=
Use the Fundamental Counting Principle to solve.
A television programmer is arranging the order that three movies will be seen between the hours of 6 P.M. and 12
A.M. Two of the movies have a G rating, and they are to be shown in the first two time blocks. One of the movies is
rated NC-17, and it is to be shown in the last of the time blocks, from 10 P.M. until 12 A.M. Given these restrictions,
in how many ways can the three movies be arranged during the indicated time blocks?


The number of different ways the three movies can be arranged during the indicated time blocks is.
(Type a whole number.)

Answer:

15 quarters, 10 nickels

Step-by-step explanation:

[tex]q+n=25[/tex]

[tex]0.25q+0.05n=4.25[/tex]

multiply the first equation by -0.25 and add the second equation to it

[tex]-0.25q-0.25n=-6.25\\0.25q+0.05n=4.25[/tex]
________________
                [tex]-0.2n=-2[/tex]

[tex]n=\frac{-2}{-0.2} =10[/tex]   has 10 nickels

[tex]q=25-n=25-10=15[/tex]  has 15 quarters

10(.05) +15(0.25) = 0.5 + 3.75 = 4.25

Hope this helps

Answer:

x ≤ −8

Step-by-step explanation:

Let's solve your inequality step-by-step.

3 - 3x/4 ≥ 9

Step 1: Simplify both sides of the inequality.

-3/4x + 3 ≥ 9

Step 2: Subtract 3 from both sides.

-3/4x + 3 - 3 ≥ 9 − 3

-3/4x ≥ 6

Step 3: Multiply both sides by 4/(-3).

(4/-3) * (-3/4x) ≥ (4/-3) * (6)

x ≤ −8

Answer:

x ≤ −8

Answer:

[tex]y=\frac{-3}{4}x-3[/tex]

Step-by-step explanation:

This is a linear graph so the equation will be in the format of y=mx+b.

b, the y-intercept, will be -3, as that is where the line crosses the y-axis.

m, the slope, is [tex]\frac{-3}{4}[/tex], calculated from the change in y over the change in x from points (-4,0) to (0,-3).

Now substitute the values in our equation:

[tex]y=\frac{-3}{4}x-3[/tex]

Answer:

y= -3/4+[-3]

Step-by-step explanation:

Answer:

The answer is 4900, because the next number is less than five, due to that we don't have to add a number (+1).

---

Thank you so much for participate in the community of Brainly.com

I hope help you :)

Kind regards, Antonio.

Answer: [tex]x=-5[/tex]

Step-by-step explanation:

[tex]x+5=0 \implies x=-5[/tex]

1. Observe that

[tex]\nabla \dfrac{x^2y^2w^2}2 = \langle xy^2w^2, x^2yw^2, x^2y^2w\rangle[/tex]

is a gradient field, so the gradient theorem holds and the integral in question is indeed path-independent. Its value is

[tex]\dfrac{x^2y^2w^2}2\bigg|_{x=1,y=3,w=2}^{x=2,y=4,w=1} = 32 - 18 = \boxed{14}[/tex]

2. [tex]dz[/tex] is an exact differential if we can find a scalar function [tex]z=f(x,y)[/tex] such that

[tex]\dfrac{\partial f}{\partial x} = 3x^2 (x^2+y^2) = 3x^4 + 3x^2y^2[/tex]

[tex]\dfrac{\partial f}{\partial y} = 2y(x^3+y) = 2x^3y + 2y^2[/tex]

Integrating both sides of the first equation with respect to [tex]x[/tex] yields

[tex]f(x,y) = \dfrac35 x^5 + x^3 y^2 + g(y)[/tex]

Differentiating with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y} = 2x^3y + \dfrac{dg}{dy} = 2x^3y + 2y^2 \\\\ \implies \dfrac{dg}{dy} = 2y^2 \implies g(y) = \dfrac23y^3 + C[/tex]

and we ultimately find

[tex]f(x,y) = \boxed{z = \dfrac35 x^5 + x^3y^2 + \dfrac23 y^3 + C}[/tex]

(We can also use the same method here to determine the scalar function in part (1).)

Then the integral is path-independent, and its value is

[tex]f(2,1) - f(1,2) = \dfrac{418}{15} - \dfrac{149}{15} = \boxed{\dfrac{269}{15}}[/tex]

Answer:

addition property

Step-by-step explanation:

See attached image.

Using the future value formula, it is found that you would need to deposit $272.95 in the account each month.

It is given by:

[tex]V(n) = P\left[\frac{(1 + r)^{n-1}}{r}\right][/tex]

In which:

For this problem, considering that there are monthly compoundings, the parameters are:

r = 0.08/12 = 0.0067, V(n) = 300000, n = 25 x 12 = 300.

Hence we solve for P to find the monthly payment.

[tex]V(n) = P\left[\frac{(1 + r)^{n-1}}{r}\right][/tex]

[tex]300000 = P\left[\frac{(1.0067)^{299}}{0.0067}\right][/tex]

1099.12P = 300000

P  = 300000/1099.12

P = $272.95.

You would need to deposit $272.95 in the account each month.

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Answer:

421/6 degrees

Step-by-step explanation:

The acute angles of a right triangle are complementary, so

[tex]10x+6+2x+7=90 \\ \\ 12x+13=90 \\ \\ 12x=77 \\ \\ x=\frac{77}{12}[/tex]

So, the acute angles measures 421/6 degrees and 119/6 degrees, and thus the larger angle is 421/6 degrees.

Answer:

b = 133, a = 47

Step-by-step explanation:

If the two lines are parallel, then angle b is equal to 180-47 = 133 (same side interior angles), and angle is equal to angle 0 (alternate interior angles)

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

[tex]\qquad❖ \: \sf \:a = \theta[/tex]

( by Alternate interior angle pair )

[tex]\qquad \therefore\: \sf \:a = 47 \degree[/tex]

Next,

[tex]\qquad❖ \: \sf \:b = 180 - \theta[/tex]

( by co - interior angle pair )

[tex]\qquad❖ \: \sf \:b = 180 - 47[/tex]

[tex]\qquad \therefore \: \sf \:b = 133 \degree[/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

[tex]\qquad❖ \: \sf \: \:a = 47 \degree[/tex]

and

[tex]\qquad❖ \: \sf \:b = 133 \degree[/tex]

Answer:

3

Step-by-step explanation:

6.01 has 3 sig fig

I think that's how I do it

Applying the Pythagorean theorem, the length of the radius is: 35.3 units.

According to the Pythagorean theorem, the sum of the squares of the shorter sides of a right triangle equals the square of the longest side. For example, if a and b are two smaller legs of a right triangle, and c is the longest leg (hypotenuse) of the right triangle, then the Pythagorean theorem states that:

a² + b² = c².

Given the following parameters of the right triangle:

Triangle ABC is a right triangle with angle ACB as the right angle according to the tangent theorem since segment BC is tangent to Circle A

Segment AB = 58 (hypotenuse of the right triangle)

Segment BC = 46 (small leg of the right triangle)

Radius = AC (small leg of the right triangle)

Using the Pythagorean theorem, we have:

AC = √(AB² - BC²)

AC = √(58² - 46²)

AC = 35.3 units (to the nearest tenth).

Therefore, by using the Pythagorean theorem, the length of the radius of the circle, which is segment AC, to the nearest tenth is: 35.5 units.

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Answer: None of these are correct

Step-by-step explanation:

I believe that the equation is showing transitive property.

Assuming you had a bill of $120 when you went to the restaurant with your three friends, the 20% tip would come out to $24.

First assume an amount that the meal you ate cost. Let's assume that amount if $120.

If the tip rate if 20%, it means that you need to find out what 20% of the bill of $120 is.

If you don't want to use a calculator, there are other ways you can calculate the tip.

An easy method would be to find 10% of the bill. To find 10%, all you have to do is to move the decimal point from right to left.

The decimal point in $120 is at the far right so the 10% amount would be:
= $12.0

Now that you have the 10% amount, remember that 20% is simply twice the amount of 10% so:

= 12 x 2

= $24

The total amount then becomes:

= 120 + 24

= $144

In conclusion, your tip amount would be $24.

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Answer:

Option 3

Step-by-step explanation:

[tex]\frac{5}{3}=\frac{AD}{7-AD} \\ \\ 35-5AD=3AD \\ \\ 35=8AD \\ \\ AD=4 \frac{3}{8}[/tex]

Step-by-step explanation:

(2^2x²/xy²) for x=3 and y=2

we have

2^2(3)²/((3)(2²))

2^2(9)/((3)(4))

2^18/12

262144/12

21845.3

The value of  expression [tex]2^{2} x^{2}/xy^{2}[/tex]  for x=3 and y=2 is, 3

An expression is a sentence with at least two numbers or variables having mathematical operation. Math operations can be addition, subtraction, multiplication, division.

For example, 2x+3

Given that,

The expression,  [tex]2^{2} x^{2}/xy^{2}[/tex]

The value of expression when, x = 3 & y = 2

⇒  2²×3²/3×2²

⇒  3

Hence, The value is 3

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The 7th term of an Arithmetic Progression 50+45+40 is 20

The Arithmetic Progression is given as:

50+45+40

In the above Arithmetic Progression, we have

First term, a= 50

Common difference, d = 45 - 50 = -5

The nth term of the Arithmetic Progression is calculated as:

Tn = a + (n - 1)d

Substitute the known values in the above equation

Tn = 50 + (n - 1) * -5

Substitute 7 for n in the above equation

Tn = 50 + (7 - 1) * -5

Evaluate the product

Tn = 50 - 30

Evaluate the difference

T7 = 20

Hence, the 7th term of an Arithmetic Progression 50+45+40 is 20

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The single payment six months from now is required to settle the debt will be $4042.

The interest rate monthly will be 8/1200.

The number of time period given is 6 months.

Therefore, the future value will be:

= 2000(1 - 8/1200)^6

= 2081.35

The present value will be:

= 2000(1 - 8/1200)^3

= 1960.53

Therefore, the single payment will be:

= 2081.35 + 1960.53

= 4042

In conclusion, the correct amount is $4042.

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Answer: [tex]\Large\boxed{3.52~;~66.88}[/tex]

Step-by-step explanation:

Given information

One number is 19 times the other

The smaller number is less than 5

Sum = 70.4

Set variables

Let x be the smaller number

Let 19x be the larger number

Set equation according to the given information

Smaller number + Larger number = Sum

x + 19x = 70.4

Simplify by addition

20x = 70.4

Divide 20 on both sides

20x / 20 = 70.4 / 20

[tex]\Large\boxed{x=3.52}[/tex]

Substitute values into the variables to find the larger number

Larger number = 19x

                         = 19 (3.52)

                         [tex]\Large\boxed{=66.88}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

smaller number 3.52

greater number 66.88

Step-by-step explanation:

Let x and y be two numbers ,where x < 5 and x ≤ y.

x + y = 70.4

y = 19x

By replacing y by 19x in the first equation we get :

x + 19x = 70.4

Then

20x = 70.4

Then

x = 70.4÷20

  = 3.52

y = 19x

Then

y = 19 × 3.52  (just replace x by 3.52)

  = 66.88

Answer:

f(g(x)) = f(2x - 3) = 3(2x - 3) + 5 = 6x - 9 + 5 = 6x - 4

f(g(2)) = f(2*2 - 3) = f(4 - 3) = f(1) = 3*1 + 5 = 3 + 5 = 8

The composite function values:

f(g(x)) = 6x- 4

f(g(2)) = 8.

Any function that is created by combining two or more other functions is referred to as a composite function. To put it another way, given two functions, f, and g, the composite function of f and g, denoted as f(g(x)), is a new function that applies g to the input x before applying f to the output of g. (x).

To find the value of fg(x), we need to compute g(x) first and then substitute it into f(x).

g(x) = 2x - 3

Substitute g(x) into f(x):

f(g(x)) = 3(2x - 3) + 5

f(g(x)) = 6x - 4

Therefore, fg(x) = 6x - 4.

To find the value of fg(2), substitute 2 into fg(x):

fg(2) = 6(2) - 4

fg(2) = 8

Therefore, fg(2) = 8.

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Answer: 4.6 (rounded to the nearest tenth)

Step-by-step explanation:

I assume that 1.6 is the coefficient of x, which is what we're trying to find out in this case.

Therefore, dividing 1.6 on both sides we get

from

1.6x = 7.37

to

x = 7.37/1.6 = 4.6 (approximately, rounded to the nearest tenth).

There are different ways to construct an angle. The steps used in making a line segment are; First, you have to put the compass at one specific end of the line segment. Then you shift the compass slowly a little bit so it will be longer than half the length of the line segment. Then you have to draw arcs up and down the line. while using the same compass width, you then draw arcs from one specific end of the line. and thereafter you put your ruler at the point where the arcs cross, and you then draw the line segment. The difference is that to bisect an angle, one has to divide the shape or angle into two congruent parts while in the construction of a line segment, there are differences in length. What is the construction process of Angles? This entails a good construction process in making angles. They are step-by-step processes used to produce detailed and exact geometric figures.

well, keeping in mind that a year has 365 years, so let's see

[tex]\stackrel{Oct}{10}~~ + ~~\stackrel{Nov}{30}~~ + ~~\stackrel{Dec}{31}~~ + ~~\stackrel{Jan}{31}~~ + ~~\stackrel{Feb}{21}\implies 123~days[/tex]

so the 3600 were borrowed for only 123 days of a year, that'd be 123/365 of a year, thus

[tex]~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$3694.63\\ P=\textit{original amount deposited}\dotfill & \$3600\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &\frac{123}{365} \end{cases}[/tex]

[tex]3694.63=3600[1+(\frac{r}{100})(\frac{123}{365})]\implies \cfrac{3694.63}{3600}=1+\cfrac{123r}{36500} \\\\\\ \cfrac{3694.63}{3600}-1=\cfrac{123r}{36500}\implies \left( \cfrac{36500}{123} \right)\left( \cfrac{3694.63}{3600}-1 \right)=r\implies \stackrel{\%}{7.8}\approx r[/tex]

7.8% is the simple interest rate was charged if Robin repaid $3,694.63.

Simple interest is based on the principal amount of a loan or the first deposit in a savings account.

A=P(1+rt)

A is final amount, P is principle amount, r is rate of interest and t is time.

We know that in a year we have 365 days.

But the time here is from oct 21 to feb 21, which has 123 days.

so the 3600 were borrowed for only 123 days of a year, that'd be 123/365 of a year, thus

Apply these values in formula

A=P(1+rt)

3694.63=3600(1+r/100(123/365))

3694.63=3600(1+123r/36500)

3694.63/3600=1+123r/36500

3694.63/3600-1=123r/36500

(3694.63-3600)/3600=123r/36500

94.63/3600=123r/36500

Apply cross multiplication

3453995=442800r

7.80=r

Hence 7.8% is the simple interest rate was charged if Robin repaid $3,694.63.

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It will take 6 weeks for the number of infected people to reach 1200

The function is given as:

y equals 15 times 2 to the power of t

Rewrite the function properly as follows

y = 15 * 2^t

When the population is 1200, the equation becomes

15 * 2^t = 1200

Divide both sides by 15

2^t = 80

Take the logarithm of both sides

t * log(2) = log(80)

Divide both sides by log(2)

t = 6.3

Approximate

t = 6

Hence, it will take 6 weeks for the number of infected people to reach 1200

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The number of students that applied for all universities is 3, the number of students that applied for at least two of the universities is 20, the number of students that applied for at most two of the universities is 53, and the number of students that applied for Addis Ababa but not Bahir Dar University is 5.

Given that there are 60 students out of which 30 students had applied for Addis Ababa University, 25 students applied for Bahir Dar University and 24 students applied for Wachemo University. 11 students applied for both Addis Ababa and Bahir Dar Universities, 6 applied for both Addis Ababa and Wachemo Universities, 9 applied for both Wachemo and Bahir Dar Universities while 4 applied neither of the aforementioned universities.

Let A, B, and W denote the sets of students apply to Addis Ababa Uni (A), Bahir Dar Uni (B), or Wachemo Uni (W). Let U denote the universal set of all students in the class.

We're given the cardinalities of several sets:

total number of students n(U)=60, A applicants is n(A)=30, B applicants n(B)=25, W applicants n(W)=24, A and B applicants n(A∩B)=11, A and W applicants n(A∩W)=6, B and W applicants n(B∩W)=9 non-applicants n(U\(A∪B∪W))=4

The last cardinality tells us n(A∪B∪W),60-4=56 students applied anywhere at all.

We want to find n(A∩B∩W), the number of students that applied to each of the three universities.

By the inclusion/exclusion principle,

n(A∪B∪W)=n(A)+n(B)+n(W)-n(A∩B)-n(A∩W)-n(B∩W)+n(A∩B∩W)

56=30+25+24-11-6-9+n(A∩B∩W)

n(A∩B∩W)=3

Now, we will find the number of students that applied for at least two of the universities.

n(A∩B)=n(A∩B∩W)+n(A∩B∩W')

11=3+n(A∩B∩W')

8=n(A∩B∩W')

Similarly, we will find

n(A∩B'∩W)=3

n(A'∩B∩W)=6

n(A∩B'∩W')=16

n(A'∩B∩W')=8

n(A'∩B'∩W)=12

then the total number of students applied for at least two students is

n(A∩B∩W')+n(A∩B'∩W)+n(A'∩B∩W)+n(A∩B∩W)=20

Now, we will find the number of students that applied for atmost two universities, we get

n(A∩B∩W')+n(A∩B'∩W)+n(A'∩B∩W)+n(A∩B'∩W')+n(A'∩B∩W')+n(A'∩B'∩W)=53

now, we will find the number of students that applied for Addis Ababa but not Bahir Dar University is

n(A∩B')=n(A)-n(B)

n(A∩B')=30-25

n(A∩B')=5

hence, the total students is 60 and the number of students that applied for all universities is 3, the number of students that applied for at least two of the universities is 20, the number of students that applied for at most two of the universities is 53, and the number of students that applied for Addis Ababa but not Bahir Dar University is 5.

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Using the normal distribution, the indicated probability is given as follows:

P(7 ≤ x ≤ 11) = 0.6247.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The mean and the standard deviation are given, respectively, by:

[tex]\mu = 8, \sigma = 2[/tex]

P(7 ≤ x ≤ 11) is the p-value of Z when X = 11 subtracted by the p-value of Z when X = 7, hence:

X = 11:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (11 - 8)/2

Z = 1.5

Z = 1.5 has a p-value of 0.9332.

X = 7:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (7- 8)/2

Z = -0.5

Z = -0.5 has a p-value of 0.3085.

0.9332 - 0.3085 = 0.6247.

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The slope of a line that is parallel to y=5x+3 is 5x.

Two lines must have the same slope to be parallel to each other.

Answer:

39

Step-by-step explanation:

In a triangle, the sum of any two side lengths must exceed the length of the remaining third side. Therefore these 3 inequalities must be true.

5 + 19 > x

5 + x > 19

19 + x > 5

We can ignore the third inequality because, for any positive value of x, the inequality is true.

x < 26

x > 14

Now we know that x, the length of the third side, must be greater than 14 but less than 26. Since we are asked for the smallest whole number possible, the third side would be length 15. Therefore the perimeter is 5 + 19 + 15 = 39.

The perimeter of the bigger octagon that is similar to the smaller one is: 238 units.

An octagon can be described as a polygon that has 8 sides and 8 angles.

What are Similar Octagons?

Octagons are referred to as similar to each other if they have the same angle measure but corresponding sides that are proportional to each other. That is, they have the same shape but different sizes.

How to Find the Perimeter of Similar Polygons?

Given the two octagons in the diagram are similar, and:

Perimeter of the smaller octagon = 34 units

A side of the smaller octagon = 4 units

Perimeter of the bigger octagon = ?

A corresponding side of the bigger octagon = 28 units

Therefore:

Perimeter of the smaller octagon/Perimeter of the bigger octagon = side of the smaller octagon/corresponding side of the bigger octagon

Plug in the values

34/Perimeter of the bigger octagon = 4/28

Perimeter of the bigger octagon = (28 × 34)/4

Perimeter of the bigger octagon = 238 units.

In summary, the perimeter of the bigger octagon is: 238 units.

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