\\)

\(& =11.29-(2.83)^2 \\\)

= 3.2811

h) Standard deviation of dogs you will walk this week

SD (x) = \(\sqrt{3.2811} \\\)

= 1.8113

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You start a dog-walking business. Define X to be a random variable denoting how many dogs you walk

this week. The probability mass function (pmf), f(x), of X is defined as follows:

Table attached at end of the solution

(a) What is the probability that you will walk at least 2 dogs this week?

(b) What is the probability that you will walk at most 2 dogs this week?

(c) What is the probability that you will walk 3 dogs this week?

(d) What is the probability that the number of dogs you will walk this week not be less than 5 ?

(e) What is the expected number of dogs you will walk this week?

(f) What is the expected value of \($X^2$\) ?

(g) What is the variance of dogs you will walk this week?

(h) What is the standard deviation of dogs you will walk this week?

Answer:

I think it's 56 is the answer

a and b must be on opposite sides of zero​ (option D)

Explanation:

if a × b = c

and c is less than zero.

c can be less than zero if a < 0 and b > 0

c can be less than zero if b < 0 and a > 0

From the above, we can say a nd b must be on opposite sides this is because the left hand side of 0 shows > 0. While the right side of 0 shows > 0.

Hence, for a × b = c where c < 0,

a and b must be on opposite sides of zero​ (option D)

The argument presented is an example of a fallacy known as affirming the consequent, which incorrectly assumes that if the consequent of a conditional statement is true, then the antecedent must also be true. It fails to consider other possibilities and overlooks the fact that not all square shapes are cubes.

The argument presented is an example of a fallacy known as affirming the consequent, which is a formal logical fallacy. This fallacy occurs when one assumes that if the consequent (the "then" part) of a conditional statement is true, then the antecedent (the "if" part) must also be true.

In this case, the argument incorrectly assumes that because all cubes are square (if cube, then square) and a dice is a cube, then the dice must be square. However, this reasoning is flawed. While it is true that all cubes are square, not all square shapes are cubes. Therefore, even though a dice is a cube, it does not necessarily mean that it must be square.

The argument fails to consider other possibilities, such as the fact that cubes can have other shapes besides squares, like rectangular cubes. This logical error highlights the importance of careful reasoning and avoiding fallacies when making arguments.

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The change in potential energy of the pair of charges is approximately -3.997 J.

To calculate the change in potential energy of the pair of charges, we can use the formula:

ΔU = U_f - U_i

where ΔU is the change in potential energy, U_f is the final potential energy, and U_i is the initial potential energy.

Since the first charge is stationary, its potential energy does not change. Therefore, the initial potential energy (U_i) is zero.

The potential energy (U_f) of the second charge can be calculated using the formula:

U_f = k * (q1 * q2) / r

where k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2), q1 is the charge of the first particle (+2.40 μC), q2 is the charge of the second particle (-4.30 μC), and r is the distance between the charges.

First, we need to calculate the distance between the two points. Using the distance formula:

r = √[(x2 - x1)^2 + (y2 - y1)^2]

where (x1, y1) are the coordinates of the first point (0.140 m, 0) and (x2, y2) are the coordinates of the second point (0.230 m, 0.230 m).

Plugging in the values:

r = √[(0.230 - 0.140)^2 + (0.230 - 0)^2]

r = √[(0.09)^2 + (0.230)^2]

r = √[0.0081 + 0.0529]

r = √0.061

r ≈ 0.247 m

Now we can calculate the final potential energy (U_f):

U_f = (8.99 x 10^9 N m^2/C^2) * ((2.40 x 10^-6 C) * (-4.30 x 10^-6 C)) / 0.247 m

U_f ≈ -3.997 J (Joules)

Therefore, the change in potential energy (ΔU) is:

ΔU = U_f - U_i

ΔU = -3.997 J - 0 J

ΔU ≈ -3.997 J (Joules)

Now let's calculate the work done by the electric force on q2.

The work done (W) by the electric force can be calculated using the formula:

W = ΔU = U_f - U_i

Since the initial potential energy (U_i) is zero, the work done is equal to the change in potential energy (ΔU):

W ≈ -3.997 J (Joules)

Therefore, the work done by the electric force on q2 is approximately -3.997 J.

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The temperature of the soda after 20 minutes is approximately -18 degrees Celsius. To find the initial temperature of the soda, we can evaluate the function T(x) at x = 0.

Substitute x = 0 into the function T(x):

T(0) = -19 + 39e^(-0.45*0).

Simplify the expression:

T(0) = -19 + 39e^0.

Since e^0 equals 1, the expression simplifies to:

T(0) = -19 + 39.

Calculate the sum:

T(0) = 20.

Therefore, the initial temperature of the soda is 20 degrees Celsius.

To find the temperature of the soda after 20 minutes, we substitute x = 20 into the function T(x):

Substitute x = 20 into the function T(x):

T(20) = -19 + 39e^(-0.45*20).

Simplify the expression:

T(20) = -19 + 39e^(-9).

Use a calculator to evaluate the exponential term:

T(20) = -19 + 39 * 0.00012341.

Calculate the sum:

T(20) ≈ -19 + 0.00480599.

Round the answer to the nearest degree:

T(20) ≈ -19 + 1.

Therefore, the temperature of the soda after 20 minutes is approximately -18 degrees Celsius.

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INCOMPLETE QUESTION

A can of soda is placed inside a cooler. As the soda cools, its temperature Tx in degrees Celsius is given by the following function, where x is the number of minutes since the can was placed in the cooler. T(x)= -19 +39e-0.45x. Find the initial temperature of the soda and its temperature after 20 minutes. Round your answers to the nearest degree as necessary.

Answer:

52

Step-by-step explanation:

f(n) is purely based on the previous value of f(n), or f(n-1), so we can start with f(1) and work our way up. We know f(1) = 2, so to find f(2), we plug f(1) into

f(n-1)²+3 to get

f(1)²+3 = 2²+3 = 4+3=7

Thus, f(2) =7. Similarly,

f(3) = f(3-1)²+3 = f(2)² + 3 -= 7² + 3= 52

False. if you are given a graph with two shif table lines, the correct answer will always require you to move both lines.

In a graph with two shiftable lines, the correct answer may or may not require moving both lines. It depends on the specific scenario and the desired outcome or conditions that need to be met.

When working with shiftable lines, shifting refers to changing the position of the lines on the graph by adjusting their slope or intercept. The purpose of shifting the lines is often to satisfy certain criteria or align them with specific points or patterns on the graph.

In some cases, achieving the desired outcome may only require shifting one of the lines. This can happen when one line already aligns with the desired points or pattern, and the other line can remain fixed. Moving both lines may not be necessary or could result in an undesired configuration.

However, there are also situations where both lines need to be shifted to achieve the desired result. This can occur when the relationship between the lines or the positioning of the lines relative to the graph requires adjustments to both lines.

Ultimately, the key is to carefully analyze the graph, understand the relationship between the lines, and identify the specific criteria or conditions that need to be met. This analysis will guide the decision of whether one or both lines should be shifted to obtain the correct answer.

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The answer as a decimal to the nearest hundredth is -3.45.

For the given question;

The expression is given as;

= -3.7 + 1/4

Now, the fraction 1/4 can be written in decimal as 0.25.

Put the value of 1/4.

=  -3.7 + 0.25

The subtraction of larger negative to smaller positive gives negative number.

= -3.45

Thus, the answer as a decimal to the nearest hundredth is -3.45.

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Check the picture below.

Answer:

y= −1/2x + 3/4

Step-by-step explanation:

Write in slope-intercept form, y=mx +b.

Problem: 6x+12y=9

Hopefully this helps you!

Still stuck? Get 1-on-1 help from an expert tutor now.

Step-by-step explanation:

-1/4, -1/5, -1/6, -1/7, -1/8, 1/8, 1/7, 1/6, 1/5, 1/4

Answer: 20

Step-by-step explanation:

To answer these types of questions, you have to think about what number minus 7 would equal 13. You can find out by simply adding 13+7. Then check your answer.

20-17=13 is accurate

If a random sample of 35 cars are loaded onto the ferry, the probability that the maximum safe weight will be exceeded is 0.014 or 1.4%.

Let X be the weight of a single passenger car. The mean weight of a car is μ = 1.7 tons and the standard deviation is σ = 0.8 tons.

The total weight of 35 cars is:

W = 35X

By the central limit theorem, the distribution of W is approximately normal with mean μ_W = 35μ = 59.5 tons and standard deviation σ_W = sqrt(35)σ = 3.44 tons.

The probability that the maximum safe weight of 67 tons will be exceeded is the same as the probability that W is greater than 67 tons:

P(W > 67) = P((W - μ_W) / σ_W > (67 - μ_W) / σ_W)

= P(Z > (67 - 59.5) / 3.44)

= P(Z > 2.18)

where Z is a standard normal random variable.

Using a standard normal distribution table or calculator, we can find that P(Z > 2.18) = 0.014.

Therefore, the probability that the maximum safe weight of 67 tons will be exceeded when 35 cars are loaded onto the ferry is approximately 0.014 or 1.4%.

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The idea that group work theories are appropriate for all clients all the time is a myth (Third option).

About Group Process:

The term "group process" describes how people collaborate within an organization to complete tasks. Typically, organizations invest a lot of time and effort into defining and achieving goals, but little thought is usually given to how those goals are affecting the most valuable resource of a group - its people.

Since different clients have different demands and priorities, group work might not be suitable for many, but not for all. Different types of clients require different types of collaboration, type of group or individuals, and working strategies that work for their goals the best.

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By using the concept of division we can find the number of students assigned per teacher.

In mathematics, division is the process of dividing a number into equal parts and calculating the maximum number of equal parts that may be formed. It is one of the fundamental operations in mathematics.

Number of teachers=29

Number of students in sixth grade=406

The number of students in assigned for each teacher

=\(\frac{Total number of students}{number of teachers}\)

=\(\frac{406}{29}\)

=14

The number of students assigned per teacher is 14.

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The Scale factor was used to enlarge the picture is 7/ 31.5.

A scale factor is a numerical value that can be used to alter the size of any geometric figure or object in relation to its original size. It is used to find the missing length, area, or volume of an enlarged or reduced figure as well as to draw the enlarged or reduced shape of any given figure. It should be remembered that the scale factor only affects how big a figure is, not how it looks.

Given:

A picture is 7 inches wide and 9 inches long.

A photographer enlarges it so it is 31.5 inches wide and 40.5 inches long.

So, scale factor is

= Original dimension / Enlarged dimension

= 9/ 40.5

= 90/ 405

= 10/45

= 2/9

Now, simplifying the scale factor 7/31.5 because it has Nr < Dr.

= 7/ 3.15

= 70/31.5

= 2/9

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