A local high school has 1,203 students with 367 freshmen, 382 sophomores, 249 juniors, and 205 seniors. What is the relative frequency of students who are not
seniors at this high school?
The relative frequency is %.
(Type an integer or decimal rounded to the nearest tenth as needed.)

Answer:

A = { x:x >3 ,where x --> R}

B = { x:x <4, where x --> R}

Y = { x:x is month of year with more than 30 days}

W = { x:x is all days of a week}

Answer:

15 degrees

Step-by-step explanation:

Answer:

The earth turns 15 degrees in one hour. Hope this helps!

Step-by-step explanation:

Answer:

6/x2

Step-by-step explanation:

Answer:

[tex]\mathsf {\frac{6}{x^{2}}}[/tex]

Step-by-step explanation:

[tex]\mathsf {\frac{2}{x^{2}} + \frac{4}{x^{2}} }[/tex]

[tex]\mathsf {\frac{2+4}{x^{2}}}[/tex]

[tex]\mathsf {\frac{6}{x^{2}}}[/tex]

The correct answer are as follows:

A. False

B. True

C. False

D. False

The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input x.

A. Since x-intercepts indicate the amount of each bacteria at the start of the experiment, there was more of bacteria B than bacteria A at the start.

False, it is the y-intercept of the function that indicates the amount at the start of the experiment.

B. Since y-intercepts indicate the amount of each bacteria at the start of the experiment, there was more of bacteria B than bacteria A at the start.

True, the y-intercept is given when x = 0, indicating the initial value of the function.

C. Since the maximum value in the table for bacteria A is greater than the maximum value in the table for bacteria B, bacteria A has a faster growth rate than bacteria B.

False, because the maximum value of each table is given in different times, and also the initial value of each table is different.

D. Since the minimum value in the table for bacteria A is less than the minimum value in the table for bacteria B, bacteria A has a slower growth rate than bacteria B.

False, the growth rate is not given by the initial value. If we model both tables with an exponential function, the count of bacteria A quadruped in two days, and the count of bacteria B doubled in one day, so they have the same growth rate.

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Answer:

x-intercepts: -5 +/-[tex]\sqrt{21}[/tex]. y-intercept: 4

Step-by-step explanation:

For x-intercepts use the quadratic equation

[tex]x= (-b +/-\sqrt{b^2-4ac} )/2a[/tex]

Fill in your values:

[tex]x= (-10 +/-\sqrt{100-16} )/2[/tex]

And solve

[tex]x=-5 +/- \sqrt{21}[/tex]

For y-intercepts set x=0 and solve. In this case, setting x= 0 gives you y=4.

When h = 0 ,  [f(x + h) - f(x)]/h  = 2x +2 for the given expression.

The missing expression is  f(x) = x² + 2x.

f(x+h)-f(x)/h is called the formula of difference quotient.

When h---> 0 ,

the expression f'(x) = (f(x-h) - f(x)) / h is the equation for tangent to the line

Here the the expression is

f(x) = x² + 2x

Determining f(x + h) by substituting x = x + h on both sides of the given f(x).

Then f(x + h) = (x + h)² + 2(x + h)

= x² + 2xh + h² + 2x + 2h

the difference f(x + h) - f(x).

f(x + h) - f(x) = [x²  + 2xh + h²  + 2x + 2h] - [x² + 2x]

= 2xh + h² + 2h

Divide the difference from h as in the formula of difference coefficient

[f(x + h) - f(x)]/h = (2xh + h² + 2h) / h

= 2x + h + 2

So, when h = 0 ,  [f(x + h) - f(x)]/h  = 2x +2

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Answer:

angle M = angle P ( 90°)

so MN is parallel to pq

MN makes 90° with UT so it is perpendicular

We conclude that the line that passes through (0, 0) and (2, -3) is perpendicular to line l.

Remember that two lines are perpendicular only if the slope of one of the lines is equal to the opposite of the inverse of the slope of the other line.

So, if line l has the slope 2/3.

Then the perpendicular lines have a slope equal to -3/2.

Now, remember that if a line goes through two points (x₁, y₁) and (x₂, y₂), then the slope of that line is:

[tex]a = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

So here we just need to find two points (x₁, y₁) and (x₂, y₂) such that the slope is equal to -3/2.

If we define (x₁, y₁) = (0, 0), then the other point must be:

[tex]a = \frac{y_2 - 0}{x_2 - 0} = -3/2\\\\y_2/x_2 = -3/2[/tex]

Then we can write the other point as (2, -3).

So we conclude that the line that passes through (0, 0) and (2, -3) is perpendicular to line l.

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The length of CD is 7-3=4.

Therefore, if we dilate by a scale factor of 3, the length is 4(3) = 12

Answer:

b) 12

Step-by-step explanation:

the length of the image is 12. i just know. trust me.

The volume of the block after cutting out the smaller cubes is  [tex]\frac{63}{125}[/tex] cubic units.

Given that, Sarah has a solid wooden cube with a length of 4/5 centimetres. From each of its 8 corners, she cuts out a smaller cube with a length of 1/5 centimetre.

We need to find the volume of the block after cutting out the smaller cubes.

The volume of a cube is defined as the total space enclosed by the cube in a three-dimensional space. The formula to find the volume of a cube is a³, where a=edge of a cube.

Now, the volume of a solid wooden cube with a length of 4/5 centimetre

[tex]=(\frac{4}{5} )^{3} =\frac{4}{5} \times \frac{4}{5}\times \frac{4}{5}=\frac{64}{125}[/tex] cubic units.

The volume of a smaller cube with a length of 1/5 centimetre

[tex]=(\frac{1}{5} )^{3} =\frac{1}{5} \times \frac{1}{5}\times \frac{1}{5}=\frac{1}{125}[/tex] cubic units.

The volume of the block after cutting out the smaller cubes[tex]=\frac{64}{125}-\frac{1}{125}=\frac{63}{125}[/tex]

Therefore, the volume of the block after cutting out the smaller cubes is  [tex]\frac{63}{125}[/tex] cubic units.

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Answer:

the answer is 64/125 <3

Differentiate using the Quotient Rule –

[tex]\qquad[/tex][tex]\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\[/tex]

According to the given question, we have –

Let's solve it!

[tex]\qquad[/tex][tex]\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\[/tex]

[tex]\qquad[/tex][tex]\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\[/tex]

[tex]\qquad[/tex][tex]\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\[/tex]

[tex]\qquad[/tex][tex] \pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\[/tex]

[tex]\qquad[/tex][tex]\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\[/tex]

[tex]\qquad[/tex][tex]\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\[/tex]

[tex]\qquad[/tex][tex]\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\[/tex]

[tex]\qquad[/tex][tex]\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\[/tex]

Answer:

(1,4)

Step-by-step explanation:

Answer:

the answer is L

Step-by-step explanation:

rounding up to any number/decimal requires the digit before the specified digit to be 5 or greater.
rounding down requires the digit to be 4 or less.
if the estimated time to the nearest 0.01 of a second is 35.20 then the actual race time could be from 35.195 to 35.205 since any number between could be rounded to 35.20

The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.

To solve the problem, we must:

Know how many 30-minutes have passed during the time period.

1 day = 24 hours

1 hour = 60 minutes = 2 × (30 minutes)

1 day = 24 hours × 2 × (30 minutes)

1 day = 48 × (30 minutes)

Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.

Friday = 48 × [tex]\frac{1}{2}[/tex] = 24

Saturday = 48

Sunday = 48

Monday = 48 × [tex]\frac{3}{4}[/tex] = 36

TOTAL = 24 + 48 + 48 + 36 = 156

Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.

Divide the number of total seconds late by the number of 30-minutes passed.

That is, the number of total seconds late= 468 seconds ÷ 156

= 3 seconds  

Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

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Answer:

3x²-2x+1=?

3(4²)-2(4)+1=?

3(16)-8+1=?

48-8+1=?

40+1=41

The function g(x) is obtained by the transformation of the function f(x) is translated towards left by 1 unit.

A statement, principle, or policy that creates the link between two variables is known as a function.

The function f(x) is given below.

f(x) = 2ˣ

Then the function g(x) is given as

[tex]\rm g(x) = 2^{x + 1}[/tex]

The function g(x) is obtained by the transformation of the function f(x) is translated towards left by 1 unit.

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Answer:

216

Step-by-step explanation:

4x+8yz+3y

Let x=3 y=4 z=6

Substitute the values into the expression

4(3)+8(4)(6)+3(4)

Multiply first

12+192+12

Then add

216

The maximum number of spherical balloons that can be held in the canopy will be 30.

The canopy is the fabric cover that hangs above a bed. It is a covering that is fastened to or carried over a dignitary or a holy item.

The base layer of spheres in a square pyramidal arrangement has n2 balls, where n is the number of balls that make up a square's side.

n=4 inch

The total number of spheres in the pyramid is;

⇒n(n+1)(2n+1)/6

⇒4(4+1)(2×4+1)/6

⇒(4×5×9)/6

⇒30

Hence, in the canopy, 30 spherical balloons are the most that may be accommodated.

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[tex]\displaystyle\\|\Omega|=\binom{15}{3}=\dfrac{15!}{3!12!}=\dfrac{13\cdot14\cdot15}{2\cdot3}=455[/tex]

a)

[tex]\displaystyle\\|A|=\binom{10}{3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{2\cdot3}=120\\\\P(A)=\dfrac{120}{455}=\dfrac{24}{91}\approx26.4\%[/tex]

b)

[tex]\displaystyle\\|A|=\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdot5}{2}=10\\\\P(A)=\dfrac{10}{455}=\dfrac{2}{91}\approx2.2\%[/tex]

c)

[tex]\displaystyle\\|A|=\binom{10}{2}\cdot5=\dfrac{10!}{2!8!}\cdot5=\dfrac{9\cdot10}{2}\cdot5=225\\\\P(A)=\dfrac{225}{455}=\dfrac{45}{91}\approx49.5\%[/tex]

d)

[tex]A[/tex] - at least one is spoilt

[tex]A'[/tex] - none is spoilt

[tex]P(A)=1-P(A')[/tex]

We calculated [tex]P(A')[/tex] in a).

Therefore

[tex]P(A)=1-\dfrac{24}{91}=\dfrac{67}{91}\approx73.6\%[/tex]

Answer:

What is Poly in gender. Gender is Gender "Female" "Non-binary" "Male". Poly is Triangle gender which means random gender.

Answer: Explanation: To find the degree of the polynomial, add up the exponents of each term and select the highest sum.

Step-by-step explanation:

Answer:

the answer is A

Step-by-step explanation:

The line is the middle point of the graph/ the middle number of all therefore it's the median.

The solution for the first inequality is a ≥ 2 or a ∈ [2, ∞), and for the second inequality the solutions are b < -5 or b ∈ (-∞, -5)

It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than known as inequality.

We have two inequalities:

a.) -12a +7 ≤31

b.) -9 > 3b +6

a)  -12a +7 ≤ 31

-12a ≤ 31 - 7

-12a ≤ 24

-a ≤ 2

a ≥ 2  (sign changed because multiplied by a negative number)

a ∈ [2, ∞)

b.) -9 > 3b +6

-15 > 3b

-5 > b

or

b < -5

b ∈ (-∞, -5)

Thus, the solution for the first inequality is a ≥ 2 or a ∈ [2, ∞), and for the second inequality the solutions are b < -5 or b ∈ (-∞, -5)

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Part 1: Finding [tex](g\circ h)(x)[/tex]

Note that [tex](g\circ h)(x)=g(h(x))[/tex]

In this case, it is [tex]g(\sqrt{x+4})=(\sqrt{x+4})^{2}-3=x+4-3=\boxed{x+1}[/tex]

Part 2: Domain

For the domain, we need to make sure the radicand of h(x) is greater than or equal to 0, so we get [tex]x+4 \geq 0 \longrightarrow \boxed{x \geq -4}[/tex]

The composition (g ο h) = x+ 1 and the domain of the composition (g ο h) is = (-∞ , ∞).

The composition (g ο h) means that the function g(x) composes of h(x) that is g(h(x)). So we have to put the the values of h(x) in the function g(x).

Now here it is given that

g(x) = [tex]x^{2} -3[/tex] and h(x) = [tex]\sqrt[2]{x + 4}[/tex]

so to get (g ο h) we have to replace the x with the value of h(x)

That gives :

g(h(x)) = [tex]\sqrt[2]{x+4} ^{2} - 3[/tex]

g(h(x)) = x + 4 - 3 = x +1

So the composition (g ο h) = x+ 1

The domain of the function is the values of x for which the function is defined.

so the domain of the composition (g ο h) is :

g(h(x)) = x+ 1

This function is undefined for no value of x .

So its domain is (-∞ , ∞).

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The wire breaks if the load [tex]X[/tex] exceeds some amount. So what you're asked to do is find [tex]P(X\ge74)[/tex] and [tex]P(X\ge89)[/tex].

In either case, transform [tex]X[/tex] to the random variable [tex]Z[/tex] that's normally distributed with expected value 0 and standard deviation 1.

[tex]P(X\ge74) = P\left(\dfrac{X-80}3 \ge \dfrac{74-80}3\right) \\ = P(Z \ge -2) \\ = 1-P(Z < -2) \approx 0.9773[/tex]

[tex]P(X\ge89) = P\left(\dfrac{X-80}3 \ge \dfrac{89-80}3\right) \\= P(Z \ge 3) \\= 1 - P(Z < 3) \approx 0.0013[/tex]

An equation is formed of two equal expressions. The value of x that maximizes the area of the printed region of the billboard is 9.655 ft.

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

Given x is the left-right width of the billboard and y is the height of the billboard. Therefore,

The total area of the billboard, A= x·y

The total printed area of the billboard, [tex]A_p=(x-2)(y-8)[/tex]

Given in problem that the area of the billboard is 3600 ft².

x·y = 3600

y = (3600)/x

Substituting the value of y in the equation of the total printed area of the billboard,

[tex]A_p = (x-2)(\dfrac{3600}{x}-8)\\\\A_p = 3600 -8x -\dfrac{7200}{x} + 16\\\\A_p =3616-8x - \dfrac{7200}{x}[/tex]

Now, the value of x is needed to be minimum, therefore, differentiating the given function,

[tex]\dfrac{d}{dx}A_p =\dfrac{d}{dx}3616-8x - \dfrac{7200}{x}\\\\\dfrac{d}{dx}A_p =-8 - \dfrac{7200}{x^2}[/tex]

Equate the differentiated function with 0,

[tex]0=-8x - \dfrac{7200}{x^2}\\\\8x = - \dfrac{7200}{x^2}\\\\x^3 = 900\\\\x = 9.655 ft.[/tex]

Hence, the value of x that maximizes the area of the printed region of the billboard is 9.655 ft.

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Answer: Rectangles

Step-by-step explanation:

Whenever rectangles are rotated about an axis, a cylinder is formed.