Z decreased by 15%
Express as algebra

Answer:

34. (c) 12

35. (a) -12

36. (a) 51

37. (a) 13

Step-by-step explanation:

34.)

[tex] \implies \: \sf\dfrac{4}{xx + 2} = \dfrac{6}{2xx - 3} \\ \\ \implies \: \sf4(2xx - 3) = 6(xx + 2) \\ \\ \implies \: \sf 8xx - 12 = 6xx + 12 \\ \\ \implies \: \sf 8xx - 6xx = 12 + 12 \sf \\ \\ \implies \: \sf 2xx = 24 \\ \\ \implies \: \sf xx = \dfrac{24}{2} \\ \\ \implies \: \sf xx = 12 \\ [/tex]

Hence, Required answer is option (c) 12.

35.)

[tex] \implies \: \sf \dfrac{xx - 2}{2} = \dfrac{3xx + 8}{4} \\ \\ \implies \: \sf2(3xx + 8) = 4(xx - 2) \\ \\ \sf 6xx + 16 = 4xx - 8 \\ \\ \implies \: \sf 6xx - 4xx = - 8 - 16 \\ \\ \implies \: \sf 2xx = - 24 \\ \\ \implies \: \sf xx = \dfrac{ - 24}{2} \\ \\ \implies \: \sf xx = - 12 \\ [/tex]

Hence, Required answer is option (a) -12.

36.)

[tex] \implies \: \sf\sqrt{xx - 2} = 7 \\ \\ \implies \: \sf xx - 2 = {(7)}^{2} \\ \\ \implies \: \sf xx - 2 = 49 \\ \\ \implies \: \sf xx = 49 + 2 \\ \\ \implies \: \sf xx = 51[/tex]

Hence, Required answer is option (a) 51.

37.)

[tex] \implies \: \sf \sqrt{2xx - 10} = 4 \\ \\ \implies \: \sf 2xx - 10 = {(4)}^{2} \\ \\ \implies \: \sf 2xx - 10 = 16 \\ \\ \implies \: \sf 2xx = 16 + 10 \\ \\ \implies \: \sf 2xx = 26 \\ \\ \implies \: \sf xx = \dfrac{26}{2} \\ \\ \implies \: \sf xx = 13 \\ [/tex]

Hence, Required answer is option (a) 13.

Answer:

Step-by-step explanation:

We can start by assuming that the relationship between the ticket price and attendance is linear, so we can write the equation for the line that connects the two data points we have:

Point 1: (9, 23000)

Point 2: (7, 30000)

The slope of the line can be calculated as:

slope = (y2 - y1) / (x2 - x1)

slope = (30000 - 23000) / (7 - 9)

slope = 3500

So the equation for the line is:

y - y1 = m(x - x1)

y - 23000 = 3500(x - 9)

y = 3500x - 28700

Now we can use this equation to find the attendance for any ticket price. To maximize revenue, we need to find the ticket price that generates the highest revenue. Revenue is simply the product of attendance and ticket price:

R = P*A

R = P(3500P - 28700)

R = 3500P^2 - 28700P

To find the ticket price that maximizes revenue, we need to take the derivative of the revenue equation and set it equal to zero:

dR/dP = 7000P - 28700 = 0

7000P = 28700

P = 4.10

So the ticket price that would maximize revenue is $4.10. However, we need to make sure that this price is within a reasonable range, so we should check that the attendance at this price is between 23,000 and 30,000:

A = 3500(4.10) - 28700

A = 5730

Since 23,000 < 5,730 < 30,000, we can conclude that the ticket price that would maximize revenue is $4.10.

Answer:

The derivative does not exist at the extremum (-2, 0).

Step-by-step explanation:

Given function:

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

To differentiate the given function, use the chain rule and the power rule of differentiation.

[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule of Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Power Rule of Differentiation}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=nx^{n-1}$\\\end{minipage}}[/tex]

[tex]\begin{aligned}\textsf{Let}\;u &= x+2& \implies f(u) &= u^{\frac{2}{3}}\\\\\implies \dfrac{\text{d}u}{\text{d}{x}}&=1 &\implies \dfrac{\text{d}y}{\text{d}u}&=\dfrac{2}{3}u^{(\frac{2}{3}-1)}=\dfrac{2}{3}u^{-\frac{1}{3}}\end{aligned}[/tex]

Apply the chain rule:

[tex]\implies f'(x) = \dfrac{\text{d}y}{\text{d}{u}} \cdot \dfrac{\text{d}u}{\text{d}{x}}[/tex]

[tex]\implies f'(x) = \dfrac{2}{3}u^{-\frac{1}{3}} \cdot1[/tex]

[tex]\implies f'(x) = \dfrac{2}{3}u^{-\frac{1}{3}}[/tex]

Substitute back in u = x + 2:

[tex]\implies f'(x) = \dfrac{2}{3}(x+2)^{-\frac{1}{3}}[/tex]

[tex]\implies f'(x) = \dfrac{2}{3(x+2)^{\frac{1}{3}}}[/tex]

An extremum is a point where a function has a maximum or minimum value. From inspection of the given graph, the minimum point of the function is (-2, 0).

To determine the value of the derivative at (-2, 0), substitute x = -2 into the differentiated function.

[tex]\begin{aligned}\implies f'(-2) &= \dfrac{2}{3(-2+2)^{\frac{1}{3}}}\\\\ &= \dfrac{2}{3(0)^{\frac{1}{3}}}\\\\&=\dfrac{2}{0} \;\;\;\leftarrow \textsf{unde\:\!fined}\end{aligned}[/tex]

As the denominator of the differentiated function at x = -2 is zero, the value of the derivative at (-2, 0) is undefined.  Therefore, the derivative does not exist at the extremum (-2, 0).

Answer: Below :)

Step-by-step explanation:

(a) The percentage of students and teachers who bought tickets to go on the trip is 65%. So the number of students and teachers who bought tickets can be found by multiplying the total number of students and teachers by 65%:

65/100 x 360 = 234

Therefore, 234 students and teachers bought tickets to go on the trip.

(b) Let the number of teachers who bought tickets be x. Then the number of male students who bought tickets is 3x, and the number of female students who bought tickets is 5x.

The total number of students and teachers who bought tickets is:

x + 3x + 5x = 9x

We know that 234 students and teachers bought tickets, so we can set up the following equation:

9x = 234

Solving for x, we get:

x = 26

So the number of teachers who bought tickets is 26, the number of male students who bought tickets is 78 (3x), and the number of female students who bought tickets is 130 (5x).

However, 4 of the female students who bought tickets did not go on the trip, so the actual number of female students who went on the trip is:

130 - 4 = 126

(c) The ratio of the number of male students who went on the trip to the number of female students who went on the trip is:

78/126

Simplifying this ratio by dividing both numerator and denominator by 6, we get:

13/21

Therefore, the ratio of the number of male students who went on the trip to the number of female students who went on the trip is 13:21.

Step-by-step explanation:

a probability is always the ratio

desired cases / totally possible cases

(a)

the experimental probability is just using the actual experience to predict any future results.

the total number of cases was 20, and the number of desired cases (yellow) was 12.

so, the experimental probability of landing on yellow is

12/20 = 3/5 = 0.600

(b)

the theoretical probability of a totally fair spinner landing on yellow is 2 out of 5 possibilities, so

2/5 = 0.4000

(c)

the correct statement is the first one.

with a more or less balanced (fair) spinner the experimental numbers should get closer and closer to the theoretical numbers, the more spins we make.

Answer:

5 as a percentage of 100 is 5/100 which is 5%

Answer:2

Step-by-step explanation:

Answer:

2 Tons

Step-by-step explanation:

Homer’s car weighs 2 tons because there are 2,000 pounds in a ton and 4,000 divided by 2,000 equals 2

According to the given information, the equation of the ellipse is [tex](x+8)^2/256 + (y+1)^2/784 = 1.[/tex]

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that deals with the study of geometric shapes using algebraic principles. It involves the use of coordinates to represent points, lines, curves, and other geometric figures on a plane or in space.

we need to know the coordinates of its foci, co-vertices, and the center. We can start by finding the center of the ellipse, which is the midpoint of the line segment joining the foci:

Center = ( (-8 + (-8))/2 , (14 + (-16))/2 ) = (-8,-1)

Next, we can find the distance between the foci, which is given by:

[tex]distance between foci = 2c = sqrt[(14 - (-16))^2 + (-8 - (-8))^2] = 30[/tex]

where c is the distance from the center to either focus.

We also know that the distance between the co-vertices is given by:

distance between co-vertices = 2a = |-16 - 0| = 16

where a is the distance from the center to either co-vertex.

Finally, we can use the standard form equation for an ellipse centered at the origin:

[tex](x^2/a^2) + (y^2/b^2) = 1[/tex]

where b is the distance from the center to either vertex.

To find b, we can use the Pythagorean theorem:

[tex]b^2 = c^2 - a^2 \\b^2 = 30^2 - 16^2\\b^2 = 784\\b = 28[/tex]

Now we have all the information we need to write the equation of the ellipse:

[tex](x+8)^2/16^2 + (y+1)^2/28^2 = 1[/tex]

Therefore, according to the given information, the equation of the ellipse is [tex](x+8)^2/256 + (y+1)^2/784 = 1.[/tex]

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Answer:

We can write the diameter and circumferance of base as -

D = 2√(750ρ/πh)

C = 2π√(750ρ/πh)

Step-by-step explanation:

What is function?

A function is a relation between a dependent and independent variable.

Mathematically, we can write → y = f(x) = ax + b.

Given is to find the diameter and height of the tin can.

Assume the density of coffee as {ρ}. We can write the volume of the tin can as -

Volume = mass x density

Volume = 750ρ

We can write -

πr²h = 750ρ

r = √(750ρ/πh)

D = 2r

D = 2√(750ρ/πh)

Now, we can write the circumferance as -

C = 2πr

C = 2π√(750ρ/πh)

Therefore, we can write the diameter and circumferance of base as -

D = 2√(750ρ/πh)

C = 2π√(750ρ/πh)

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Answer:

Acquired resources

Step-by-step explanation:

Acquired resources

The matrix A of the linear transformation T(M) with respect to the standard basis for U2×2 is given by:

T([1000]) = [8 0]

[0 0]

T([0010]) = [0 0]

[0 9]

T([0001]) = [0 1]

[0 0]

To find the matrix A of the linear transformation T(M), we need to apply T to each basis vector of U2×2 and express the result as a linear combination of the basis vectors for U2×2. We can then arrange the coefficients of each linear combination as the columns of the matrix A.

Let's begin by finding T([1000]). We have:

T([1000]) = [8097][1000][8097]^-1

= [8 0]

[0 0]

To express this result as a linear combination of the basis vectors for U2×2, we need to solve for the coefficients c1, c2, and c3 such that:

[8 0] = c1[1000] + c2[0010] + c3[0001]

Equating the entries on both sides, we get:

c1 = 8

c2 = 0

c3 = 0

Therefore, the first column of the matrix A is [8 0 0]^T.

Next, we find T([0010]). We have:

T([0010]) = [8097][0010][8097]^-1

= [0 0]

[0 9]

Expressing this as a linear combination of the basis vectors for U2×2, we get:

[0 0] = c1[1000] + c2[0010] + c3[0001]

Equating the entries on both sides, we get:

c1 = 0

c2 = 0

c3 = 0

Therefore, the second column of the matrix A is [0 0 0]^T.

Finally, we find T([0001]). We have:

T([0001]) = [8097][0001][8097]^-1

= [0 1]

[0 0]

Expressing this as a linear combination of the basis vectors for U2×2, we get:

[0 1] = c1[1000] + c2[0010] + c3[0001]

Equating the entries on both sides, we get:

c1 = 0

c2 = 1

c3 = 0

Therefore, the third column of the matrix A is [0 1 0]^T.

Putting all of this together, we have:

A = [8 0 0]

[0 0 1]

[0 0 0]

Therefore, the matrix A of the linear transformation T(M) is:

T([1000]) = [8 0]

[0 0]

T([0010]) = [0 0]

[0 9]

T([0001]) = [0 1]

[0 0]

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The bases for the row space and null space of A, we put A into reduced row echelon form and solve for the null space. The dot product of basis vectors shows they are orthogonal.

To find the bases for the row space and null space of A, we perform row operations on A until it is in reduced row echelon form:

[ 1 -1  3 |  5 ]    [ 1 -1  3 |  5 ]

[ 2  1 -5 | -9 ] -> [ 0  3 -11 | -19]

[-1 -1  2 |  2 ]    [ 0  0  0  |  0 ]

[ 1  1 -1 | -1 ]    [ 0  0  0  |  0 ]

The reduced row echelon form of A tells us that there are two pivot columns, corresponding to the first and second columns of A. The third and fourth columns are free variables. Therefore, a basis for the row space of A is given by the first two rows of the reduced row echelon form of A:

[ 1 -1  3 |  5 ]

[ 0  3 -11 | -19]

To find a basis for the null space of A, we solve the system Ax = 0. Since the third and fourth columns of A are free variables, we can express the solution in terms of those variables. Setting s = column 3 and t = column 4, we have:

x1 - x2 + 3x3 + 5x4 = 0

2x1 + x2 - 5x3 - 9x4 = 0

-x1 - x2 + 2x3 + 2x4 = 0

x1 + x2 - x3 - x4 = 0

Solving for x1, x2, x3, and x4 in terms of s and t, we get:

x1 = -3s - 5t

x2 = s + 2t

x3 = s

x4 = t

Therefore, a basis for the null space of A is given by the vectors:

[-3  1  1  0]

[ 5  2  0  1]

To verify that every vector in the row space of A is orthogonal to every vector in the null space of A, we compute the dot product of each basis vector for the row space with each basis vector for the null space:

[ 1 -1  3 |  5 ] dot [-3  1  1  0] = 0

[ 1 -1  3 |  5 ] dot [ 5  2  0  1] = 0

[ 0  3 -11 | -19] dot [-3  1  1  0] = 0

[ 0  3 -11 | -19] dot [ 5  2  0  1] = 0

Since all dot products are equal to zero, we have verified that every vector in the row space of A is orthogonal to every vector in the null space of A.

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_____The given question is incomplete, the complete question is given below:

in excercises 7 and 8 find bases for the row space and null space of a. verify that every vector in the row(a) is orthogonal to every vector in null(a). a = [ 1 -1 3   5 2 1   0 1 -2   -1 -1 1]

Answer: A

y < 2x + 1

y > 1/2x + 2

The value for mean, variance, and standard deviation for the given set of data is 63.22, 794.04, and 28.17, respectively.

The method to calculate the various operations are:

Mean:

= (79 + 52 + 64 + 99 + 75 + 48 + 52 + 24 + 76) / 9 = 63.22

Mean is a measure of central tendency found by adding all the observations and dividing the result by the number of frequency or the total number of data set values.

Variance:

= ((79 - 63.22)² + (52 - 63.22)² + (64 - 63.22)² + (99 - 63.22)² + (75 - 63.22)² + (48 - 63.22)² + (52 - 63.22)² + (24 - 63.22)² + (76 - 63.22)² / (9-1) = 794.04

(Here, 63.22 is the mean calculated earlier)

Standard deviation:

= √(Variance)

=√(794.04) = 28.17

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Complete question is:

For each of the given sample data sets below, calculate the mean, variance, and standard deviation. (a) 79, 52, 64, 99, 75, 48, 52, 24, 76 mean =_______

variance = __________

standard deviation =_________

The probability that the average of 49 marathons is between 143 and 147 minutes is 0.5980. The 60th percentile is 148.25 minutes, and the median is 146 minutes.

The average of a sample of 49 marathons will be approximately normally distributed with mean = 146 minutes and standard deviation = 15/sqrt(49) = 15/7.

To find the probability that the average of the sample will be between 143 and 147 minutes, we can standardize the values:

z1 = (143 - 146) / (15/7) = -1.4

z2 = (147 - 146) / (15/7) = 0.4667

Then, using a standard normal distribution table or calculator, we find:

P(-1.4 < Z < 0.4667) = P(Z < 0.4667) - P(Z < -1.4)

= 0.6788 - 0.0808

= 0.5980

So the probability that the average of the sample will be between 143 and 147 minutes is 0.5980.

To find the 60th percentile for the average of these 49 marathons, we need to find the z-score such that the area to the left of the z-score is 0.6. Using a standard normal distribution table or calculator, we find:

P(Z < z) = 0.6

z = 0.25

Then, we can solve for the corresponding value of X:

0.25 = (X - 146) / (15/7)

X = 148.25

So the 60th percentile for the average of these 49 marathons is 148.25 minutes.

To find the median of the average running times, we note that the median of a normal distribution is equal to its mean. Therefore, the median of the average running times is 146 minutes.

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(a) Therefore after t = 1.75 seconds the diver will hit the water.

(b) The velocity the diver hit the water is 56 ft/s.

From the given condition we have d(t) = 16t²

and the height is 49ft

(a) Now when the diver hit the water the equation become

16t² = 49

t² = 49/16

t = ±7/4

t = ±1.75

since time can not be negative so t = 1.75

Therefore after t = 1.75 seconds the diver will hit the water.

(b)

Now differentiating d(t) with respect to t we get

d'(t) = 32t

now putting t=7/4 we get

the velocity d'(7/4) = 32*7/4

d'(7/4) = 56ft/s

Therefore the velocity the diver hit the water is 56 ft/s.

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The complete question is :

A cliff diver plunges from a height of 49ft above the water surface. The distance the diver falls in t seconds is given by the function d(t)=16t²ft

(a) After how many seconds will the diver hit the water?

(b) With what velocity (in ft/s ) does the diver hit the water?

To obtain the required equation we divide the equation by sin²Ø.

The first six functions are trigonometric, with the domain value being the angle of a right triangle and the range being a number. The angle, expressed in degrees or radians, serves as the domain and the range of the trigonometric function (sometimes known as the "trig function") of f(x) = sin. Like with all other functions, we have the domain and range. In calculus, geometry, and algebra, trigonometric functions are often utilised.

The given equation is:

sin²Ø+cos²Ø=1

To obtain the required equation we divide the equation with sin²Ø:

sin²Ø/sin²Ø +cos²Ø/ sin²Ø = 1/sin²Ø

1 + cot²Ø = csc²Ø

Hence, to obtain the required equation we divide the equation by sin²Ø.

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Answer:

area of semi circle =pi r^2/2

3.14*6*6/2=56.2

area of rectangle=lb

=20*12=240

240+56.2=296.2

rounding it it will become 300 ft sqr

perimeter of rectangle without including 4th side=20+12+20=52

perimeter of semicircle=pi r+d (d is not needed here)

3.14*6=18.84

so total perimeter=52+18.84=70.84ft

Step-by-step explanation:

Answer:

-4(x - 5) + 8x = 9x - 3

Simplifying the left side:

-4x + 20 + 8x = 9x - 3

4x + 20 = 9x - 3

Subtracting 4x from both sides:

20 = 5x - 3

Adding 3 to both sides:

23 = 5x

Dividing both sides by 5:

x = 23/5

Therefore, the equation equivalent to the given equation is:

5x - 23 = 0

Rhiannon arrived home approximately 17 minutes earlier than Alexander.

This is the arithmetic mean and is calculated by adding a group of numbers and then dividing by the count of those numbers. For example, the average of 2, 3, 3, 5, 7, and 10 is 30 divided by 6, which is 5.

To solve this problem, we can use the formula:

time = distance / speed

a) The time it took Alexander to get home is:

time_Alexander = 14 km / 20 km/h = 0.7 hours

The time it took Rhiannon to get home is:

time_Rhiannon = 10 km / 24 km/h = 0.41667 hours

Since Rhiannon's time is smaller than Alexander's, Rhiannon arrived home earlier.

b) The time difference between their arrivals is:

time_difference = time_Alexander - time_Rhiannon = 0.7 hours - 0.41667 hours = 0.28333 hours

To convert this to minutes, we can multiply by 60:

time_difference_in_minutes = 0.28333 hours x 60 minutes/hour ≈ 17 minutes

Therefore, Rhiannon arrived home approximately 17 minutes earlier than Alexander.

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Since the p-value (0.0803) exceeds the significance threshold (0.05), the null hypothesis cannot be ruled out.

The mean in mathematics is a measurement of a collection of numerical data's central tendency. It is determined by adding up all of the values in the set and dividing the result by the total number of values. This value is frequently referred to as the average value. The mean (or mathematical mean) is calculated as follows: (Sum of Values) / Mean (number of values)

given

The null hypothesis states that the mean number of units generated during the day and night shifts is the same. The contrary hypothesis (Ha) states that more units are created on average on the night shift than on the day shift.

"day" + "night"

Bravo! Night precedes day.

b. The following method can be used to calculate the test statistic:

t = sqrt(1/n night + 1/n day) * sqrt(x night - x day)

where s p is the pooled standard deviation and x night and x day are the sample averages, n night and n day are the sample sizes, and s p is represented by:

Sqrt(((n night - 1)*s night2 + (n day - 1)*s day2) / (n night + n day - 2)) yields the value s p.

S p is equal to sqrt(((74 - 1)*35 + (68 - 1)*28) / (74 + 68 - 2)), which equals 31.88.

t = (358 - 352) / (31.88 * sqrt(1/74 + 1/68)) = 1.19

1.19 is the test result.

the p-value is 0.0803 as a result (rounded to 4 decimal places).

Since the p-value (0.0803) exceeds the significance threshold (0.05), the null hypothesis cannot be ruled out.

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The complete question is :- Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. The mean number of units produced by a sample of 68 day-shift workers was 352. The mean number of units produced by a sample of 74 night-shift workers was 358. Assume the population standard deviation of the number of units produced is 28 on the day shift and 35 on the night shift.

Using the 0.05 significance level, is the number of units produced on the night shift larger?

a. State the null and alternate hypotheses.

O : Day/Night: H:

Day Night

b. Compute the test statistic. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)

c. Compute the p-value. (Round your answer to 4 decimal places.) p-value

The area of this triangle is 30 m².

Area is a surface measure, that is, it is the amount of space that a geometric figure occupies on a flat surface.

To calculate the area of a triangle, we can use the formula:

Area = (base x height) / 2

In the case of the given triangle, we can choose the measure of 5m as the base and the measure of 12m as the height, since the height forms a right angle with the base and is perpendicular to it.

So, we have:

Area = (b*h)/2

Area = (5m * 12m) / 2

Area = 30m²

if we make quarterly deposits and invest them at an nominal annual interest rate of 8% compounded quarterly for 40 years, we will have $143,004.54 at the end of the 40 years.

The first step in solving this problem is to calculate the amount of each quarterly deposit. We know that the first deposit is $400, and each subsequent deposit decreases by 1% from the previous deposit. This means that each deposit is 99% of the previous deposit. To calculate the size of each deposit, we can use the following formula:

deposit_ n = deposit_(n-1) * 0.99

Using this formula, we can calculate the size of each quarterly deposit as follows:

deposit_1 = $400

deposit_2 = deposit_1 * 0.99 = $396.00

deposit_3 = deposit_2 * 0.99 = $392.04

deposit_4 = deposit_3 * 0.99 = $388.12

...

We can continue this pattern for 40 years (160 quarters) to find the size of each quarterly deposit.

Next, we need to calculate the future value of these deposits using an nominal annual interest rate of 8% compounded quarterly. We can use the formula for compound interest to calculate the future value:

[tex]FV = PV * (1 + r/n)^(n*t)[/tex]

where FV is the future value, PV is the present value (which is zero since we are starting with deposits), r is the nominal annual interest rate (8%), n is the number of times the interest is compounded per year (4 since we are compounding quarterly), and t is the number of years (40).

We can substitute the values into the formula and solve for FV:

[tex]FV = $400 * (1 + 0.08/4)^(440) + $396.00 * (1 + 0.08/4)^(439) + $392.04 * (1 + 0.08/4)^(4*38) + ... + $1.64 * (1 + 0.08/4)^4[/tex]

After solving this equation, we get a future value of $143,004.54, rounded to two decimal places. This means that if we make quarterly deposits and invest them at an nominal annual interest rate of 8% compounded quarterly for 40 years, we will have $143,004.54 at the end of the 40 years.

This calculation highlights the power of compound interest over long periods of time. By making regular contributions and earning interest on those contributions, our investment grows exponentially over time. It also shows the importance of starting early and consistently contributing to an investment over time in order to achieve long-term financial goals.

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Answer:

I'm pretty confident that the answer is E

Answer:

77.7

Step-by-step explanation:

if you know, you know. laso make sure your calculator is on degrees and not radians

the terminal point of the angle determined by sin(0) < 0 and cos(0) > 0 is in the fourth quadrant.

why it is and what is trigonometry?

If sin(0) < 0 and cos(0) > 0, then we know that the angle 0 is in the fourth quadrant of the unit circle.

In the unit circle, the x-coordinate represents cos(θ) and the y-coordinate represents sin(θ). Since cos(0) > 0, we know that the terminal point of the angle is to the right of the origin. And since sin(0) < 0, we know that the terminal point is below the x-axis.

The fourth quadrant is the only quadrant where the x-coordinate is positive and the y-coordinate is negative, so that is the quadrant where the terminal point of the angle lies.

Therefore, the terminal point of the angle determined by sin(0) < 0 and cos(0) > 0 is in the fourth quadrant.

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It focuses on the study of the functions of angles and their applications to triangles, including the measurement of angles, the calculation of lengths and areas of triangles, and the analysis of periodic phenomena.

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so here we get two outcomes one is 2 and other is 4.

so there is total 2 outcomes.

total no. of possibility is 5

so the probability of waiting between 2 and 4 minutes to use the ATM is 2/5.

Set of sophomores taking discrete math = A ∩ B. Set of sophomores not taking discrete math = A - B or A ∩ B^c. Set of students who are sophomores or in discrete math = A ∪ B. Set of students who are not sophomores or not in discrete math = (A ∩ B)^c or A ∪ B^c.

The set of sophomores taking discrete math at your school is the intersection of the set of sophomores A and the set of students in discrete math B. So, it can be expressed as A ∩ B.

The set of sophomores at your school who are not taking discrete math is the difference between the set of sophomores A and the set of students in discrete math B. So, it can be expressed as A - B or A ∩ B^c, where B^c is the complement of B (i.e., the set of students who are not in discrete math).

The set of students at your school who either are sophomores or are taking discrete math is the union of the set of sophomores A and the set of students in discrete math B. So, it can be expressed as A ∪ B.

The set of students at your school who either are not sophomores or are not taking discrete math is the complement of the intersection of the set of sophomores A and the set of students in discrete math B.

This can be expressed as (A ∩ B)^c or as A ∪ B^c, where B^c is the complement of B (i.e., the set of students who are not in discrete math). Note that this set includes all students who are either juniors, seniors, or not enrolled in discrete math.

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