here are the options



∠2and∠4



∠1and∠5



∠3and∠6

Answer:

30%

Step-by-step explanation:

Answer: D

Step-by-step explanation:

just got it correct.

Answer:

the answer is D

hope it help

Answer: D, x²-6x+9

Step-by-step explanation:

(x-3)² is the same as (x-3)(x-3).

So you would do:

x times x=x²

x times -3=-3x

-3 times x=-3x

-3 times -3=9

The equation would look like this:

x²-3x-3x+9

Then you would have to collect the like terms. Like terms are terms that have the same variables and powers. So it would look like this:

x²-6x+9

Hope this helps :)

Answer:

its 62!

p-by-step explanation:

Answer:

62.

Step-by-step explanation:

A=2(wl+hl+hw)=2·(3·5+2·5+2·3)=62

Answer:

this is true but what do we have to do?

Step-by-step explanation:

Answer: x = 35, angle 1 = 85, angle 2 = 70, angle 3 = 25

Step-by-step explanation:

Finding x: 85+2x+x-10=180

3x= 105

x= 35

x-10 = angle 3 due to a property of parallels lines so put x in to get that angle 3 is 25 degrees

angle 2x = 70 degrees

70 + 25+angle 1 = 180

angle 1 = 85

angle 2 = 180-85-25 = 70

Answer:

y=3x+2

Step-by-step explanation:

y=mx+b

b=2

m=3

plug in

y=3x+2

Answer:

The answer to the question provided is 14.

Step-by-step explanation:

》The Distance Formula:

[tex] d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} } [/tex]

》Plug in.

[tex]d = \sqrt{( - 5 - 9)^{2} + (3 - 3) ^{2} } \\ d = \sqrt{( - 14)^{2} + (0) ^{2} } \\ d = \sqrt{196 + 0} \\ d = \sqrt{196} \\ d = 14[/tex]

Based on the given data, a 95% confidence interval estimate of the mean number falls between 12.22 and 16.78.

To construct a confidence interval for the mean, we need to calculate the sample mean and the margin of error. The formula for the margin of error is:

Margin of Error = Z * [tex]\frac{standard deviation}{\sqrt{n} }[/tex]

where Z is the critical value corresponding to the desired confidence level (for 95% confidence level, Z ≈ 1.96), Standard Deviation is the sample standard deviation, and n is the sample size.

From the given data, we calculate the sample mean to be 14.33 and the sample standard deviation to be 1.91. Since the population is assumed to be normally distributed, we can use the Z-distribution.

Using the formula for the margin of error, we find:

Margin of Error = 1.96 * (1.91 / √9) ≈ 1.39

The confidence interval is calculated by subtracting and adding the margin of error to the sample mean:

Confidence Interval = (14.33 - 1.39, 14.33 + 1.39) = (12.94, 15.72)

Rounded to at least two decimal places, the 95% confidence interval estimate of the mean number is approximately (12.22, 16.78). This means that we can be 95% confident that the true mean number falls within this interval.

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the conditional probability pr[x>1000 | x>500] is P(x > 500) = 1 - CDF(500).

Given that x has an exponential likelihood thickness work with β = 500, we are able to compute the likelihood that x is more noteworthy than 500, i.e., P(x > 500).

For an exponential conveyance with parameter β, the likelihood thickness work (PDF) is given by:

f(x) = (1/β) * e^(-x/β), where x ≥ 0.

To discover P(x > 500), we got to coordinate the PDF from 500 to limitlessness:

P(x > 500) = ∫[500, ∞] (1/β) * e^(-x/β) dx.

Let's calculate this likelihood:

P(x > 500) = ∫[500, ∞] (1/500) * e^(-x/500) dx.

To calculate this indispensably, ready to utilize the truth that the necessity of the PDF over its whole extent is rise to 1. So, ready to revamp the likelihood as:

P(x > 500) = 1 - P(x ≤ 500).

Since the exponential conveyance is memoryless, P(x ≤ 500) is break even with the total conveyance work (CDF) at 500.

P(x > 500) = 1 - CDF(500).

The CDF of the exponential dissemination is given by:

CDF(x) = ∫[0, x] (1/β) * e^(-t/β) dt.

To calculate P(x > 500), we have to assess CDF(500) and subtract it from 1.

Presently, let's calculate P(x > 500):

P(x > 500) = 1 - CDF(500)

= 1 - ∫[0, 500] (1/500) * e^(-t/500) dt.

To calculate the conditional probability P(x > 1000 | x > 500), we have to consider the occasion that x > 500 is our modern test space. The conditional probability is at that point given by:

P(x > 1000 | x > 500) = P(x > 1000, x > 500) / P(x > 500).

Since x takes after an exponential conveyance, it is memoryless, which implies the likelihood of x > 1000 given x > 500 is the same as the likelihood of x > 500. Hence, we have:

P(x > 1000 | x > 500) = P(x > 500) = 1 - CDF(500).

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Answer:

838.4

Step-by-step explanation:

From the question,

Total worth of the property = 10480.

Eugenia share: Her 10 brothers share = 1:4

Total ratio = 1+4 = 5.

Her 10 brother's share = (4/5)(10480)

Her 10 brother's share = 41920/5

Her 10 brother's share = 8384.

If the 10 brothers shared their portion equally,

Each brother's share = 8384/10

Each brother's share = 838.4

Answer:

There are 10 odd numbers: {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} There are 4 factors of 8: {1, 2, 4, 8} There is 1 number in both lists: {1} Probability of an odd or a factor of 8 = (10 + 4 - 1)/20 = 13/20

The quantity f divided by g of 8 is 5, which is answer choice D.

To find f divided by g of 8, we need to plug in x = 8 into both f(x) and g(x), then divide f(8) by g(8).

f(x) = 3x + 1

f(8) = 3(8) + 1

f(8) = 25

g(x) = x - 3

g(8) = 8 - 3

g(8) = 5

Now, we can divide f(8) by g(8) to get:

f(8) / g(8) = 25 / 5

f(8) / g(8) = 5

Alternatively, we can find f divided by g of x generally by first applying the functions to x:

f(x) / g(x) = (3x + 1) / (x - 3)

We can then plug in x = 8 and simplify:

f(8)  g(8) = (3(8) + 1) / (8 - 3)

f(8) / g(8) = 25 / 5

f(8) / g(8) = 5

Regardless of the method we use, we find that f divided by g of 8 is 5.(option-d)

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Answer:

2

Step-by-step explanation:

14 / (7)

If a student obtained the standard value z= 1.8, then the exam grade is 680.

Given, a placement exam has a measure of x = 500 and a standard deviation of s = 100, and a student obtained the standard value z = 1.8, and we are to find the exam grade.

In order to find the exam grade, we can use the formula, z = (x - μ) / σ where x is the score, μ is the mean, and σ is the standard deviation.

Substituting the given values, we get1.8 = (x - 500) / 100

Multiplying both sides by 100, we get180 = x - 500

Adding 500 to both sides, we get680 = x

Therefore, the exam grade is 680.So, the correct option is d. 680.

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Answer:

area = 175 cm^2

Step-by-step explanation:

area = base * height

area = 5 cm + 35 cm

area = 175 cm^2

The value of the expression 6x(y^2 - z) when x = 0, y = -1, and z = 2 is 0.

To find the value of the algebraic expression 6x(y^2 - z) when x = 0, y = -1, and z = 2, we substitute the given values into the expression.

First, let's evaluate the inner expression (y^2 - z):

Substituting y = -1 and z = 2, we have (-1)^2 - 2 = 1 - 2 = -1.

Now, we substitute x = 0 and the result of the inner expression (-1) into the outer expression:

6x(y^2 - z) = 6(0)(-1) = 0.

Therefore, when x = 0, y = -1, and z = 2, the value of the expression 6x(y^2 - z) is 0.

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Answer:

The answer is D

Step-by-step explanation:

They are using 135 times the weeks, but the car already had 100 miles on it to start.  Juan buys a car that has 40 miles originally on it and he drives it 150 miles per week. So, the word problem is D

y=-1x+4

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hope it helps

sorry I had no idea how to explain

the answer are 4kg ,12kg and 20 kg

Answer:

x = 5

x = 2

Step-by-step explanation:

2(2x + 4) = 8x - 12

2x + 4 = 4x - 6

x + 2 = 2x - 3

x = 5

Question 17

8/4 = (x + 2)/(2x - 2)

2(2x - 2) = x + 2

4x - 4 = x + 2

3x = 6

x = 2

Given:Standard deviation, s = $2000Confidence level = 96%Margin of error, E = $500We have to find the sample size, n.

Sample size formula is given as:\[n={\left(\frac{z\text{/}2\times s}  {E}\right)}^{2}\]Where, z/2 is the z-score at a 96% confidence level. Using the standard normal table, we can get the value of z/2 as follows:z/2 = 1.750Incorporating all the values in the formula, we get:\[n={\left(\frac{1.750\times 2000}{500}\right)}^{2}\] Simplifying,\[n=21\]Therefore, a sample size of 21 is required if we wish to be 96% confident that our mean will be within $500 of the true mean salary of college instructors.

To determine the sample size needed to be 96% confident that the mean salary will be within $500 of the true mean salary, we can use the formula for sample size in a confidence interval.

The formula is:

n = (Z * σ / E)^2

Where:

n is the required sample size

Z is the z-score corresponding to the desired confidence level (in this case, 96% confidence level)

σ is the standard deviation of the population (given as $2000)

E is the maximum error tolerance (given as $500)

First, we need to find the z-score corresponding to a 96% confidence level. The remaining 4% is split evenly between the two tails of the distribution, so we look up the z-score that corresponds to the upper tail of 2% (100% - 96% = 4% divided by 2).

Using a standard normal distribution table or a calculator, the z-score for a 2% upper tail is approximately 2.05.

Now we can substitute the values into the formula:

n = (Z * σ / E)^2

n = (2.05 * 2000 / 500)^2

Calculating this expression:

n = (4100 / 500)^2

n = 8.2^2

n = 67.24

Rounding up to the next whole number, the required sample size is approximately 68.

Therefore, a sample size of 68 is needed to be 96% confident that the mean salary will be within $500 of the true mean salary of college instructors.

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Given information: The starting salaries of college instructors have a standard deviation (SD) of $2000. The sample size that is needed if we want to be 96% confident that our mean will be within $500 of the true mean salary of college instructors is to be calculated.

Hence, 49 is the sample size that is needed if we want to be 96% confident that our mean will be within $500 of the true mean salary of college instructors.

The formula for the sample size required is as follows:

[tex]n = [(Z \times \sigma) / E]^{2}[/tex]

Here, Z is the value from the normal distribution for a given confidence level, σ is population standard deviation, E is the maximum error or the margin of error, which is [tex]\$500n = [(Z \times \sigma) / E]^2[/tex]

On substituting the given values, we get:

[tex]n = [(Z \times \sigma) / E]^2[/tex]

[tex]n= [(Z \times \$2000) / \$500]^2[/tex]

[tex]n = [(1.7507 \times \$2000) / \$500]^2[/tex]

[tex]n = (7.003 \times 7.003)[/tex]

n = 49 (rounded off to the next whole number)

Hence, 49 is the sample size that is needed if we want to be 96% confident that our mean will be within $500 of the true mean salary of college instructors.

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Answer:

i dont know

Step-by-step explanation:

Answer:

1.A,B,C,D

2. AB, CD

3. AC, BD

4. Line AD (don't take my word for this one)

Step-by-step explanation:

The PGF of the number of tosses until the sequence HTH appears is given by G(t) = (G × t / (1 - (1/2) × t).

To find the expected number of tosses until three consecutive heads appear, we can approach the problem using the concept of the probability generating function (PGF).

Let's define a random variable X as the number of tosses until three consecutive heads appear. We want to find E(X), the expected value of X.

To determine the PGF of X, we consider the possible outcomes at each toss. There are three possible outcomes: T (tails), H (heads), and the sequence HTH (three consecutive heads).

At the first toss, the possible outcomes are T and H. The PGF for this situation is given by:

[tex]G1(t) = Pr(X = 1) \times t^1 + Pr(X = 2) \times t^2[/tex]

Since we can have either T or H on the first toss, we have Pr(X = 1) = 1/2 and Pr(X = 2) = 1/2. Therefore:

[tex]G1(t) = (1/2) \times t + (1/2) \times t^2[/tex]

Now, let's consider the situation after the first toss:

If the first toss resulted in T, we are back to the starting point. Therefore, the PGF is G(t).

If the first toss resulted in H, we are one step closer to our goal (HTH). The PGF for this situation is G(t) × t.

Combining these two cases, we have:

[tex]G(t) = (1/2) \times t + (1/2) \times t^2 \times G(t)[/tex]

Simplifying the equation, we get:

[tex]G(t) = (1/2) \times t / (1 - (1/2) \times t^2)[/tex]

Next, let's consider the situation after the second toss:

If the second toss resulted in T, we are back to the starting point. Therefore, the PGF is G(t).

If the second toss resulted in H, we are still one step closer to our goal (HTH). The PGF for this situation is G(t) × t.

Combining these two cases, we have:

G(t) = (1/2) × t + (1/2) × t × G(t)

Simplifying the equation, we get:

G(t) = (1/2) × t / (1 - (1/2) × t)

Finally, we can calculate the expected value E(X) using the PGF:

E(X) = G'(1)

To find the derivative of G(t), we can use the quotient rule:

G'(t) = [(1 - t) × 1 - t × (-1/2)] / (1 - (1/2) × [tex]t)^2[/tex]

Simplifying the equation, we get:

G'(t) = 1 / (1 - (1/2) × [tex]t)^2[/tex]

Evaluating G'(1), we have:

[tex]E(X) = G'(1) = 1 / (1 - (1/2) \times 1)^2 = 1 / (1 - 1/2)^2 = 1 / (1/2)^2 = 1 / (1/4) = 4[/tex]

Therefore, the expected number of tosses until three consecutive heads appear is 4.

Additionally, the PGF of the number of tosses until the sequence HTH appears is given by:

G(t) = (G × t / (1 - (1/2) × t)

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Answer: The zeros of the  quadratic function  are: 5/2+(1/2)*sqrt(31), 5/2-(1/2)*sqrt(31)

Step-by-step explanation:

Answer:

x ≈ 36.09

Step-by-step explanation:

Complete question

Use △DEF, shown below, to answer the question that follows: Triangle DEF where angle E is a right angle. DE measures x. DF measures 55. Angle D measures 49 degrees. What is the value of x rounded to the nearest hundredth? Type the numeric answer only in the box below.

According to the diagram

m<E = 90 degrees

m<D = 49 degrees

DF = 55 = hypotenuse

DE = x = adjacent

EF = opposite

According to SOH CAH TOA identity;

Cos theta = adj/hyp

Cos m<D = DE/DF

Cos 49 = x/55

x = 55cos49

x = 55(0.6561)

x = 36.0855

x ≈ 36.09 (to the nearest hundredth)

Answer:

x ≈ 36.09

correct on khan

Step-by-step explanation:

Answer:

86 degrees

Step-by-step explanation:

180 - (74 + 20) = 86

Answer:

I. Length = 600 meters

II. Width = 30 meters

Step-by-step explanation:

Given the following data;

Length = 120m

Width = 6m

Drawing scale = 1:5

To find the length and width using the given drawing scale;

This ultimately implies that, with this drawing scale, the length and width of the rectangle would increase by a factor of 5 (multiplied by 5) i.e the rectangle is 5 times bigger in real-life than on the diagram.

For the length;

120 * 5 = 600 meters

For the width;

6 * 5 = 30 meters

Therefore, the dimensions of the rectangle using the given drawing scale is 600 meters by 30 meters.