Can someone who's good at geometry pls help with my Khan Academy lil brainlist you

Answer: $50.99

Step-by-step explanation:

Given

The computer game is Priced at \(\$59.99\)

It is marked down by \(15\%\)

Final price is

\(\Rightarrow 59.99-15\%\ \text{of}\ 59.99\\\Rightarrow 59.99(1-0.15)=59.99\times 0.85\\\Rightarrow \$50.99\)

Answer:

A. (8,14)

Step-by-step explanation:

When looking at the chart pairs there is no other matching option than (8,14).

You have to look in at each column to see the pairs (ex. 8,16 and 4,10).

Answer:

C is the correct answer

Answer:

D

Step-by-step explanation:

Answer:

The answer is C

Step-by-step explanation:

I may be wrong but you need to line the decimal

Answer:

c 9+4= 13 and -7+-3 is -10 and 15 is 15 XD

Answer:

25802

Step-by-step explanation:

In algebra When something is in a ( ) You multiply what is outside the box so 679 (38) is another way of saying 679 x 38 = 25802

Answer:

A, C, and D have been simplified correctly

Answer:

Each book cost $12.50 dollars.

Step-by-step explanation:

1,187.50 / 95 = 12.50

The growth rate from the first year to the second year will be option : c) 33.1%

We need to calculation of the growth rate from the first year to the second year, since we know that there are 146 ferrets are found in a population in the first year. In the next year, there are 183 ferrets in the same population, so therefore, we can now calculate the growth rate, the growth rate is = (183 - 146)/146 = 33.1%

We need to basically take the difference of the population and then we have to divided it by the first-year population so that the growth rate could come as the solution.

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Answer:

Probability = 0.500

Step-by-step explanation:

If it is given that the reader is a teen, we have a total of 80 readers, where 40 of those readers have Rowling as their favorite.

So the probability that the reader's favorite is Rowling given that the reader is a teen is the number of teen readers that have Rowling as favorite (40) over the total number of teen readers (80)

Probability = 40 / 80 = 0.500

Substituting the values ∂c_j / ∂ℓ(θ) = (1/N) ∑_i=1^N E[h∼p(h|v_i)][h_j] - (1/N) ∑_i=1^N E(v,h)∼p(v,h)[h_j].

To show that ∂c_j / ∂ℓ(θ) = (1/N) ∑_i=1^N E[h∼p(h|v_i)][h_j] - (1/N) ∑_i=1^N E(v,h)∼p(v,h)[h_j], we'll first derive the expression for the gradient of the log-likelihood function ℓ(θ) with respect to c_j. Then, we'll use the concept of the expectation under the RBM distribution to further simplify the expression.

Deriving the gradient: We have ℓ(θ) = (1/N) ∑_i=1^N log p_θ(v_i), where p_θ(v) = ∑_h p_θ(v, h). Taking the gradient of ℓ(θ) with respect to c_j:

∂ℓ(θ) / ∂c_j = (1/N) ∑_i=1^N ∂log p_θ(v_i) / ∂c_j. Using the energy-based model, p_θ(v, h) = 1/Z Σ_e^-E(v,h), where E(v,h) = -v^TWh - b^Tv - c^Th. Taking the derivative of log p_θ(v_i) with respect to c_j:

∂log p_θ(v_i) / ∂c_j = ∂log (1/Z ∑_h e^-E(v_i, h)) / ∂c_j

= 1/Z ∑_h ∂(-E(v_i, h)) / ∂c_j

= -1/Z ∑_h [∂(v_i^TWh) / ∂c_j + ∂(b^Tv_i) / ∂c_j + ∂(c^Th) / ∂c_j]

Notice that ∂(c^Th) / ∂c_j = [h_j]. Applying this to the expression above:

∂log p_θ(v_i) / ∂c_j = -1/Z ∑_h [∂(v_i^TWh) / ∂c_j + ∂(b^Tv_i) / ∂c_j + [h_j]]

= -1/Z ∑_h [0 + 0 + [h_j]]

= -1/Z ∑_h [h_j]

= -E[h∼p(h|v_i)][h_j]

Therefore, ∂ℓ(θ) / ∂c_j = (1/N) ∑_i=1^N -E[h∼p(h|v_i)][h_j] = - (1/N) ∑_i=1^N E[h∼p(h|v_i)][h_j]. Using the expectation under the RBM distribution:

Using the expectation under the RBM distribution, we can write E[h∼p(h|v_i)][h_j] = E(h_j) = ∑_h (h_j * p(h_j=1|v_i)). Similarly, E(v,h)∼p(v,h)[h_j] = E(h_j) = ∑_v,h (h_j * p(v,h)). Substituting these values into the previous expression, we get:

∂ℓ(θ) / ∂c_j = - (1/N) ∑_i=1^N E[h∼p(h|v_i)][h_j] + (1/N) ∑_i=1^N E(h_j) = (1/N) ∑_i=1^N E(h_j) - (1/N) ∑_i=1^N E(h_j)

= (1/N) ∑_i=1^N E(h_j) - (1/N) ∑_i=1^N E(h_j)

= 0.

Hence, ∂c_j / ∂ℓ(θ) = (1/N) ∑_i=1^N E[h∼p(h|v_i)][h_j] - (1/N) ∑_i=1^N E(v,h)∼p(v,h)[h_j]. Assuming we already know the optimal parameters W and b, the contrastive divergence algorithm to find the optimal c involves the following steps:

Initialize c with some initial values.

For each training example v_i:

Sample a hidden vector h_i from p(h|v_i).

Compute the positive gradient contribution: ∂c_j / ∂ℓ(θ) = E[h∼p(h|v_i)][h_j].

Sample a reconstructed vector v'_i from p(v|h_i).

Sample a hidden vector h'_i from p(h|v'_i).

Compute the negative gradient contribution: ∂c_j / ∂ℓ(θ) = - E(v'_i, h'_i)∼p(v,h)[h_j].

Update c using a learning rate: c_j = c_j + learning_rate * (∂c_j / ∂ℓ(θ)).

Repeat steps 2 and 3 for multiple iterations until convergence.

This algorithm iteratively adjusts the values of c to maximize the log-likelihood function ℓ(θ) by updating c in the direction that reduces the difference between the positive and negative gradient contributions.

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Answer:

(20/4)

Step-by-step explanation:

The term "quotient" implies division, which is present in the (20/4) part of the expression.

The statement is true. If we take any two distinct prime numbers, and the number "1", these numbers will always be pairwise relatively prime.

Two numbers are said to be pairwise relatively prime if their greatest common divisor (GCD) is equal to 1. In this case, since we are considering distinct prime numbers, their GCD will always be 1 since prime numbers have no common factors other than 1 and themselves. Therefore, any two distinct prime numbers will be relatively prime.

Similarly, when we consider the number "1", it is relatively prime to any other number because its only divisor is 1. Hence, when "1" is paired with any prime number, the GCD will be 1.

In summary, whether we consider two distinct prime numbers or pair them with the number "1", the resulting numbers will always be pairwise relatively prime.

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To determine if a data point is considered an outlier for a data set, we need to calculate the interquartile range (IQR) and use it to define the outlier boundaries. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1). The correct option is (B).

We have that Q1 = 9 and Q3 = 17, we can calculate the IQR as follows:

IQR = Q3 - Q1 = 17 - 9 = 8

To identify outliers, we can use the following rule:

- Any data point that is less than Q1 - 1.5 * IQR or greater than Q3 + 1.5 * IQR is considered an outlier.

Using this rule, we can evaluate each data point:

A. 27: This data point is greater than Q3 + 1.5 * IQR = 17 + 1.5 * 8 = 29. It is considered an outlier.

B. 17: This data point is not an outlier because it is equal to the third quartile (Q3).

C. 3: This data point is less than Q1 - 1.5 * IQR = 9 - 1.5 * 8 = -3. It is considered an outlier.

D. 41: This data point is greater than Q3 + 1.5 * IQR = 17 + 1.5 * 8 = 29. It is considered an outlier.

Therefore, the outliers in the data set are A (27) and D (41).

As for when to use the mean, it is generally recommended to use the mean as a measure of central tendency when the data is symmetric and does not have extreme values.

Therefore, the correct option would be B. When the data is symmetric.

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The unit tangent vector, unit normal vector, and the binormal vector of r(t) = sin(2t)i 3tj 2 sin2(t) k can be obtained using the formulae:T(t) = r'(t) / ||r'(t)||N(t) = T'(t) / ||T'(t)||B(t) = T(t) x N(t) where r(t) is the position vector at time t, ||r'(t)|| is the magnitude of the derivative of r(t) with respect to time, i.e. the speed, and x denotes the cross product of two vectors.

Given r(t) = sin(2t)i + 3tj + 2 sin2(t) k

The derivative of r(t) is given by r'(t) = 2 cos(2t) i + 3 j + 4 sin(t) cos(t) k

The magnitude of the derivative of r(t) with respect to time is ||r'(t)|| = √(4cos2(2t) + 9 + 16sin2(t)cos2(t))

= √(13 + 3cos(4t))

Thus,T(t) = r'(t) / ||r'(t)||= [2 cos(2t) i + 3 j + 4 sin(t) cos(t) k] / √(13 + 3cos(4t))

N(t) = T'(t) / ||T'(t)|| where T'(t) is the derivative of T(t) with respect to time.

We obtain T'(t) = [-4 sin(2t) i + 4 sin(t)cos(t) k (13 + 3cos(4t))3/2 - (2cos(2t)) (-12 sin(4t)) / (2(13 + 3cos(4t))]j (13 + 3cos(4t))3/2

= [-4 sin(2t) i + 12cos(t)k] / √(13 + 3cos(4t))

Thus,N(t) = T'(t) / ||T'(t)||= [-4 sin(2t) i + 12cos(t)k] / √(16sin2(t) + 144cos2(t))

= [-sin(2t) i + 3 cos(t) k] / 2B(t) = T(t) x N(t)

= [2 cos(2t) i + 3 j + 4 sin(t) cos(t) k] x [-sin(2t) i + 3 cos(t) k] / 2

= [3 cos(t)sin(2t) i + (6 cos2(t) - 2 cos(2t)) j + 3 sin(t)sin(2t) k] / 2

Therefore, the unit tangent vector, unit normal vector, and the binormal vector of r(t) = sin(2t)i + 3tj + 2 sin2(t) k are:

T(t) = [2 cos(2t) i + 3 j + 4 sin(t) cos(t) k] / √(13 + 3cos(4t))N(t)

= [-sin(2t) i + 3 cos(t) k] / 2B(t) = [3 cos(t)sin(2t) i + (6 cos2(t) - 2 cos(2t)) j + 3 sin(t)sin(2t) k] / 2

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The reasonable estimate of the  probability that it is sunny tomorrow is option D: 25%

The concept of probability is determining the chance or likelihood of a certain event occurring. When we lack certainty regarding the result of a happening, we can discuss the chances of particular consequences, which denote their level of probability. The value of probability is one that ranges form  0-1

From the graph,

Total numbers of days of weather = days of weather that is  sunny + days of weather that is rainy + days of weather that is snowy

= 3 + 5 + 2 + 2 = 12

So, the probability of days of weather to be sunny + 3.12 = 1/4

= 0.25

Hence there is a 25 % chance that the weather will be sunny.

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The following bar graph summarizes the weather conditions in Crayonton for each day this month so far. Based on this data, what is a reasonable estimate of the probability that it is sunny tomorrow? Choose the best answer.

a 14

b 17

c 21

d 25

So sorry I replied so late but I'll leave the answer here just in case.

Answer:

7. Domain: [0,30]  Range: [0,7]

8. f(20) = 4.67 or 4 2/3; After 20 seconds, the squirrel is 4.67 meters away from the camera.

9. x = 1.25 and x = 21.43; The quirrel is 4 meters away from the camera at 1.25 seconds and 21.43 seconds.

10. At 2.5 and 19.29 seconds

11. 10 seconds

12. The squirrel went to the back of the camera, away from its line of view.

Step-by-step explanation:

7. The line is between 0 and 30 on the x-axis and is between 0 and 7 on the y-axis.

8. The slope of the third part of the line is not a whole number, so you need to calculate it. The rise is -7 meters and run is 15 seconds, so the slope is -7/15. Plugin -7/15 into the slope-intercept form (y = mx+b) and the equation is f(x) = -7/15 x+14. Plugin x=20 into the equation:

f(20) = -7/15(20)+14

f(20) = -9.33+14

f(20) = 4.67

9. The first slope of the graph will also need to be calculated. Rise is 7-3 (4) and the run is 5. Slope = rise/run = 4/5 = 0.8. Plugin the slope into y = mx+b with the y-intercept, and the equation is f(x) = 0.8x+3. Replace f(x) with 4 and solve the equation.

4 = 0.8x+3

1 = 0.8x

x = 1.25

There is another x value that matches with f(x) = 4 in the third part of the graph, so we'll use the slope-intercept form for the previous question for this one.

4 = -7/15 x+14

-10 = -7/15 x

x = 21.43

10. The squirel is 5 meters away from the camera twice, so we'll be using the same equations for the previous question.

5 = 0.8x+3

2 = 0.8x

x = 2.5

5 = -7/15 x+14

-9 = -7/15 x

x = 19.29

11. The squirrel is 7 meters away between 5 and 15 seconds. 15-5 = 10.

12. The squirrel most likely went somehwere where the camera can't see it, and it's probably the back of it.

The total number of seeds that Josiah would plant would be = nR×S

To determine the total number of seeds that Josiah will plant will be to add the seeds in the total number of rooms he planted.

Let each row be represented as = nR

Where n represents the number of rows planted by him.

Let the seed be represented as = S

The total number of seeds he planted = nR×S

Therefore, the total number of seeds that was planted Josiah would be = nR×S.

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The density of a piece of lead if it has a mass of 3.5 kg and occupies a volume of 3,097 x 10 m³ 0.113.

Suppose that a finite amount of substance is there having its properties as:

mass of substance = m kg

density of substance = d kg/m³

volume of that substance = v m³

Then, they are related as:

\(d = \dfrac{m}{v}\)

We are given that;

Volume of lead= 3.097 x 10 m³

Mass= 3.5kg

Now,

Density= 3.5/ 3.097 x 10

=3.5/30.97

=0.113

Therefore, by density formula the answer will be 0.113.

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The complete question is;

1. Determine the density of a piece of lead if it has a mass of 3.5 kg and occupies a volume

of 3,097 x 10 m³.

Data

Formula(s)

Substitution

Result

Answer: If you multiply six positives, it would be positive. If you multiply six negatives, it would be positive.

Step-by-step explanation:

If you see, 2x2x2x2x2x2=64, it would be positive.

If you multiply -2x-2x-2x-2x-2x-2x, it would be positive. A negative plus a negative is better. Hope you learned

It will take the town 21 years 4 months to reach 4600 in population

a rate is a ratio that compares two different quantities which have different units. For example, if we say John types 50 words in a minute, then his rate of typing is 50 words per minute. The word "per" gives a clue that we are dealing with a rate.

initial population = 3000

population increase every year = 2.5% of 3000

which is 2.5/100 x 3000 = 75

let the time taken to reach 4600 be d

increment for d number of years = 75d

Total population after d years is 75d + 3000

so 75d + 3000 = 4600

75d = 4600  - 3000

75d = 1600

d = 1600/75

d = 21.33 years which is 21 years 4 months

In conclusion 21 years 4 months is the time taken for the ton tto get tto 4600

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The argument is valid. The argument is valid based on the direct truth-table method.

To test the validity of the argument, we can use the direct truth-table method. Let's break down the argument and construct the truth table for the given premises and the conclusion:

Premises:

G ⊃ H

R ≡ G

~H v G

Conclusion:

R • H

Constructing the truth table:

We have three propositions: G, H, and R. Each proposition can have two truth values, true (T) or false (F). Therefore, we need 2^3 (8) rows in the truth table to evaluate all possible combinations.

By evaluating the truth table, we find that in all rows where the premises (1, 2, 3) are true, the conclusion (R • H) is also true. There is no row where the premises are true, but the conclusion is false. Therefore, the argument is valid.

The argument is valid based on the direct truth-table method. This means that if the premises (G ⊃ H, R ≡ G, ~H v G) are true, then the conclusion (R • H) must also be true.

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D.

It's an upside-down ln(x) function that's been shifted 3 units to the left.

Answer:

6 * 4/15

Well you have this and we can make it multiply like

6/1 * 4/15

So you just multiply normally, with the same for numerator and denonminator

So 6*4=24

1*15=15

So 24/15

24/15 is 1 9/15 in mixed

1 9/15 can be simplified so

1 3/5

So 1 3/5 is the simplest form

\(\large\underline{\sf{Solution-}}\)

\( \sf \: 6 \: of \frac{4}{15} \\ \)

\( \sf \: = 6 \times \frac{4}{15} \\ \)

\(\sf \: = \frac{6 \times 4}{15} \\ \)

\(\sf \: = \frac{24}{15} \\ \)

\(\sf \: = \frac{24 \div 3}{15 \div 3} \\ \)

\(\sf \: = \frac{8}{5} =1\frac{3}{5} \approx 1.6 \\ \)

You can put 14 boxes of candy weighing 3/4 lb each in the bin.

To determine how many 3/4 lb boxes of candy can fit in a bin, we divide the total weight of the bin by the weight of each box.

First, let's convert the mixed number 10 1/2 lbs to an improper fraction.

10 1/2 lbs = (10 * 2 + 1) / 2 = 21/2 lbs

Next, we divide the total weight of the bin (21/2 lbs) by the weight of each box (3/4 lb):

(21/2 lbs) / (3/4 lb) = (21/2) * (4/3) = (21 * 4) / (2 * 3) = 84/6 = 14

As a result, you can fill the bin with 14 boxes of sweets that each weigh 3/4 lb.

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Answer: it's \(\sqrt{5}\)

Step-by-step explanation:

The chain rule is a formula used in calculus to find the derivative of a composition of functions. It is an important tool for solving problems in many areas of mathematics and science.

The chain rule formula can be stated as follows:

If y = f(g(x)), then the derivative of y with respect to x is given by:

dy/dx = (df/dg) * (dg/dx)

In other words, the derivative of y with respect to x is equal to the derivative of f with respect to g, multiplied by the derivative of g with respect to x.

Here, f and g are functions of x, and y is a function of g. The chain rule formula tells us how to find the derivative of y with respect to x, by taking into account the effect of both f and g on the function y.

The chain rule can be extended to more complex compositions of functions, by applying the formula repeatedly. For example, if y = f(g(h(x))), then the chain rule can be applied twice, as follows:

dy/dx = (df/dg) * (dg/dh) * (dh/dx)

This formula tells us how to find the derivative of y with respect to x, by taking into account the effects of all three functions f, g, and h on the function y.

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