help pls ahdjsjfjdjcjdj

Answer:

30%

Step-by-step explanation:

75% of 40 is 30

The percentage of total applicants that accepts offers of admission is = 30%

The percentage of acceptance for students that apply for entrance is = 40%

The percentage of admitted applicants that accept the offer= 75%

This means that it is only 75% of 40% of the accepted students actually accept the offer of admission.

That is, 75/100 × 40%

= 3000/100

= 30%

Therefore, the percentage of total applicants that accepts offers of admission is = 30%

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Answer:

\(y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x+2)\)

OR

\(y+2=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)\)

Step-by-step explanation:

Hi there!

Point-slope form: \(y-y_1=m(x-x_1)\) where \((x_1,y_1)\) is a point and \(m\) is the slope

1) Determine the slope

\(m=\frac{\displaystyle y_2-y_1}{\displaystyle x_2-x_2}\) where two given points are \((x_1,y_1)\) and \((x_2,y_2)\)

Plug in the given points (-2, 6) and (3,-2):

\(m=\frac{\displaystyle -2-6}{\displaystyle 3-(-2)}\\\\m=\frac{\displaystyle -8}{\displaystyle 3+2}\\\\m=-\frac{\displaystyle 8}{\displaystyle 5}\)

Therefore, the slope of the line is \(-\frac{\displaystyle 8}{\displaystyle 5}\). Plug this into \(y-y_1=m(x-x_1)\):

\(y-y_1=-\frac{\displaystyle 8}{\displaystyle 5}(x-x_1)\)

2) Plug in a point \((x_1,y_1)\)

\(y-y_1=-\frac{\displaystyle 8}{\displaystyle 5}(x-x_1)\)

We're given two points, (-2, 6) and (3,-2), so there are two ways we can write this equation:

\(y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x-(-2))\\\\y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x+2)\)

OR

\(y-(-2)=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)\\y+2=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)\)

I hope this helps!

Answer:

C

Step-by-step explanation:

2+4=6

6+4=10

and

18+36=54

54+36=90

you just add hope this helps :))

Step-by-step explanation:

your correct I'm pretty sure

Answer:

Its actually the 3rd option

r = 41/2

Answer:

y = r / (k+v)

Step-by-step explanation:

First, you use the distributive property and take out y: y(k+v)=r

Then you divide both sides by (k+v): y(k+v)/(k+v) = r/(k+v)

Finally you get y = r / (k+v)

Hence the correct option is to translate down 2 unit

Transformations can be non-rigid (when the preimage's size or shape is unaltered) or stiff (where the size is changed but the shape remains the same). These are the fundamental guidelines that this concept abides by. It is a straightforward approach to alter 2D forms.

Planar transformations and spaces can be distinguished from one another using the dimensions of the operand sets.

We have the parent function f(X)= x

and transformed function is g(x)= -3f(x+5)-2

the parent function f(x) is translated dawn 2 units and stretched by 5 .

Hence the correct option is to translate down 2 unit

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Answer:

x = 2.35

Step-by-step explanation:

Answer:

2.35 for K12

Step-by-step explanation:

Answer:

which out of those you want answer for? happy to help

Step-by-step explanation:

The operation to the given figure is translation. The correct option is C.

There are four possible transformations of a point-line or geometric figure, each of which has an impact on the object's location and structure.

Additionally, while translation does not change the direction it faces or its size, rotation revolves an object around a fixed point while moving it only left, right, up, or down. A dilation enlarges something while a reflection creates a mirror image.

Picture, after transformation, refers to the object's final structure and location. From the refers to the object's initial condition.

The translation only alters the location of the geometry rather than its size and shape.

As a result, Figure ABC is translated to A'B'C' on the right and bottom sides.

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The image of the figure is attached with the answer below.

Answer:

The walkway should be 2ft wide.

Step-by-step explanation:

Let x be the width of the walkway surrounding the pool.

The area including the pool is 560ft².

Let's solve for the value of x.

The length plus twice the width of the pool is multiplied by the width plus twice the width of the pool. This makes up the whole area.

A = lw

560 = (24 + 2x)(16 + 2x)

560 = 384 + 48x + 32x + 4x²

4x² + 80x + 384 - 560 = 0

4x² + 80x - 176 = 0

Divide the whole equation by 4

x² + 20x - 44 = 0

(x + 22)(x -2) = 0

x = -22; x = 2

Since we are dealing with dimensions, take the one with the positive value.

x = 2ft

Let's check

560ft² = (24ft + 2x)(16ft + 2x)

560ft² = (24ft + 2(2ft)) (16ft + 2(2ft))

560ft² = (24ft + 4ft)(16ft + 4ft)

560ft² = (28ft)(20ft)

560ft² = 560ft² ✔

Answer:

1/2

Step-by-step explanation:

The class width of each class is 17.

Class width is the distance between a class's upper and lower bounds (category). It may also refer to one of the following more precisely, depending on the author:

the difference between the upper and lower limits of two (nearby) consecutive classes, or the lowest and higher limits of two classes.

The width of each class in a frequency distribution table must be the same. Due to the fact that you are only working with a single width, this makes calculating the class width relatively simple (as opposed to varying ones).

To find the class width in this, we subtract the lower limits of two consecutive class. i.e.

47 - 30 = 17

64 - 47 = 17

81 - 64 = 17

98 - 81 = 17

So that, The class width of each class is 17.

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Answer:

The correct answer is true.

Answer:

3

Step-by-step explanation:

Set the denominator in  

3

x

2

−

2

x

−

15

equal to  

0

to find where the expression is undefined.

x

2

−

2

x

−

15

=

0

Solve for  

x

.

Tap for more steps...

x

=

5

,

−

3

The domain is all values of  

x

that make the expression defined.

Interval Notation:

(

−

∞

,

−

3

)

∪

(

−

3

,

5

)

∪

(

5

,

∞

)

Set-Builder Notation:

{

x

|

x

≠

−

3

,

5

}

Answer:

5220

Step-by-step explanation:

Answer:

the equation is y=-x+8

Answer:

6.834 x 10^-6 meters

Step-by-step explanation:

Answer: 6.834 × 10-6

Step-by-step explanation: To get 0.000006834 we have to times 6.834 by 10 with a negative exponent. Since there are 6 zeros we know it will be 10^-6.

The given system of equations has no solution.

Given that a system of equations, 12x + 36y = −3 and −12x − 36y = 0, we need to solve it,

We know for a system of equations =

a₁x + b₁y = c₁ and a₂x + b₂y = c₂

If a₁ / a₂ = b₁ / b₂ = c₁ / c₂ then the system has infinite solutions.

If a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂ then the system has no solutions.

If a₁ / a₂ ≠ b₁ / b₂ then the system has a unique solution.

comparing the coefficients given,

-12/12 = -1

-36/36 = -1

-3/0 = -3

We get here the condition of a₁ / a₂ ≠ b₁ / b₂, which means that the system has no solution.

Hence the given system of equations has no solution.

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The area of the figure in this problem is given as follows:

A. 30 in².

The area of a triangle is given by half the multiplication of the base length by the height of the triangle.

Applying the Pythagorean Theorem, we can obtain the missing side length, as follows:

4² + x² = 6²

x² = 36 - 16

x² = 20

x = square root of 20

x = 4.47.

For a right triangle, one side is considered the height of the other, hence the sides are:

Hence the area of the triangle is then obtained as follows:

Area = 0.5 x 4 x 12.47

Area = 25 in².

Which is closest to 30 in².

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Answer:

102

Step-by-step explanation:

25 x 3 - 5 + 32

75 - 5 + 32

70 + 32

=102

Answer:

102

Step-by-step explanation:

25(3)-5+32

75-5+32

70+32

102

Answer:

120.75 miles apart

Step-by-step explanation:

The statements b and d are incorrect to express the fact that a function f is onto.

A function is a relation from  A  to  B  with the condition that for every element in the domain, there exists a unique image in the codomain (this is really two conditions: existence of an image and uniqueness of an image). We denote it  f(x) , which is pronounced as “  f  of  x .”

b. f is onto every element in its domain has a corresponding image in its co-domain.

This statement is incorrect.

d. f is onto V x E X,3y EY such that f (x)-y e. f is onto the range of f is the same as the co-domain of f.

This statement in incorrect.

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Whats the question?......................

Answer:  

For a;  

The answer will be 3  

For b;  

The answer will be -4  

Step-by-step explanation:  

To find the gradient of the lines, you must pick 2 points on the line and then find  \(dy/dx\)  

I hope this helped!  

Can I have brainliest?  

Have a great day/night/afternoon!  

(I already went to the other question you posted, do not worry.)

- ❤ 7272033Alt ❤

Using the Empirical Rule and the Central Limit Theorem, we have that:

It states that, for a normally distributed random variable:

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation \(s = \frac{\sigma}{\sqrt{n}}\).

In this problem, the standard deviation of the distribution of sample means is:

\(s = \frac{0.657}{\sqrt{50}} = 0.09\)

68% of the means are within 1 standard deviation of the mean, hence the bounds are:

99.7% of the means are within 3 standard deviations of the mean, hence the bounds are:

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Answer:

About 68% of the sample means fall within the interval $1.64 and $1.82.

About 99.7% of the sample means fall within the interval $1.45 and $2.01.

Step-by-step explanation:

To verify the given intervals, we need to calculate the standard error of the mean (SE) for a sample size of 50 using the population mean and standard deviation provided.

The standard error of the mean (SE) can be calculated as:

SE = population standard deviation / √(sample size)

Given that the population mean is $1.73 and the population standard deviation is $0.657, and the sample size is 50:

SE = $0.657 / √50 ≈ $0.09299 (rounded to 5 decimal places)

Now, we can calculate the intervals:

For the interval where about 68% of the sample means fall:

Interval = (Mean - 1 * SE, Mean + 1 * SE)

Interval = ($1.73 - $0.09299, $1.73 + $0.09299)

Interval ≈ ($1.63701, $1.82299)

So, about 68% of the sample means fall within the interval $1.64 and $1.82, which matches the given statement.

For the interval where about 99.7% of the sample means fall:

Interval = (Mean - 3 * SE, Mean + 3 * SE)

Interval = ($1.73 - 3 * $0.09299, $1.73 + 3 * $0.09299)

Interval ≈ ($1.54703, $1.91297)

So, about 99.7% of the sample means fall within the interval $1.55 and $1.91, which is different from the given statement.

The correct interval for about 99.7% of the sample means, rounded to the nearest hundredth, is $1.55 and $1.91, not $1.45 and $2.01 as mentioned in the statement.

Answer:

x = 14

Step-by-step explanation:

<LOJ = 3x

<KOJ = (2x + 12)°

<LOK = 82°

m<LOJ + m<KOJ = m<LOK (angle addition postulate)

3x + 2x + 12 = 82 (substitution)

5x + 12 = 82

5x = 82 - 12 (Subtraction property of equality)

5x = 70

x = 70/5 (division property of equality)

x = 14

Polynomial: 6x^3 + 12x^2 - 18x

Form 1: 6(x^3 + 2x^2 - 3x)

Explanation: This form was created by factoring out the GCF of 6 from each term in the polynomial.

Form 2: 6x(x^2 + 2x - 3)

Explanation: This form was created by factoring out the GCF of 6x from each term in the polynomial.

In both forms, the GCF of 6 allows us to factor out common factors and simplify the polynomial expression while preserving its original meaning.

To create a factorable polynomial with a greatest common factor (GCF) of 6, we can start by considering the common factors of the terms. Let's construct a polynomial:

Polynomial: 6x^3 + 12x^2 - 18x

To rewrite this polynomial in two other equivalent forms, we can factor out the GCF from each term and simplify.

Form 1: 6(x^3 + 2x^2 - 3x)

In this form, we factored out the GCF of 6 from each term: 6x^3/6 = x^3, 12x^2/6 = 2x^2, and -18x/6 = -3x.

Form 2: 6x(x^2 + 2x - 3)

In this form, we factored out the GCF of 6x from each term: 6x^3/6x = x^2, 12x^2/6x = 2x, and -18x/6x = -3.

We created each form by identifying the common factors in the polynomial and factoring them out. This process allows us to rewrite the polynomial in equivalent forms that emphasize different aspects.

In Form 1, we factored out the GCF of 6, leaving the remaining polynomial inside the parentheses. This form highlights the fact that the polynomial can be written as the product of the GCF and the remaining terms.

In Form 2, we factored out the GCF of 6x, again leaving the remaining polynomial inside the parentheses. This form emphasizes the fact that the polynomial can be written as the product of the GCF and another factor, in this case, x.

By rewriting the polynomial in these two forms, we have created equivalent expressions that emphasize the presence of the GCF and demonstrate the polynomial's factorability.

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