Which expressions are equivalent to when x0? Check all that apply.

Answers

Answer 1

Answer: we have that

$((x+4))/(3) / (6)/(x) = (x*(x+4))/(3*6) \n \n = (( x^(2) +4x))/(18)$

therefore

case a)
$((x+4))/(3) * (x)/(6)$
Is equivalent

case b)
$(6)/(x) * ((x+4))/(3)$
Is not equivalent

case c)
$(x)/(6) * ((x+4))/(3)$
Is equivalent

case d)
$((2 x^(2) +4x))/(6)$
Is not equivalent

case e)
$((2 x^(2) +4x))/(18)$
Is equivalent

Hence

the answer is

$((x+4))/(3) * (x)/(6)$

$(x)/(6) * ((x+4))/(3)$

$((2 x^(2) +4x))/(18)$

Answer 2

Answer:

The correct representation of the expression are:

$(x+4)/(3)((x)/(6))$

$((x)/(6))(x+4)/(3)$

$(x^2+4x)/(18)$

Step-by-step explanation:

We are given an algebraic expression as:

$(x+4)/(3)$ ÷ $(6)/(x)$

This expression could also be written as:

$=((x+4)/(3))/((6)/(x))$

We know that any expression of the form:

$((a)/(b))/((c)/(d))$ is given by:

$((a)/(b))/((c)/(d))=(a* d)/(b* c)}$

Hence, we get the given expression as:

$=((x+4)/(3))/((6)/(x))=((x+4)* x)/(3* 6)\n\n=(x+4)/(3)* (x)/(6)$

Also, on solving we get:

$=((x+4)/(3))/((6)/(x))=(x^2+4x)/(18)$

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In his study of pea plants, Gregor Mendel used which method to produce offspring?a.cross-pollination, by using parents that had identical traits b.cross-pollination, by using parents that had different traits c.self-pollination, by using one parent d.random pollination, by using both identical- and different-trait matings

The area formed by a plane cutting through a solid perpendicular to its axis is aA.cross section
B.Longitudinal section
C.Cylindrical Section
D.Middle section

Answers

Answer:

cross section

Step-by-step explanation:

The area formed by a plane cutting through a solid perpendicular to its axis is a cross section.

A cross section area is the area obtained from a three dimensional object like sphere when it is sliced at any axis perpendicular at a point. For example cross section of a cylinder sliced parallel to its base is circle.

Domain is the set of all input values, while range is the set of all:

Answers

all output values. This is because the domain of a function is the set of all possible x-values that you will put into the equation while the range is the set of all possible y-values, the solution.

What is the rate of change of the function enter as a fraction

marking brainlist

Answers

Answer:

The rate of function is: 5/2

Step-by-step explanation:

As we know that the rate of change of a function is a ratio that compares the difference between the output values to the difference between the corresponding input values

rate of change = change in y-values / change in y-values

So, all we need is to find the slope between any two points as the slope is basically termed as the rate of change.

$\mathrm{Slope}=(y_2-y_1)/(x_2-x_1)$

$\left(x_1,\:y_1\right)=\left(-5,\:18\right),\:\left(x_2,\:y_2\right)=\left(-3,\:23\right)$

$m=(23-18)/(-3-\left(-5\right))$

$m=(5)/(2)$

Thus, the rate of function is: 5/2

3(2y-4) - 2 ) = 0.4 (1+y)

Answers

6y-12-2=0.4+0.4y
6y-10-0.4=0.4-0.4+0.4y
6y-6y-9.6=0.4y-6y
(-9.6=-5.6y)/5.6
1.7143=y

HELP PLEASE!!!1.) True or False: "The sum of all the probabilities of outcomes of an event is equal to 100.

2.) If I roll a fair die 100 times, about how many times would I roll a 4?
A.) 25 times
B.) 67 times
C.) 17 times
D.) There is not enough information to answer.

Answers

Answer:

1. false

2. 17 times

Step-by-step explanation:

The sum of all the probabilities of outcomes of an event is equal to 1

2. A die has 1,2,3,4,5,6

P(4) = number of 4's/ total numbers

= 1/6

Multiply the number of roll's by the probability of getting a 4

100 *1/6 = 50/3 = 16 2/3

This is approximately 17

Compare the mean and standard deviation of Set A and Set B.Set A: 7, 3, 4, 9, 2
Set B: 5, 8, 7, 6, 4

Answers

Set A: {7, 3, 4, 9, 2}
Finding the Mean of Set A: $\bar{x} = (7 + 3 + 4 + 9 + 2)/(5)$
   $\bar{x} = (25)/(5)$
   $\bar{x} = 5$

Finding the Standard of Set A: $\sigma = \sqrt{\frac{(\bar{x} - x_(1))^(2) + (\bar{x} - x_(2))^(2) + (\bar{x} - x_(3))^(2) + (\bar{x} - x_(4))^(2) + (\bar{x} - x_(5))^(2)}{n}}$
   $\sigma = \sqrt{((5 - 7)^(2) + (5 - 3)^(2) + (5 - 4)^(2) + (5 - 9)^(2) + (5 - 2)^(2))/(5)}$
   $\sigma = \sqrt{((-2)^(2) + (2)^(2) + (1)^(2) + (-4)^(2) + (3)^(2))/(5)}$
   $\sigma = \sqrt{(4 + 4 + 1 + 16 + 9)/(5)}$
   $\sigma = \sqrt{(34)/(5)}$
   $\sigma = √(6.8)$
   $\sigma \approx 2.6$

Finding the Mean of Set B: $\bar{x} = (5 + 8 + 7 + 6 + 4)/(5)$
   $\bar{x} = (30)/(5)$
   $\bar{x} = 6$

Finding the Standard Deviation of Set B: $\sigma = \sqrt{\frac{(\bar{x} - x_(1))^(2) + (bar{x} - x_(2))^(2) + (\bar{x} - x_(3))^(2) + (\bar{x} - x_(4))^(2) + (\bar{x} - x_(5))}{n}}$
$\sigma = \sqrt{((6 - 5)^(2) + (6 - 8)^(2) + (6 - 7)^(2) + (6 - 6)^(2) + (6 - 4)^(2))/(5)}$
$\sigma = \sqrt{((1)^(2) + (-2)^(2) + (-1)^(2) + (0)^(2) + (2)^(2))/(5)}$
$\sigma = \sqrt{(1 + 4 + 1 + 0 + 4)/(5)}$
$\sigma = \sqrt{(10)/(2)}$
$\sigma = √(5)$
$\sigma \approx 2.236$

The mean and standard deviation of Sets A and B are different.

Final answer:

Mean of Set A is 5 and Set B is 6. Standard deviation of Set A is approximately 2.83, and for Set B, it's approximately 1.67. This indicates that values in Set B are generally closer to their mean than values in Set A to their mean.

Explanation:

To compare the mean and standard deviation of Set A and Set B, we first need to calculate these for each set. Mean is the average of the numbers and standard deviation is a measure of the amount of variation or dispersion of a set of values.

First, calculate the mean by adding the numbers in each set and dividing by the total number of values. For Set A, the mean is (7+3+4+9+2)/5 = 5. For Set B, the mean is (5+8+7+6+4)/5 = 6.

The standard deviation is a bit more complex, as it involves subtracting the mean from each value, squaring the result, finding the mean of these squares, and then taking the square root of that mean. For Set A, these steps result in a standard deviation of approximately 2.83. For Set B, these steps result in a standard deviation of approximately 1.67.

In conclusion, Set B has a higher mean and a lower standard deviation compared to Set A which means values in Set B are generally closer to the mean of Set B than values in Set A are to the mean of Set A.

Learn more about Mean and Standard Deviation here:

brainly.com/question/35095365

#SPJ12

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