The entries in Table I are the probabilities that a random variable having the standard normal distribution will take on a value between 0 andz. They are given by the area of the gray region under the curve in the figure. TABLET NORMAL-CURVE AREAS 2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.1 0.2 0.3 0.4 0.5 0.0000 0.0398 0.0793 0.1179 0.1554 0.1915 0.0080 0.0478 0.0871 0.1255 0.1628 0.1985 0.0120 0.0517 0.0910 0.1293 0.1664 0.2019 0.2357 0.2673 0.2967 0.3238 0.3485 0.0160 0.0557 0.0948 0.1331 0.1700 0.2054 0.2389 0.2704 0.2995 0.3264 0.3508 0.0199 0.0596 0.0987 0.1368 0.1736 0.2088 0.0239 0.0636 0.1026 0.1406 0.1772 0.2123 0.0279 0.0675 0.1064 0.1443 0.1808 0.2157 0.0319 0.0714 0.1103 0.1480 0.1844 0.2190 0.0359 0.0753 0.1141 0.1517 0.1879 0.2224 0.2549 0.2852 0.3133 0.3389 0.3621 0.2257 0.0040 0.0438 0.0832 0.1217 0.1591 0.1950 0.2291 0.2611 0.2910 0.3186 0.3438 0.3665 0.3869 0.4049 0.4207 0.4345 0.4463 0.4564 0.4648 0.4719 0.4778 0.6 0.7 0.8 0.9 1.0 0.2580 0.2881 0.3159 0.3413 0.2324 0.2642 0.2939 0.3212 0.3461 0.2422 0.2734 0.3023 0.3289 0.3531 0.2454 0.2764 0.3051 0.3315 0.3554 0.2517 0.2823 0.3106 0.3365 0.3599 0.2486 0.2794 0.3078 0.3340 0.3577 0.3790 0.3980 0.4147 0.4292 0.4418 1.1 1.2 1.3 1.4 1.5 0.3749 0.3944 0.4113 0.4265 0.4394 0.3770 0.3962 0.4131 0.4279 0.4406 0.3810 0.3997 0.4162 0.4306 0.4429 0.3643 0.3849 0.4032 0.4192 0.4332 0.4452 0.4554 0.4641 0.4713 0.4772 0.3686 0.3888 0.4066 0.4222 0.4357 0.4474 0.4573 0.4656 0.4725 0.4783 0.3708 0.3907 0.4082 0.4236 0.4370 0.4484 0.4582 0.4664 0.4732 0.4788 0.3729 0.3925 0.4099 0.4251 0.4382 0.4495 0.4591 0.4671 0.4738 0.4793 0.3830 0.4015 0.4177 0.4319 0.4441 0.4545 0.4633 0.4706 0.4767 0.4817 1.6 1.7 1.8 1.9 2.0 0.450S 0.4599 0.4678 0.4744 0.4798 0.4515 0.4608 0.4685 0.4750 0.4803 0.4525 0.4616 0.4692 0.4756 0.4808 0.4535 0.4625 0.4699 0.4761 0.4812 2.1 2.2 2.3 2.4 2.5 0.4826 0.4864 0.4896 0.4920 0.4940 0.4830 0.4868 0.4898 0.4922 0.4941 0.4834 0.4871 0.4901 0.4925 0.4943 0.4850 0.4884 0.4911 0.4932 0.4949 0.4854 0.4887 0.4913 0.4934 0.4951 0.4857 0.4890 0.4916 0.4936 0.4952 0.4821 0.4861 0.4893 0.4918 0.4938 0.4953 0.4965 0.4974 0.4981 0.4987 0.4838 0.4875 0.4904 0.4927 0.4945 0.4959 0.4969 0.4977 0.4984 0.4988 0.4842 0.4846 0.4878 0.4881 0.4906 0.4909 0.4929 0.4931 0.4946 0.4948 0.4960 0.4961 0.4970 0.4971 0.4978 0.4979 0.4984 0.4985 0.4989 0.4989 2.6 2.7 2.8 2.9 3.0 0.4955 0.4966 0.4975 0.4982 0.4987 0.4956 0.4967 0.4976 0.4982 0.4987 0.4957 0.4968 0.4977 0.4983 0.4988 0.4962 0.4972 0.4979 0.4985 0.4989 0.4963 0.4973 0.4980 0.4986 0.4990 0.4964 0.4974 0.4981 0.4986 0.4990 Also, for 3 - 4.0, 5.0 and 6.0, the areas are 0.49997, 0.4999997, and 0.499999999. ^ The entries in Table II are values for which the area to their right under the 1 distribution with given degrees of freedom (the gray area in the figure) is equal toa. TABLE II VALUE OF d.f. Fo.050 os 0.005 d.. 63.657 9.925 1 2 3 4 6.314 2.920 2.353 2.132 2.015 12.706 4.303 3.182 2.776 2.571 0.010 31.821 6.965 4.541 3.747 3.365 1 2 3 4 5.841 4.604 4.032 S S 6 7 6 7 8 1.943 1.895 1.860 1.833 1.812 2.447 2.365 2.306 2.262 2.228 3.143 2.998 2.896 2.821 2.764 3.707 3.499 3.355 3.250 3.169 8 9 9 10 10 11 12 1.796 1.782 1.771 1.761 1.753 13 14 15 2.201 2.179 2.160 2.145 2.131 2.718 2.681 2.650 2.624 2.602 3.106 3.055 3.012 2.977 2.947 11 12 13 15 16 16 17 18 19 20 1.746 1.740 1.734 1.729 1.725 2.120 2.110 2.101 2.093 2.086 2.583 2.567 2.552 2.921 2.898 2.878 2.861 2.845 17 18 19 2.539 2.528 20 21 22 23 24 25 1.721 1.717 1.714 1.711 1.708 2.080 2.074 2.069 2.064 2.060 2.518 2.508 2.500 2.831 2.819 2.807 2.797 2.787 21 22 23 24 25 2.492 2.485 2.056 2.052 26 27 28 29 Inf. 1.706 1.703 1.701 1.699 1.645 2.479 2.473 2.467 2.462 2.326 2.779 2.771 2.763 2.048 26 27 28 29 Inf. 2.045 2.756 2.576 1.960 Question 2 (20 marks) A tutorial school has been running IELTS mock examination for many years. Below is the summary presented by the tutorial school related to the IELTS mock examination in one of the many online classes selected randomly in year 2022. IELTS score in mock examination Number of students <4 0 5 2 6 7 5 15 8 10 3 9 A report presented by this tutorial school 5 years ago indi that the population mean score of IELTS mock examination was 7.05 points and the population proportion of students got 7 points or above was 0.67. (a) Construct a 95% confidence interval estimate of population mean score a student get in the IELTS mock examination in 2022. (b) Test, at 1% level of significance, if the population proportion of students get 7 points or above in 2022 is higher than that in year 2017. (The report must include the (i) null hypothesis and alternative hypothesis, (ii) rejection region(s), (iii) calculation of test statistics, and (iv) conclusion.) (c) If the confidence interval estimate in part (a) is constructed at 99% instead of 95%, would the interval be (1) wider, (II) narrower, or (III) no change in width? (State your answer, no explanation is needed in part (C).)

The value of Tan A + B is 6.009.

[tex]sin^2 A + cos^2 A = 1sin^2 A = 1 - cos^2 Asin A = sqrt(1 - cos^2 A)sin A = sqrt(1 - (28/45)^2)sin A = sqrt(1 - 784/2025)sin A = sqrt(1241/2025)cos A = 28/45\geq[/tex]

We have to solve for Tan B

[tex]sin^2 B + cos^2 B = 1sin^2 B = 1 - cos^2 Bsin B = sqrt(1 - cos^2 B)sin B = sqrt(1 - (13/12)^2)sin B = sqrt(55/144)Therefore, tanB = sinB / cosB = (sqrt(55/144)) / (13/12) = (12 sqrt(55)) / 143[/tex]

[tex]tan(A+B) = (3207 \sqrt{55}) + 1804 \sqrt{124})) / (3115 - 207 \sqrt{(341)}[/tex]

= 6.009.

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The difference of p and z times the sum of p and z is p² - 2pz - z².

An algebraic expression in mathematics is an expression which is made up of variables and constants, along with algebraic operations such as addition, subtraction, division and multiplication.

Hence, the difference of p and z times the sum of p and z can be represented as follows:

Therefore,

(p - z) × (p + z) = (p - z) (p + z)

(p - z) (p + z)  = p² - pz - pz - z²

p² - pz - pz - z² = p² - 2pz - z²

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Answer:

Step-by-step explanation:The system of equations represented by the given points is:1x + 0y = 1   (equation 1)

1x + 6y = 1   (equation 2)

8x + 26y = 1  (equation 3)

8x - 22y = 1  (equation 4)

We can use the first equation to solve for x in terms of y:1x + 0y = 1

1x = 1

x = 1

Then we can substitute x=1 into equations 2-4 to obtain three equations in terms of y:1(1) + 6y = 1    =>   6y = 0   =>   y = 0

8(1) + 26y = 1   =>   26y = -7/8   =>   y = -7/208

8(1) - 22y = 1   =>   -22y = -7/8 - 7   =>   y = 15/44

Therefore, the solution to the system is x = 1 and y = -7/208 or y = 0 or y = 15/44.

However, we are given that x = -2, which contradicts our solution of x = 1. Therefore, there is no solution to the system with the given values.

Evaluating the given function, we can see that the intensity at 35 cm is 0.816

The intensity at a distance d in centimeters is given by the function:

I = 100*d⁻²

Here we want to find the intensity at 35 centimeters from the source, then we need to evaluate the function above at d = 35, doing that we will get:

I = 100*35⁻²

I = 100/(35²)

I = 0.816

The intensity at 35 centimeters from the source is 0.816.

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The range, difference of maximum and minimum values, is larger for Class C with a range of 10 compared to Class D with a range of 3. The data provided is not sufficient to calculate the IQR (Interquartile Range) for either class.

To begin, let's understand what the terms 'range' and 'IQR' (Interquartile Range) refer to in statistics. The range is simply the difference between the largest and smallest data values. The IQR is the range of the middle 50% of the values when ordered from lowest to highest.

For Class D, the lowest absence number is 0 and the highest is 3, thus the range is 3-0 = 3. For Class C, the lowest absence number is 0 and the highest is 10, so the range is 10-0 = 10. Thus, the range for the distribution of absences is larger for Class C.

Calculating IQR requires determining the 1st Quartile (Q1) and 3rd Quartile (Q3) values and subtracting Q1 from Q3. The data given doesn't include enough information to directly calculate these values and hence, it is impossible to definitively say which class has a larger IQR based on the provided information.

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Log₁₆(256b) in terms of the logarithm of b, which is the factor that was multiplied by 256 inside the logarithm is  2 + log₁₆(b)

We can use the product rule of logarithms, which states that the logarithm of a product is equal to the sum of the logarithms of the factors.

Therefore, we can write

log₁₆(256b) = log₁₆(256) + log₁₆(b)

We can simplify log₁₆(256) as follows:

log₁₆(256) = log₁₆(16^2) = 2

Therefore, we have:

log₁₆(256b) = log₁₆(256) + log₁₆(b)

= 2 + log₁₆(b)

So, we have rewritten log₁₆(256b) in terms of the logarithm of b, which is the factor that was multiplied by 256 inside the logarithm.

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The area of triangle ABC is :  33.44 cm²

The area of a triangle is the region enclosed within the sides of the triangle. The area of a triangle varies from one triangle to another depending on the length of the sides and the internal angles. The area of a triangle is expressed in square units, like, m2, cm2, in2, and so on.

We have the information from the question is:

In triangle ABC :

AB=6 cm,

AC=15 cm, and

m ∠A=48°.

We have to find the area of Triangle ABC

Area of triangle is : 1/2 × bc × (SinA)

Area of Triangle = 1/2 × 6 × 15 × (Sin 48° )

We know the value of Sin 48° is 0.7431

Area of Triangle = (1/2) × 66.88

Area of triangle = 33.44 cm²

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Answer:

Step-by-step explanation:

[tex]\frac{x}{18} =cot~30\\x=18cot~30=18 \times \sqrt{3} =18\sqrt{3} \\base=36\sqrt{3} \\area~of~triangle=3\times \frac{1}{2} \times36\sqrt{3} \times 18=972\sqrt{3} \approx 1683.55~mm^2[/tex]

The lengths of the legs, the circumference and the radius of the circles are;

First part; The diameter = 18 units

The second leg = 9 units

Second part; The circumference = 34·π units

Third part; The radius is r = 36 cm

The radius of a circle is the distance from the center to the circumference.

First part;

cos(30°) = 9·√3 ÷ The diameter = (√3)/2

The diameter = (9·√3) ÷ ((√3)/2) = 18

The length of the other leg, using the relationship between special triangles is 18/2 = 9

Second part;

The lengths of the legs of the right triangle indicates that we get;

The square of the diameter, D² = 30² + 16² = 1156

The length of the diameter, D = √(1156) = 34

The formula for finding the circumference of a circle, C, based on the diameter can be presented as follows;

C = π × D

Therefore;

The circumference of the circle, C = π × 34

The circumference of the circle = 34·π units

Third part

The length of the 20° arc = 4·π cm

Therefore;

4·π = 2·π·r × 20/360

Where;

r = The radius of the circle

Therefore;

r = (360/20) × 4·π ÷ (2·π) = 36

The radius of the circle, r = 36 cm

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Since ΔJKL was transformed to create ΔMNO, the sequence of transformations that were used include the following:

Reflection over the y-axis.

(x, y) → (x, y - 9).

In Geometry, a reflection over or across the y-axis or line x = 0 is represented and modeled by this transformation rule (x, y) → (-x, y).

By applying a reflection over the y-axis to the coordinate of the given triangle JKL, we have the following coordinates:

Coordinate J = (3, 7)   →  Coordinate J' = (-(3), 7) = (-3, 7).

Next, we would apply a vertical translation by shifting triangle J'K'L' down by 9 units to produce ΔMNO;

(x, y)                   →         (x, y - 9).

J' (-3, 7)               →        (-3, 7 - 9) = M' (-3, -2).

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Note that the difference in the radii of the two circular tracks is about 13.68 ft

recall that the circumference of a circle is denoted by the expression

C = 2πr

In this case, C = Circumference
and r   radius

Since we have two tracks

Let ra = radius of the purple track and

rb = radius of the yello track

So

265 = 2πra

170 = 2πrb

making ra and rb subject of the expression we have

ra = 256/(2π)  =40.7436654315 ≈ 40.74

rb = 170 / (2π) = 27.0563403256 ≈ 27.06

Hence, the difference is

40.74 - 27.06 = 13.68ft

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Full Question:

at a fair Daniel and Claire went on a ride that has two separate circular tracks. Daniel rode in a purple car that travels a total distance of 265 feet around the track . Ciara rode in a yellow car that travels a total distance of 170 feet around the track. They drew drew a sketch of the ride. What is the difference ofthe radii of the two circle tracks

Answer:

it is b got it right on my test

Step-by-step explanation:

The required graph that shows Ellen's steps where her mean steps for a week is 10000 is graph 2 where the data are as,

Day 1 = 11000 , Day 2 = 9000, Day 3 = 10000, Day 4 = 12000, Day 5 = 8000, Day 6 ≈ 10000, Day 7 ≈ 10000

Mean of ungrouped data is the average of all the values of the observation in the data set.

Mean = (Summation of all values in data set)/ (Number of observations)

Here there are 7 observations and the mean number of steps is 10000.

For graph 1 the values for each day of the week are as,

Day 1 = 8000 , Day 2 = 9000, Day 3 = 12000, Day 4 = 2000, Day 5 = 6000, Day 6 = 10000, Day 7 = 10000

Thus, the mean number of steps is = (8000+ 9000+ 12000+ 2000+ 6000+ 10000 + 10000) / 7

= 8143 (approximately) ≠ 10000

For graph 2 the values for each day of the week are as,

Day 1 = 11000 , Day 2 = 9000, Day 3 = 10000, Day 4 = 12000, Day 5 = 8000, Day 6 ≈ 10000, Day 7 ≈ 10000

Thus, the mean number of steps is = (11000+ 9000+ 10000+ 12000+ 8000+ 10000 + 10000) / 7

= 10000

For graph 3 the values for each day of the week are as,

Day 1 = 7000 , Day 2 = 14000, Day 3 = 11500, Day 4 = 10000, Day 5 = 4500, Day 6 = 6000, Day 7 = 11000

Thus, the mean number of steps is = (7000+ 14000+ 11500+ 10000+ 4500+ 6000 + 11000) / 7

= 9143 (approximately) ≠ 10000

For graph 4 the values for each day of the week are as,

Day 1 = 8000 , Day 2 = 9000, Day 3 = 9500, Day 4 = 10000, Day 5 = 9000, Day 6 = 7000, Day 7 = 11000

Thus, the mean number of steps is = (8000+ 9000+ 9500+ 10000+ 9000+ 7000 + 11000) / 7

= 9071 (approximately)  ≠ 10000

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The value of the expression when x = 4 is 12.

The value of the expression when x = -8 is 6.

We have,

To evaluate the expression 10 + x/2 for a given value of the variable x, we simply substitute the value of x into the expression and simplify.

For example:

If x = 4:

10 + x/2 = 10 + 4/2

= 10 + 2

= 12

So when x = 4, the value of the expression is 12.

If x = -8:

10 + x/2 = 10 + (-8)/2

= 10 - 4

= 6

So when x = -8, the value of the expression is 6.

Thus,

The value of the expression when x = 4 is 12.

The value of the expression when x = -8 is 6.

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The two-way table constructed from the data showing the numbers of boys and girls that choose each dance category should look like this:

                         Boys      Girls         Total

Fall Fling            82          81             163

Spring Social     62         75             137

Total                 144        156            300

A two-way table is a data display tool that displays the frequencies for two different categories collected from a single group.

A two-way table use rows to represent one category variable and the columns to represent a second category variable, while the cells display the proportions or frequencies of each row and column combination.

The total number of boys in eighth grade = 144

The number of boys who choose to attend the Fall Fling = 82

Therefore, the number of boys who attend the Spring Social = 62 (144 - 82)

The total number of girls in eighth grade = 156

The number of girls who choose to attend the Spring Social = 75

Therefore, the number of girls who attend the Fall Fling = 81 (156 - 75)

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Answer:

62,  82, 163, 300

Step-by-step explanation:

I got this on study island

Answer:

Step-by-step explanation:

Answer:

2.38 s

4.62 s

Step-by-step explanation:

The height of a projectile is given by the function:

[tex]h(t) = -t^2 + 7t + 9[/tex]

where:

To determine the time(s) at which the rocket is 20 ft above the ground, set h(t) = 20 and solve for t.

[tex]-t^2+7t+9=20[/tex]

To solve the equation for t, complete the square.

Subtract 9 from both sides:

[tex]-t^2+7t=11[/tex]

Divide both sides by -1:

[tex]t^2-7t=-11[/tex]

Add the square of half the coefficient of the term with the variable "t" to both sides of the equation:

[tex]t^2-7t+\left(\dfrac{-7}{2}\right)^2=-11+\left(\dfrac{-7}{2}\right)^2[/tex]

Simplify:

[tex]t^2-7t+\dfrac{49}{4}=\dfrac{5}{4}[/tex]

We have now created a perfect square trinomial on the left side of the equation. Factor the perfect square trinomial:

[tex]\left(t-\dfrac{7}{2}\right)^2=\dfrac{5}{4}[/tex]

To solve for t, square root both sides:

[tex]t-\dfrac{7}{2}=\pm \sqrt{\dfrac{5}{4}}[/tex]

[tex]t-\dfrac{7}{2}=\pm\dfrac{\sqrt{5}}{2}[/tex]

Add 7/2 to both sides of the equation:

[tex]t=\dfrac{7}{2}\pm\dfrac{\sqrt{5}}{2}[/tex]

[tex]t=\dfrac{7\pm\sqrt{5}}{2}[/tex]

Therefore, the times at which the rocket is 20 ft above the ground is:

[tex]t=\dfrac{7-\sqrt{5}}{2}=2.38\; \rm s[/tex]

[tex]t=\dfrac{7+\sqrt{5}}{2}=4.62\; \rm s[/tex]

Answer:

B. The candidate received at least 250 votes in half of the 6 towns.

This is because the median number of votes is the middle value when all the vote counts are put in order. This means that at least three towns gave the candidate more than 250 votes, and at least three towns gave the candidate fewer than 250 votes. So, the candidate received at least 250 votes in half of the 6 towns. The total number of votes the candidate received in the election cannot be determined from this information. Similarly, the number of votes received in individual towns cannot be determined.

The system of equations in the context of this problem is given as follows:

The solution is given as follows:

h = 35, c = 55.

The variables used for the system of equations are given as follows:

Considering the Monday's earnings, the equation is given as follows:

5h + 3c = 340.

Considering the Wednesday's earnings, the equation is given as follows:

2h + 6c = 400.

Hence the system is:

Simplifying the second equation by 2, we have that:

h + 3c = 200.

Subtracting the first equation by the second, we can obtain the value of h as follows:

4h = 140

h = 140/4

h = 35.

Hence the value of c is given as follows:

c = (200 - 35)/3

c = 55.

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The number of cans brought by Joes team is J = 299 cans

Given data ,

Let the number of cans brought by Joes team be J

Now , sage team brought 224 cans

And , sage team brought 3 times cans as asifs team

And , joes team brought 4 times as asifs team

On simplifying the equation , we get

Asif's team brought 224/3 = 74.67 (rounded to the nearest whole number) cans.

Joe's team brought 74.67 x 4 = 298.68

So , J = 299 cans

Hence , the equation is solved and J = 299 cans

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Step-by-step explanation:

First group of five    total score   =    5 * 6.5 = 32.5

Group of fifteen total score         =   15 * 4.5 = 67.5

total score of  20    is    67.5 + 32.5 = 100

Average =  100 /20 = 5.0

Answer:

92.

Step-by-step explanation:

To find the range of the set of numbers 20, 3, 95, 54, 71, 66, we first need to find the difference between the largest and smallest numbers in the set.

Largest number = 95

Smallest number = 3

Range = Largest number - Smallest number

Range = 95 - 3

Range = 92

Therefore, the range of the set of numbers 20, 3, 95, 54, 71, 66 is 92.

Answer:92

Step-by-step explanation: highest-lowest

it helps to put the terms in order first

95-3=92

Answer:

C)  1/2  and 8

Step-by-step explanation:

-2x + y = 7              Eq. 1

6x + y = 11               Eq. 2

From Eq. 1:

y = 7 + 2x               Eq. 3

From Eq. 2:

y = 11 - 6x              Eq. 4

Equalyzing Eq. 3 and Eq. 4:

7 + 2x = 11 - 6x

2x + 6x = 11 - 7

8x = 4

x = 4/8

x = 1/2

From Eq. 3:

y = 7 +2* 1/2

y = 7 + 1

y = 8

Check:

From Eq. 2

6x + y = 11

6*1/2 + 8 = 11

3 + 8 = 11

In the circle the measure of arc YP as per the given measurements is equal to option D. 166°

In the circle,

Measure of arc ST = 70°

Measure of arc RS = 96°

Measure of angle YZP = 128°

Arc formed by vertically opposite angles are equal

This implies,

Measure of arc YT = Measure of arc PR

Measure of arc ST + Measure of arc RS + Measure of arc YT = 180°

⇒70° + 96° + Measure of arc YT = 180°

⇒Measure of arc YT = 180° - 166°

⇒Measure of arc YT = 14°

⇒ Measure of arc PR = 14°

Measure of arc YP = 360° - ( Measure of arc (PR + RS + ST + TY ))

⇒ Measure of arc YP = 360° - ( 14° + 96° + 70° + 14° )

⇒Measure of arc YP = 360° -194°

⇒ Measure of arc YP = 166°

Therefore, the measure of arc YP is equal to option D. 166°

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The probability of the task taking between 48 and 49.9 minutes is 0.1357 or 13.57% (rounded to four decimal places).

To calculate this probability, we first need to find the total probability of the entire range between 43 and 57 minutes. This can be found by subtracting the lower bound from the upper bound and dividing by the total range:

P(43 < X < 57) = (57 - 43) / (57 - 43) = 1

Since the probability of the entire range is 1, we can find the probability of any sub-interval by dividing the length of the sub-interval by the length of the total range. Therefore, the probability of the task taking between 48 and 49.9 minutes is:

P(48 < X < 49.9) = (49.9 - 48) / (57 - 43) = 1.9 / 14 = 0.1357 or 13.57%

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The expression 16x⁹y¹⁸√3xy is equivalent to √769x¹⁹y³⁷

If a > O and y > 0, which expression is equivalent to √769x¹⁹y³⁷

Factor and rewrite the radicand in exponential form:

√16²×3x¹⁸×xy³⁶×y

Simplify the radical expression:

16x⁹y¹⁸ √2²×3xy

Factor and rewrite the radicand in exponential form:

8x⁹y¹⁸√2²×3xy

8x⁹y¹⁸.2√3xy

Multiply the monomials:

16x⁹y¹⁸√3xy

Hence, the expression 16x⁹y¹⁸√3xy is equivalent to √769x¹⁹y³⁷

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Anna will make  $1602.71 in all over 10 weeks.

Here, the first week salary of Anna = $116

i.e., the initial salary a =  $116

She was guaranteed a raise of 7% each week.

so, her salary in the next week would be,

116 + 7% of 116

7 percent of 116 is:

116 × 7/100 = 8.12

so, her salary will be $124.12

So, the equation for the salary after 'm' weeks would be,

[tex]n = 116\times (1 + 0.07)^{m - 1}\\\\n = 116\times (1 .07)^{m - 1}[/tex]

Using this equation , the salary in the second week m = 2 would be,

n = 124.12

Salary in the third week m = 3 would be,

n = 116 × [tex](1.07)^{3-1}[/tex]

n = 132.81

Salary in the fourth week m = 4 would be,

n = 116 × [tex](1.07)^{4-1}[/tex]

n = 142.11

Salary in the fifth week m = 5 would be,

n = 116 × [tex](1.07)^{5-1}[/tex]

n = 152.05

Salary in the sixth week m = 6 would be,

n = 116 × (1.07)⁶⁻¹

n = 162.7

Salary in the seventh week m = 7 would be,

n = 116 × (1.07)⁷⁻¹

n = 174.08

Salary in the eighth week m = 8 would be,

n = 116 × (1.07)⁸⁻¹

n = 186.27

Salary in the nineth week m = 9 would be,

n = 116 × (1.07)⁹⁻¹

n = 199.31

And the salary after the tenth week m = 10 would be,

n = 116 × (1.07)¹⁰⁻¹

n = 213.26

The total money after 10 weeks would be,

T = 116 + 124.12 + 132.81+ 142.11 + 152.05 + 162.7 + 174.08 + 186.27 + 199.31 + 213.26

T = $1602.71

This is the required amount.

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10: HL

11: AAS

12: SSS

13: SSS

14: HL

15: AAS

16: AAS

17: HL

18: AAS

19: SAS

20 AAS

1: SAS

2: ASA

3: HL

4: HL

5: ASA

6: ASA

7: HL

8: SSS

9: HL