Question 5 (1 point)
|8|=

Answer:

The speed of the balloon is 0.16 m/s.

Step-by-step explanation:

CD = 300 m

Let AD = x

AB = y

time, t = 3 min

Triangle, ADC

\(tan 15 = \frac{AD}{BC}\\\\0.27 \times 300 = x \\\\x = 80.4 m\)

Triangle, BCD

\(tan 20 = \frac{BD}{BC}\\\\0.36 \times 300 = x + y \\\\x + y = 109.2 m\)

So, y = 109.2 - 80.4 = 28.8 m

Speed = 28.8/180 = 0.16 m/s

Answer:

The domain of a function is the set of all input values of the function. The range of a function is the set of all possible outputs of the function, given its domain.

N 7

Part A

In a direct variation

k=y/x

take one point from the graph

(2,4)

k=4/2

Part B

In a direct variation, the equation is

y=kx

we have

k=2

therefore

Answer:

-6

Step-by-step explanation:

-1+(-5)

You are moving down the numberline, then subtracting again because of the negative sign. -(1+5) is how I like to think of it.

So, 1+5=6, and you add the negative sign. -1+(-5)=-6.

Answer:

x would equal 8. because you would add AB and BC together and set it equal to AC

Hoped I helped

Can I get a brainliest

Answer:

0N, balanced

Step-by-step explanation:

There are equal forces on each side.

Hope this helps.

Please mark me Brainliest.

Answer: there's no orange fish

Answer:

b=99°

Step-by-step explanation:

108°+72°+81°+b=360°

b=99°

Answer:

24 full weeks needed

Step-by-step explanation:

30 + 3x = 100

3x = 70

x = 23.333

24 weeks needed

The domain restrictions of the expression q²−7q−8/q²+3q−4 are q ≠ -4 and q ≠ 1. (option c)

The denominator of the expression is q²+3q−4. To determine the values that would make the denominator equal to zero, we can set it equal to zero and solve for q:

q² + 3q - 4 = 0

Now, we can factorize the quadratic equation:

(q + 4)(q - 1) = 0

To find the values of q, we set each factor equal to zero and solve for q:

q + 4 = 0 or q - 1 = 0

Solving these equations, we get:

q = -4 or q = 1

So, the values of q that would make the denominator equal to zero are q = -4 and q = 1. These are the values we need to exclude from the domain of the expression to avoid division by zero.

Therefore, the correct answer is option c) q ≠ 1 and q ≠ -4.

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Complete Question:

What are the domain restrictions of q²−7q−8/q²+3q−4?

a) q≠1 and q≠−8

b) q≠−1 and q≠4

c) q≠1 and q≠−4

d) q≠−1 and q≠8

To begin, let's define some of the terms mentioned in the question. A random variable (RV) is a variable whose possible values are outcomes of a random phenomenon.

A non-negative RV is a random variable that can only take non-negative values (i.e. values greater than or equal to zero).

A variable is a quantity or factor that can vary in value.

Now, let's look at the problem at hand.

We are given that G is an uniform random variable on [-t,t]. This means that the probability distribution of G is uniform over the interval [-t,t].

We are also given that X is a non-negative RV that is independent of G. This means that the probability distribution of X is not affected by the values of G.

Finally, we are asked to show that for any t >= 0, it holds:

(smoothing Markov)

To prove this, we can use the definition of conditional probability.

P(X > x | G = g) = P(X > x, G = g) / P(G = g)

By independence, we know that P(X > x, G = g) = P(X > x) * P(G = g).

Since G is a uniform RV, we know that P(G = g) = 1 / (2t) for any g in [-t,t].

So, we can simplify the equation as:

P(X > x | G = g) = P(X > x) * (2t)

Now, we can use the law of total probability to find P(X > x), which is the probability that X is greater than x:

P(X > x) = ∫ P(X > x | G = g) * P(G = g) dg

where the integral is taken over the interval [-t,t].

Substituting in the equation we derived earlier, we get:

P(X > x) = ∫ P(X > x) * (2t) * 1/(2t) dg

Simplifying, we get:

P(X > x) = 2 * ∫ P(X > x) dg

Now, we can use the definition of expected value to find E(X):

E(X) = ∫ x * f(x) dx

where f(x) is the probability density function of X.

Using the same logic as before, we can find the probability that X is greater than or equal to t:

P(X >= t) = 2 * ∫ P(X >= t) dg

Substituting this into the original equation, we get:

(smoothing Markov)

Therefore, we have shown that for any non-negative RV X which is independent of G and for any t >= 0, it holds that:

(smoothing Markov)

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Answer: give me brainlist and i will tell u

Step-by-step explanation:

Inferential statistics involves using sample data to make inferences, predictions, or generalizations about a larger population, providing valuable insights and conclusions based on statistical analysis.

The term "inferential statistics" is associated with the task of making inferences or drawing conclusions about a population based on sample data.

In other words, it involves using sample data to make generalizations or predictions about a larger population.

Inferential statistics is concerned with analyzing and interpreting data in a way that allows us to make inferences about the population from which the data is collected.

It goes beyond simply describing the sample and aims to make broader statements or predictions about the population as a whole.

This branch of statistics utilizes various techniques and methodologies to draw conclusions from the sample data, such as hypothesis testing, confidence intervals, and regression analysis.

These techniques involve making assumptions about the underlying population and using statistical tools to estimate parameters, test hypotheses, or predict outcomes.

The goal of inferential statistics is to provide insights into the larger population based on a representative sample.

It allows researchers and analysts to generalize their findings beyond the specific sample and make informed decisions or predictions about the population as a whole.

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Answer:

not equivalent to meteorologists ratio

Step-by-step explanation:

meteorologists ratio is

rainy days : sunny days = 2 : 5

last months weather is

rainy days : sunny days

= 10 : 20 ( divide both parts by LCM of 10 )

= 1 : 2 ← not equivalent to 2 : 5

Hello,

\(sin(45^o)=\dfrac{\sqrt{2} }{2} \\\\sin(45^o)=\dfrac{a}{c} \\\\\dfrac{\sqrt{2} }{2}=\dfrac{a}{6} \\\\a=6*\dfrac{\sqrt{2} }{2}=3\sqrt{2}\\\)

Answer A

The number of subscribers with land-line phones for any year after 2000, taking into account the 27% annual decline rate.

According to the given information, land-line phones are decreasing at a rate of 27% each year since 2000. This means that each year, the number of subscribers with land-line phones is reduced to 73% (100% - 27%) of the previous year's value.

To create a formula that models this scenario, we start with the initial number of subscribers in the year 2000, which is given as 10,000. We can express this as y(0) = 10,000, where y(0) represents the number of subscribers in the starting year (t = 0).

Based on the decline rate of 27% per year, we can express the relationship between the number of subscribers in a given year (y) and the number of subscribers in the previous year (y-1) as follows:

y = 0.73 * y(t-1)

Here, 0.73 represents the probability of retaining a land-line phone subscription from one year to the next, given the 27% annual decline rate. Multiplying this probability by the number of subscribers in the previous year gives us the estimated number of subscribers in the current year.

By iterating this formula year after year, we can estimate the number of subscribers with land-line phones for any given year after 2000.

For example, let's calculate the number of subscribers in the year 2023, which is 23 years after 2000 (t = 23). Using the formula, we can substitute t = 23 into the equation:

y(23) = 0.73 * y(22)

To find y(22), we substitute t = 22 into the equation:

y(22) = 0.73 * y(21)

We continue this process until we reach the starting year, y(0) = 10,000.

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Complete Question:

With the growing popularity of cell phones, a local phone company had a sharp decline in the number of customers with land-line phones in their homes. Suppose land-line phones are decreasing at a rate of 27% each year since 2000. Assume that there were 10,000 subscribers with land-line phones in the year 2000.

For convenience let t = the number of years after 2000 and y = number of subscribers with land-line phones.

A formula that models this scenario is

The length of the side included between the angles A = 30° and B = 80° is :

AC = 11.2 centimeters.

To solve this problem, we need to use the formula for the area of a triangle:
A = (1/2)bh

where A is the area, b is the length of the base, and h is the height.

Let's label the triangle ABC, where angle A = 30° and angle B = 80°. We want to find the length of the side AC, which is the base of the triangle.

First, we need to find the height of the triangle. We can use the sine function to do this:

sin(80°) = h/AC

Rearranging this equation, we get:

h = AC * sin(80°)

Now we can substitute this into the formula for the area of a triangle:

50 = (1/2) * AC * h
50 = (1/2) * AC * (AC * sin(80°))
100 = AC^2 * sin(80°)
AC = sqrt(100/sin(80°))
AC ≈ 11.2 cm

So the length of the side AC is approximately 11.2 centimeters.

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In order to find the equivalent expression, let's expand the term in parenthesis:

So the correct option is A.

the partial derivative ∂z/∂y using implicit differentiation of e^z + sin(5x^2) + 2y = 0 is:
To calculate the partial derivative ∂z/∂y using implicit differentiation of e* + sin (5x^2) + 2y = 0, we first need to differentiate both sides of the equation with respect to y.

We get:

d/dy(e^z + sin(5x^2) + 2y) = d/dy(0)

Using the chain rule, the left-hand side becomes:

∂(e^z)/∂z * ∂z/∂y + ∂(sin(5x^2))/∂y + 2

We can simplify this by recognizing that ∂(sin(5x^2))/∂y = 0, since sin(5x^2) does not depend on y. Thus, we are left with:

∂(e^z)/∂z * ∂z/∂y + 2 = 0

Now, we need to solve for ∂z/∂y:

∂z/∂y = -2 / ∂(e^z)/∂z

To find ∂(e^z)/∂z, we differentiate e^z with respect to z, giving:

∂(e^z)/∂z = e^z

Substituting this into the expression for ∂z/∂y, we get:

∂z/∂y = -2 / e^z

Therefore, the partial derivative ∂z/∂y using implicit differentiation of e^z + sin(5x^2) + 2y = 0 is:

∂z/∂y = -2 / e^z

Note that we cannot simplify this any further without knowing the value of z.
To find the partial derivative ∂z/∂y using implicit differentiation for the equation e^z + sin(5x^2) + 2y = 0, we will first differentiate the equation with respect to y, treating z as a function of x and y.

Differentiating both sides with respect to y:

∂/∂y (e^z) + ∂/∂y (sin(5x^2)) + ∂/∂y (2y) = ∂/∂y (0)

Using the chain rule for the first term, we get:

(e^z) * (∂z/∂y) + 0 + 2 = 0

Now, solve for ∂z/∂y:

∂z/∂y = -2 / e^z

So, the partial derivative ∂z/∂y for the given equation is:

∂z/∂y = -2 / e^z

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The probability that the other child is also a girl, given that one of them is a girl, is 2/3 or approximately 0.6667.

To determine the probability that the other child is also a girl given that one of them is a girl, we need to consider the possibilities of the gender combinations for Alice's two children.

Let's denote the gender of the first child as G (girl) and B (boy), and the gender of the second child as G' and B'.

There are four possible combinations for the gender of the two children: GG, GB, BG, and BB.

However, we are given that one of the children is a girl. This eliminates the BB combination since we know both children cannot be boys.

Thus, we are left with three possible combinations: GG, GB, and BG.

Out of these three combinations, two of them involve at least one girl: GG and GB. This means there is a 2 out of 3 chance that the other child is a girl.

Therefore, the probability that the other child is also a girl, given that one of them is a girl, is 2/3 or approximately 0.6667.

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The value of the missing angles of the quadrilateral are:

a = 110°

b = 70°

c = 20°

Theorem sum of angles in a triangle states that they sum up to 180 degrees. Thus:

d = 180 - (78 + 32)

d = 70°

Similarly:

e = 180 - (78 + 32 + 18)

e = 20°

Sum of angles on a straight line is 180 degrees. Thus:

a = 180 - 70

a = 110°

We can see that e + b will be equal to 90 degrees because of the definition of right angles. Thus:

b = 90° - 20°

b = 70°

From sum of angles in a triangle as 180° is:

c = 180 - (90 + 70)

c = 20°

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Hence, the equation amount of dog food left in the bag is \(0.25\)

What is the equation?

The definition of an equation is a mathematical statement that shows that two mathematical expressions are equal.

Here given that,

Jada opens a \(5\)-kg bag of dog food and fills her dog's dish with \(0.25\) kg of the content once a day.

On the first time, the dog`s food left in the bag will be

\(5-0.25 ...(1)\)

On second time,  the dog`s food left in the bag will be

\(5-0.25 ...(2)\)

On the third time,  the dog`s food left in the bag will be

\(5-0.25 ...(3)\)

So in the \(n\) times the amount of the food left in the bag will be

\(W=5-0.25(n)\)

It would be in \(n\) times.

Hence, the equation amount of dog food left in the bag is \(0.25\)

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An example of the expression that is close to zero is explained below.

It should be noted that an expression means the relationship that is used to show the relationship that exists between the variables. In this case, it should be noted we are told to create an expression that is close to zero.

The expression will be:

3x - 1= 0.9

Collect like terms

3x = 0.9 + 1

3x = 1.9

x = 1.9/3

x = 0.633

Therefore, this expression is close to zero.

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The number of words he still need to write in one of his writing sessions would be = 95 minutes.

A graduate is an individual that has completed studies in an institution for a number of years and is being issued a certificate afterwards.

The number of words committed for a day by Nick = 500 words.

He writes 120 words = 30 mins

The total number of words remaining = 500 - 120 = 380 words.

Therefore the time it will take to finish the remaining 380 words in the day = X mins.

If 30 mins= 120 words

X mins = 380 words.

Make X mins the subject of formula;

X mins = 380× 30/120

X mins = 11,400/120

X mins= 95 minutes.

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There are 9 questions on the test.

Let, the number of multiple-choice questions on the test = m

the number of essay questions on the test = s

m + s = 26  .......(1)

So, We are also told that multiple choice is worth 3 points each, so the total number of points for m questions will be 3m.

As essays are worth 8 points each, so the total number of points for s questions will be 8s.

3m + 8s = 123 .......(2)

Now we will use the substitution method to solve a system of linear equations.

From equation (1) we will get,

m = 26 -s

Substituting this value in equation (2) we will get,

3 * (26- s) + 8s =123

78 - 3s + 8s = 123

5s = 123 -78

5s = 45

s = 45/5 = 9

So, there are 9 questions on the test.

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The complete question is:

Ada’s history teacher wrote a test for the class.

The test is 26 questions long and is worth 123 points.

Ada wrote two equations, where m represents the

number of multiple choice questions on the test, and

s represents the number of essay questions on the test.

m+s= 26

3m + 8s = 123

How many essay questions are on the test?

Answer:

To change from meters per day into kilometers per year, you just have to change from meters to kilometers and from days to years.  You just multiply/divide by the unit conversions (for example, 1000 meters per kilometer) in a way that cancels out the units and leaves just kilometers and years:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

b= -7

Step-by-step explanation:

First set up the equation y=mx+b.

The add what you know, so that:

y=2/5x+b

Now plug in (10,-3) to the equation to find b.

-3=2/5(10)+b

Simplify (2/5 * 10= 4)

-3=4+b

Get b alone by substracting four to each side.

b =-7

*I recommend checking my work, but this is what I put.

a) 3 - 8 + 2i - (-5i) = -5 + 2i +5i = -5 + 7i

b) 4 - 20i - 2i + 10i² = 4 - 22i + 10(\(\sqrt{-1}\))² = 4 - 22i + 10(-1) = 4 - 10 -22i

= -6 -22i

c) Multiply both numerator and denominator by i

\(\frac{i(-2-4i)}{i^{2} }\) = \(\frac{-2i - 4i^{2} }{i^{2} }\) = \(\frac{-2i - 4(\sqrt{-1})^{2} }{(\sqrt{-1})^{2}}\) = \(\frac{-2i - 4(-1)}{-1}\) = \(\frac{-2i+4}{-1}\) = 2i - 4 = -4 + 2i

d) Multiply both numerator and denominator by (3+6i)

\(\frac{(3+6i)(-3+2i)}{(3+6i)(3-6i)}\) = \(\frac{-9+6i-18i+12i^{2} }{9-36i^{2} }\) = \(\frac{-9-12i+12(\sqrt{-1} )^{2} }{9-36(\sqrt{-1} )^{2} }\) = \(\frac{-9-12i+12(-1)}{9-36(-1)}\) = \(\frac{-9-12-12i}{9+36}\)

= \(\frac{-21-12i}{45}\) = \(-\frac{21}{45} - \frac{12}{45}i\)