Solve for m∠D:
87 A B C D

The area of the forest after 13 years is  1990.08 [tex]km^2[/tex].

What is exponential decay?

The term "exponential decay" in mathematics refers to the process of a constant percentage rate reduction in an amount over time. It can be written as y=a(1-b[tex])^x[/tex], where x is the amount of time that has passed, an is the initial amount, b is the decay factor, and y is the final amount.

Here initial area a = 3500[tex]km^2[/tex] ,

Rate of decay = 4.25%= 4.25/100 = 0.0425

Now using exponential decay formula then,

=> [tex]y=a(1-b)^x[/tex] where x=13 years .

=> y = [tex]3500(1-0.0425)^{13}[/tex]

=> y = [tex]3500(0.9575)^{13}[/tex]

=> y = 1990.08 [tex]km^2[/tex]

Hence the area of the forest after 13 years is  1990.08 [tex]km^2[/tex].

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The height of the building , given the shadow that it casts would be 93 meters tall .

Utilizing proportions can be an effective method to tackle this issue. By defining ' h ' as the height of the building responsible for casting a 39.75 meter shadow, we may establish the ensuing proportion :

h / 39. 75 = 3. 2 / 1. 37

Solving for h gives us :

h / 39. 75 = 2. 336

h = 2. 336 x 39. 75

h = 92. 8 56

h = 93 meters tall

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The equation of the ellipse is:

x²/64 + y²/289 = 1

The standard form of the equation of an ellipse with centered at the origin and major axis parallel to the vertical is:

x²/b² + y²/a² = 1

Coordinates of covertices are ( ±b, 0) = (±8, 0)

Coordinates of foci are (0, ±c) = (0, ±15)

Also c²= a²-b²

a² = b² + c²

a² = 8² + 15²

a² =289

a = √289

a = 17

Using x²/b² + y²/a² = 1:

x²/8² + y²/17² = 1

x²/64 + y²/289 = 1

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Complete Question

Write an equation for an ellipse centered at the origin, which has foci at (0,±15) and co-vertices at (±8,0)

Our calculated t-value of [tex]4.22[/tex] is greater than the critical t-value of 1.684, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the mean starting salary for college graduates who major in Statistics has increased.

We can use a one-sample t-test to test the claim that the mean starting salary for college graduates who major in Statistics has increased.

The null hypothesis is that the mean starting salary is still $[tex]54000[/tex], while the alternative hypothesis is that the mean starting salary is greater than $[tex]54000[/tex].

We can calculate the t-statistic as:

t = (sample mean - null hypothesis value) / (standard error of the mean)

= [tex](63500-54000)/(10540/\sqrt{40}[/tex]

= [tex]4.22[/tex]

where [tex]\sqrt{40}[/tex] is the sample size.

The degrees of freedom for the t-test is [tex](n-1) = 39[/tex], where n is the sample size.

Using a t-table or calculator with [tex]39[/tex] degrees of freedom, a one-tailed test, and a significance level of [tex]0.05[/tex], we find the critical t-value to be [tex]1.684[/tex]

Since our calculated t-value of [tex]4.22[/tex] is greater than the critical t-value of [tex]1.684[/tex], we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the mean starting salary for college graduates who major in Statistics has increased.

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The calculated value of x from the special triangle is 17.0

From the question, we have the following parameters that can be used in our computation:

The right triangle

The right triangle is a special triangle that has a measure of 45 degrees

For a triangle that has a measure of 45 degrees, we have

Hypotenuse = Legs√2

In this case, we have

Leg = x

Hypotenuse = 24

So, we have

x√2 = 24

Divide by √2

x = 24/√2

Evaluate

x = 17.0

Hence, the calculated value of x is 17.0

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Answer: X=3

Step-by-step explanation:

a) The number of weeks that she has been saving when she needs $277 is given as follows: 17 weeks.

b) The amount she will need after 5 weeks is given as follows: $505.

The function for this problem is defined as follows:

A(w) = 600 - 19w.

The output of the function is the amount remaining after w weeks.

Hence the number of weeks that she has been saving when she needs $277 is obtained as follows:

277 = 600 - 19w

19w = 600 - 277

w = (600 - 277)/19

w = 17 weeks.

The amount that she needs after 5 weeks is given as follows:

A(5) = 600 - 19 x 5

A(5) = $505.

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If points A, B, and C are not three nonadjacent vertices of the cube or three adjacent vertices of the cube, then the cross-section will be a polygon with more than four sides, and its shape will depend on the location of the points A, B, and C on the cube.

What is meant by cross section?

The form we obtain by cutting a straight through an object is called a cross-section.

The shape of the cross section of a cube that is cut into two pieces by a single slice that passes through point A, B, and C depends on the location of these points on the cube.

If points A, B, and C are three nonadjacent vertices of the cube, then the cross section will be a triangle. Specifically, it will be an isosceles triangle with base equal to the length of the edge of the cube and legs equal to the distances between point A and the other two points B and C.

If points A, B, and C are three adjacent vertices of the cube, then the cross section will be a square. Specifically, it will be a square with side length equal to the length of the edge of the cube.

If points A, B, and C are not three nonadjacent vertices of the cube or three adjacent vertices of the cube, then the cross section will be a polygon with more than four sides, and its shape will depend on the location of the points A, B, and C on the cube.

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The probability of correctly guessing which pies were sold on the first day is  near to 2/3.

The Fundamental Counting Principle states that if there are m number of ways to do an activity and n number of ways for a second activity, then there are m x n ways to do the two activities together.

In this case, there are 7 pies and 4 of them were sold on the first day. Therefore, the probability of correctly guessing which pies were sold on the first day is 4/7.

This can be written in simplest form as 2/3.

In conclusion, the probability of correctly guessing which pies were sold on the first day is 2/3.

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question 9.

Based on a cubic modeling equation, the  prediction about this data that is  most reliable is a $70k house would have 1,354 square ft.

Option A is correct.

Question 10.

The second coordinate in a scatterplot of the linearized data and  rounding the answer to the tenth place is 0.8

Option A is correct.

Question 11.

The approximate value when x=4 and rounding the answer to the nearest whole number is 1,168

Option C is correct.

The equation is given as

Log y= 0.75821x+0.03442

We substitute x = 4 into the given equation and solve for y:

log y = 0.75821(4) + 0.03442

log y = 3.07286 + 0.03442

log y = 3.10728

We take the antilogarithm of both sides,

y = 10^(3.10728)

y = 1,168.66

Rounding off, the closest answer is option  (c) 1,168.

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Answer:

We should use 30 pints of the first drink and 120 pints of the second drink

Step-by-step explanation:

Let x represent the number of pints of the first drink used
Volume of fruit juice in x pints = 35% of x = 0.35x

Let y represent the number of pints of the second drink used
Volume of fruit juice in y pints = 85% of y= 0.85y

Since the final mixture volume is 150 pints we get
x + y = 150  ..... [1]

If the final mixture has to contain 75% pure fruit juice, the volume of fruit juice in the mixture = 0.75 x 150 = 112.5 pints

Therefore
0.35x + 0.85y = 112.5  ...[2]

We will solve this set of simultaneous equations by making the coefficients of one of the variables equal and then subtracting to eliminate that variable

Multiply equation [1] by 0.35 to get the x coefficients equal

0.35 x [1]
=>  0.35(x + y) = 0.35(150)

=> 0.35x + 035y = 52.5  ... [3]

Subtract [3] from [2]:

[3] - [2] =>

0.35x + 0.85y - (0.35x + 035y) = 112.5 - 52.5

0.35x + 0.85y - 0.35x - 0.35y = 60

0x + 0.5y = 60

0.5y = 60

y = 60/0.5

y = 120

Therefore x = 150 - 120 = 30

Therefore we should use 30 pints of the first drink and 120 pints of the second drink

Answer:

3 divide 2/5 is 7.5 since 2/5 is 0.4

The required, we can put 125 bottles of water or 8 cans of food in each box, and each box will contain the same number of items.

To find the largest number of relief supplies that can be put in each box, we need to find the greatest common factor (GCF) of 2250 and 144. We can do this by listing the prime factors of each number:

2250 = 2 * 3 * 3 * 5 * 5 * 5 * 2

144 = 2 * 2 * 2 * 2 * 3 * 3

To find the GCF, we need to find the highest power of each common prime factor that appears in both numbers. In this case, the common prime factors are 2 and 3, so we take the highest power of each:

2 * 3 * 3 = 18

So the largest number of relief supplies that can be put in each box is 18. We can check that this works by dividing each quantity by 18:

2250/18 = 125

144/18 = 8

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Working out the expression 320.041 - 47.96 gives 272.081

From the question, we have the following parameters that can be used in our computation:

320.041 - 47.96

The above expression is a difference or subtraction expression that can be done using a calculator

Using the above as a guide, we have the following:

320.041 - 47.96

Evaluate using a calculator

So, we have

320.041 - 47.96 = 272.081

Hence, teh solution of 320.041 - 47.96 is 272.081

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After 16 minutes of decay, approximately 182.3 grams of the element remains.

To calculate how much of the element remains after 16 minutes of decay, we can use the formula for exponential decay:

A = A₀ * [tex]e^{(-kt)[/tex]

where A is the amount of the element remaining after time t, A₀ is the initial amount of the element, k is the decay constant, and t is the elapsed time.

In this case, we have A₀ = 420 grams, k = 0.118 (since the element decays by 11.8% per minute), and t = 16 minutes. Plugging these values into the formula, we get:

A = 420 * [tex]e^{(-0.118*16)[/tex]

A = 182.3 grams

We round our answer to the nearest 10th of a gram, as requested in the problem.

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The centerline=  459.9

The upper control limit=  852.55.

The lower control limit = 67.25.

We determine the mean to set up the standard limits:

Mean is gotten as :

(465 + 502 + 395 + 488 + 472 + 462 + 524 + 434 + 481 + 386) / 10

mean = 459.9

we then find the Standard Deviation:

Standard Deviation = √variance

variance = ((465-459.9)^2 + (502-459.9)^2 + (395-459.9)^2 + (488-459.9)^2 + (472-459.9)^2 + (462-459.9)^2 + (524-459.9)^2 + (434-459.9)^2 + (481-459.9)^2 + (386-459.9)^2) / 9

variance = 17119.1

Standard Deviation = √(17119.1)

Standard Deviation= 130.88

Hence

Centerline = 459.9

upper control limit  = 459.9 + 3 x (130.88) = 852.55

lower control limit. = 459.9 - 3 x (130.88) = 67.25

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Answer: B) their preferences for recreation

Step-by-step explanation:

Using the quadratic equation, the solution is x = -8 and x = -17.

A second-order equation of the form ax2 + bx + c = 0 denotes a quadratic equation, where a, b, and c are real number coefficients and a 0.

The equation is x²+25x+136=0

where a, b, and c are the coefficients of the quadratic equation ax² + bx + c = 0.

For the given equation x² + 25x + 136 = 0, we can identify a = 1, b = 25, and c = 136.

Substituting these values in the quadratic formula, we get:

x = (-25 ± √(25² - 4(1)(136))) / 2(1)

x = (-25 ± √(625 - 544)) / 2

x = (-25 ± √(81)) / 2

imply them

x = (-25 + 9) / 2 or x = (-25 - 9) / 2

x = -8 or x = -17

Therefore, the solutions to the quadratic equation x^2 + 25x + 136 = 0 are x = -8 and x = -17.

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The coordinate of points K, L and M is the location of the points on a coordinate plane

Here, we have,

to determine the missing coordinates:

The given parameters are:

K = (10, )

L = ( ,10)

M = (30, )

The question has missing parameters.

So, I will assume that the line is a perfectly horizontal line.

This means that the y-coordinates of points K, L and M are equal.

The y-coordinate of point L is 10.

So, we have:

K = (10, 10)

L = ( ,10)

M = (30, 10)

Assume that point L is halfway points K and M, then we have:

K = (10, 10)

L = (20, 10)

M = (30, 10)

See attachment for the diagram of the coordinate plane showing the line

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Answer:

we have to first start converting the given volume to liters, since the answer is expected in hectoliters.

36 liters + 4,000 milliliters = 36 liters + 4 liters = 40 liters

1 hectoliter = 100 liters

Therefore, we have:

40 liters = 0.4 hectoliters

So, we have 0.4 hectoliters of fuel.

Answer:

0.4 hectoliters of fuel

Step-by-step explanation:

We Know

You have 36 liters of fuel and just added 4,000 milliliters more.

How many hectoliters of fuel do you have?

1000 milliliters = 1 liters

4000 milliliters = 4 liters

We Take

36 + 4 = 40 liters of fuel

We Convert 40 liters to hectoliters

1 liter = 0.01 hectoliters

40 liters = 40 x 0.01 = 0.4 hectoliters

So, you have 0.4 hectoliters of fuel.

We can rewrite 3^6 as the product of two whole-number powers of 3 as: [tex]3^4 . 3^2.[/tex]

To rewrite 3^6 as the product of two whole-number powers of 3, we need to find two exponents that add up to 6.

A way to do this is to use the fact that 3^4 = 81, which is close to 3^6.

Specifically, we have:

3^6 = 3^4 * 3^2

We can check that this is true by using the rule for multiplying exponents that states that a^m * a^n = a^(m+n).

Applying this rule, we get:

= 3^4 * 3^2

= 3^(4+2)

= 3^6.

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Answer:

To calculate the monthly payments for each option, we can use the PMT function in Excel or a financial calculator.

For the new car option:

Loan amount = $31,000

Interest rate (annual) = 7.19%

Loan term = 3 years

Number of payments = 3 x 12 = 36

Using the PMT function in Excel, we get:

PMT(7.19%/12, 36, -$31,000) = $966.59

For the used car option:

Loan amount = $16,000

Interest rate (annual) = 5.89%

Loan term = 4 years

Number of payments = 4 x 12 = 48

Using the PMT function in Excel, we get:

PMT(5.89%/12, 48, -$16,000) = $385.36

The difference in monthly payments between the two options is:

$966.59 - $385.36 = $581.23

Therefore, the difference in monthly payments between financing the new car and financing the used car is $581.23.

Step-by-step explanation:

The average speed of the bike is calculated as: 15.67 mph

The formula for the average speed is:

Average speed = Distance/Speed

We are given the parameters as:

Time = 180 minutes = 3 hours

Distance = 47 miles

The average speed is defined as the total distance traveled by the object in a particular time interval. The average speed is a scalar quantity. It is represented by the magnitude and does not have direction.

Thus:

Average speed = 47/3

= 15.67 mph

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7. The exact values of x is 17.

8. The exact values of x is 12.

9. The exact values of x is 10.9.

10. The exact values of y is 15.

11. The exact values of y is 15.

12. The exact values of y is 6.

The exact values of x and y is calculated by applying Pythagoras theorem as follows;

7.

x² = 8² + 15²

x² = 289

x = √289

x = 17

8.

x² = 15² - 9²

x² = 144

x = √144

x = 12

9.

x² = 12² - 5²

x² = 119

x = √119

x = 10.9

The value of y is calculated as follows;

10.

y² = 12² + 9²

y² = 225

y = √(225)

y = 15

11.

y² = 17² - 8²

y² = 225

y = √(225)

y = 15

12.

y/10 = 6/10

y = 6

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First, you need to find out how much ribbon she has total. Because there are two spools, you multiply 24 x 2. The answer for that is 48. Then you have to figure out how many bows she can make. Each bow uses 1/8 of a piece of yard of ribbon, so you can divide 48 by 8. Your answer would then be 6 bows.

Here are the equations.

[tex]24\times2 = \text{X}[/tex]

[tex]\dfrac{\text{X}}{8} =\text{B}[/tex]

Where X represents the yards of ribbon available and B represents the number of bows that Mari can make.

Complete Question:

Mari makes bows for floral arrangements. She uses 1/8 of a yard ribbon for each bow. How many bows can she make from 2 spools that each has 24 yards ribbon?

Answer:

vertical because they are right across from each with the same degrees

Step-by-step explanation:

See image

If in sample of 210 students from district and finds 35 have submitted "box-tops" this year, then

(a) Yes, data provide enough evidence regarding contest which has increased participation in "box-tops" for "education-program",

(b) the p-value is 0.0187.

Part(a) : To find if contest has increased the participation in "box-tops" for "education-program", we conduct a "hypothesis-test".

We use a "significance-level" of 0.05, which means we will accept a 5% chance of making a "Type I Error",

Let Null Hypothesis ⇒ The proportion of students submitting "box-tops" is = 12%,

Let Alternative Hypothesis ⇒ The proportion of students submitting "box-tops" for education is increased after contest,

To test above hypothesis, we use "one-sample" proportion-test.

We will use the sample proportion (p') = 35/210 = 0.1667, as estimate of  population proportion (p).

So, "test-statistic" (z) is ⇒ z = (p' - p₀)/√(p₀ × (1-p₀)/n),

where n = sample size,

In "Null-Hypothesis", the "expected-value" of the test statistic is 0, and  "standard-deviation" of the "test-statistic" is √(p₀×(1-p₀)/n),

Substituting, p₀ = 0.12 and n = 210,

We get,

⇒ z = (0.1667 - 0.12)/√(0.12 x 0.88/210) ≈ 2.08,

Now, we know that "right-tailed" critical value at 0.05 level is = 1.645,

Since, z (2.08) > 1.645, the Null-Hypothesis is Rejected.

So, above data provides enough evidence regarding contest has increased the participation in "box-tops" for "education-program".

Part(b) : We need to find the "p-value" for the test-statistic that we found in part(a),

So, p-value = P(z>2.08),

⇒ 1 - P(z ≤ 2.08),

⇒ 1 - 0.9813,

⇒ 0.0187,

Therefore, the required "p-value" is 0.0187.

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Answer:

Step-by-step explanation:

Given you have collected 70 coins and stamps, and your friend has collected 1/5 as many coins and twice as many stamps, for a total of 95 coins and stamps, you want the numbers of each.

We can let x represent the number of coins you have collected. Then 70 -x is the number of stamps you have collected. The corresponding numbers for your friend are x/5 coins and 2(70-x) stamps. Your friend's collection is ...

  x/5 +2(70 -x) = 95

  -9/5x +140 = 95 . . . . . . . . simplify

  45 = 9/5x . . . . . . . . . . . . . add 9/5x -95

  25 = x . . . . . . . . . . . . . . . . multiply by 5/9

  70 -x = 45

  1/5x = 5

  2(70 -x) = 90

You collect 25 coins and 45 stamps; your friend collects 5 coins and 90 stamps.

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