PR and os are diameters of circle T. What is the

measure of SR?

50°

* 80°

• 100°

120

Answer:

Simplifying

logx + -4 = 0

Reorder the terms:

-4 + glox = 0

Solving

-4 + glox = 0

Solving for variable 'g'.

Move all terms containing g to the left, all other terms to the right.

Add '4' to each side of the equation.

-4 + 4 + glox = 0 + 4

Combine like terms: -4 + 4 = 0

0 + glox = 0 + 4

glox = 0 + 4

Combine like terms: 0 + 4 = 4

glox = 4

Divide each side by 'lox'.

g = 4l-1o-1x-1

Simplifying

g = 4l-1o-1x-1

Answer:

[tex]x=40000[/tex]

Step-by-step explanation:

[tex]log(x)-log(4)=4\\log(\frac{x}{4})=4\\\frac{x}{4} = 10^4 \\x=4*10^4=40000[/tex]

The probability for more than 22% of the given data say their class sizes are larger than 30 is equal to 0.0864, or 8.64%.

To find the probability that more than 22% of the randomly selected teachers say their class sizes are larger than 30,

Use the binomial distribution.

Let us denote the probability of a teacher saying their class size is larger than 30 as p.

19% of the teachers responded with a class size larger than 30, we can estimate p as 0.19.

Now, calculate the probability using the binomial distribution.

find the probability of having more than 22% of the 760 teachers .

which is equivalent to more than 0.22 × 760 = 167 teachers saying their class sizes are larger than 30.

P(X > 167) = 1 - P(X ≤167)

Using the binomial distribution formula,

P(X ≤167) = [tex]\sum_{i=0}^{167}[/tex] [C(760, i) × [tex]p^i[/tex] × [tex](1-p)^{(760-i)[/tex]]

where C(n, r) represents the combination 'n choose r' the number of ways to choose r items from a set of n.

Using a statistical calculator, the probability P(X ≤ 167) is determined to be approximately 0.9136.

This implies,

The probability of having more than 22% of the randomly selected teachers say their class sizes are larger than 30 is,

P(X > 167)

= 1 - P(X ≤ 167)

≈ 1 - 0.9136

≈ 0.0864

Therefore, the probability for the given condition is approximately 0.0864, or 8.64%.

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To determine the maximum error when using [tex]$x = 0.404$[/tex] cm as an estimate of the mean diameter of bearings made by the process, we can calculate the margin of error for different confidence levels.

a) With a confidence of 95%:

For a 95% confidence level, we can use the standard normal distribution and the formula for the margin of error:

[tex]\[\text{{Margin of error (E)}} = z \times \left(\frac{{\sigma}}{{\sqrt{n}}}\right)\][/tex]

where  [tex]$z$[/tex] is the critical value corresponding to the desired confidence level, [tex]$\sigma$[/tex] is the standard deviation, and [tex]$n$[/tex] is the sample size.

Since we only have the population standard deviation [tex]($\sigma$)[/tex] and not the sample size [tex]($n$)[/tex] , we cannot calculate the margin of error without additional information. Please provide the sample size [tex]($n$)[/tex]to compute the maximum error with a 95% confidence level.

b) With a confidence of 99%:

Similarly, for a 99% confidence level, we can use the standard normal distribution and the formula for the margin of error:

[tex]\[\text{{Margin of error (E)}} = z \times \left(\frac{{\sigma}}{{\sqrt{n}}}\right)\][/tex]

Using a 99% confidence level, the critical value [tex]($z$)[/tex] is 2.576 (obtained from the standard normal distribution table).

Therefore, the maximum error with a 99% confidence level can be calculated as:

[tex]\[\text{{Margin of error (E)}} = 2.576 \times \left(\frac{{0.003}}{{\sqrt{n}}}\right)\][/tex]

Again, we need the sample size [tex]($n$)[/tex] to compute the maximum error with a 99% confidence level. Please provide the sample size to proceed with the calculation.

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A biased sample refers to a sample that is not representative of the population it is intended to represent. In a biased sample, certain characteristics or groups within the population are either overrepresented or underrepresented, leading to a distortion or skew in the data.

Bias can occur in various ways during the sampling process. Here are a few examples:

1. Selection Bias: When the method used to select the sample systematically favors or excludes certain individuals or groups from being included. This can lead to an overrepresentation or underrepresentation of specific characteristics in the sample.

2. Nonresponse Bias: When a portion of the selected sample does not participate or respond to the survey or study, resulting in a biased representation of the population.

3. Volunteer Bias: When individuals self-select to participate in a study or survey, which can introduce bias as those who volunteer may have different characteristics or motivations compared to the general population.

4. Measurement Bias: When the measurement instrument or procedure used to collect data systematically produces errors or inaccuracies that favor or exclude certain groups or characteristics.

Biased samples can lead to misleading or inaccurate conclusions about the population of interest since the sample does not accurately reflect the diversity and characteristics of the entire population. It is essential to strive for representative and unbiased samples to make valid inferences and generalizations about the population.

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Answer:

10.500

Step-by-step explanation:

step by step explenation

What's the question?

What are we supposed to do?

Answer:

I can't read the weight of the package due to the image quality, could you type it out please.

The function g(x) = (3)ˣ has a unique fixed on [0,1]

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

g(x) = (3)ˣ

The above function is an exponential function with the following features

Initial value = 1

Rate = 3

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

x = 0 in [0, 1]

So, we have

g(0) = (3)⁰

Evaluate

g(0) = 1

See that g(0) = 1 i.e. [0. 1]

Hence, the function has a unique fixed on [0,1]

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Answer:

I believe it would be 4.563 x 10^8

The coefficient of restitution between the ball and the ground is 0, indicating a completely inelastic collision.

The coefficient of restitution (e) is a measure of the elasticity or bounciness of a collision between two objects. It is defined as the ratio of the relative velocity of separation after the collision to the relative velocity of approach before the collision.

In this case, the ball is dropped vertically from a height of 1 m and returns to a height of 0.6 m. We can assume that the collision with the ground is approximately elastic, meaning that kinetic energy is conserved.

When the ball hits the ground, its initial velocity is zero, and the final velocity after the collision is also zero since it momentarily comes to rest before bouncing back up. Therefore, the relative velocity of separation is zero.

The relative velocity of the approach is the velocity just before the collision. Since the ball is dropped vertically, its velocity just before hitting the ground is given by the equation:

[tex]v = \sqrt(2gh)[/tex]

where v is the velocity, g is the acceleration due to gravity (approximately[tex]9.8 m/s^2[/tex]), and h is the initial height (1 m).

Plugging in the values:

[tex]v = \sqrt(2 * 9.8 * 1)[/tex]

[tex]= \sqrt(19.6)[/tex]

≈ 4.427 m/s

Therefore, the relative velocity of the approach is approximately 4.427 m/s.

Since the relative velocity of separation is zero, we can calculate the coefficient of restitution (e) as:

e = 0 / 4.427

= 0

Therefore, the coefficient of restitution between the ball and the ground, in this case, is 0, indicating a completely inelastic collision where the ball comes to a stop upon hitting the ground and does not bounce back.

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Answer: b im pretty sure hope this helps :)

Step-by-step explanation:

Answer:

A: x = -4

B: a = -7 (I'm assuming the '-' in front of 49 was supposed to be an '=')

Step-by-step explanation:

A: You need to group the like terms together, and isolate the variable, which in this case is x.

Take the 7 over to the other side, so the equation looks a little bit like this.

3-7 = x.

3-7 is -4, so we've gotten our answer.

x = -4.

B: -7a = 49.

When two negative numbers are multiplied together, the product becomes positive.

49 is 7x7.

-7 times -7 would be 49.

So, in this situation, a would be -7.

a = -7.

Answer: the answer is C 126°

The molarity of the diluted solution is approximately 0.220 M.

The concentration of a solute in a solution is measured by its molarity. The amount of solute that dissolves in one liter (L) of solution is the number of moles. One of the most used units of concentration is t, represented by the symbol M. Number of moles of solute contained in 1 liter of solution is how it is defined.

To calculate the molarity of a solution, you need to use the formula:

M₁V₁ = M₂V₂

Substituting these values into the formula:

(1.0 M)(121 ml) = M₂(550.0 ml)

Rearranging the equation to solve for M₂:

M₂ = (1.0 M)(121 ml) / (550.0 ml)

M₂ = 121 / 550 ≈ 0.220 M

Therefore, the molarity of the diluted solution is approximately 0.220 M.

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Answer: 4529

Step-by-step explanation:

Hope it helps Have a good Day

Just divide it

Answer:

28 in² = 28

Step-by-step explanation:

[tex]a = \frac{a + b}{2}h[/tex]

[tex]a = 6[/tex]

[tex]b = 8[/tex]

[tex]h = 4[/tex]

[tex]a = \frac{6 + 8}{2}4[/tex]

[tex] \frac{14}{2} 4[/tex]

[tex]7 \times 4[/tex]

[tex] = 28[/tex]

Answer = 28 _

The value of the integral is 15.87.

The given integral is:∫∫R 10(x+y) dAwhere R={(x,y)∣16≤x²+y²≤25,x≤0} in rectangular coordinates.In rectangular coordinates, the equation of circle is x²+y² = r², where r is the radius of the circle and the equation of the circle is given as: 16 ≤ x² + y² ≤ 25 ⇒ 4 ≤ r ≤ 5We need to evaluate the integral over the region R using rectangular coordinates and integrate first with respect to x and then with respect to y.∫∫R 10(x + y) dA = 10∫ from 4 to 5 ∫ from -√(25-y²) to -√(16-y²) (x+y) dx dy...[since x < 0]

Now, integrating ∫(x+y) dx we get ∫(x+y) dx = (x²/2 + xy)Therefore, 10∫ from 4 to 5 ∫ from -√(25-y²) to -√(16-y²) (x+y) dx dy = 10∫ from 4 to 5 ∫ from -√(25-y²) to -√(16-y²) [ (x²/2 + xy) ] dy dx= 10∫ from 4 to 5 [∫ from -√(25-y²) to -√(16-y²) (x²/2 + xy) dy] dxNow integrating with respect to y we get∫(x²/2 + xy) dy = (xy/2 + y²/2)

Putting the limits and integrating we get10∫ from 4 to 5 [∫ from -√(25-y²) to -√(16-y²) (x²/2 + xy) dy] dx = 10∫ from 4 to 5 [(∫ from -√(25-y²) to -√(16-y²) (x²/2 + xy) dy)] dx = 10∫ from 4 to 5 [(x²/2)[y]^(-√(16-x²) )_(^(-√(25-x²))] + [(xy/2)[y]^(-√(16-x²) )_(^(-√(25-x²)))] dx = 10∫ from 4 to 5 [-(x²/2)√(16-x²) + (x²/2)√(25-x²) - (x/2)(16-x²)^(3/2) + (x/2)(25-x²)^(3/2) ] dxNow integrating with respect to x, we get10∫ from 4 to 5 [-(x²/2)√(16-x²) + (x²/2)√(25-x²) - (x/2)(16-x²)^(3/2) + (x/2)(25-x²)^(3/2) ] dx = [ (10/3) [(25/3)^(3/2) - (16/3)^(3/2)] - 5√3 - (5/3)[(25/3)^(3/2) - (16/3)^(3/2) ] ]Ans: The value of the integral is 15.87.

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Answer:

The average rate of change for the depth of the river measured as feet per hour is approximately 0.3 feet/hour

Step-by-step explanation:

The depth of the river in feet with time is given by the function with the attached

From the graph, we have;

The depth of the river at hour t = 9 is f(9) = 18 feet

The depth of the river at hour t = 18 is f(18) = 21 feet

The average rate of change, A(x), for the depth of the river measured as feet per hour is given as follows;

[tex]A(X) = \dfrac{f(b) - f(a)}{b - a}[/tex]

Therefore, for the river, we have;

[tex]A(X) = \dfrac{f(18) - f(9)}{18 - 9} = \dfrac{21 - 18}{18 -9} = \dfrac{3}{9} =\dfrac{1}{3}[/tex]

The average rate of change for the depth of the river measured as feet per hour A(X) = 1/3 feet/hour

By rounding the answer to the nearest tenth, we have;

A(X) = 0.3 feet/hour.

Answer: -3

Step-by-step explanation: DeltaMath

Answer:

Since I do not know the context of the question I will list answers I think it could be based on what you asked:

1. 3.72 x 0.6 = 2.232

2. 3.72 ÷ 0.6 = 6.2

3. 3.72% of 0.6 = 0.02232

The answer is probably the first one. I can't give a definite solution without knowing the exact question being asked, sorry!

The probability that the end-of-month volume will remain greater than 18 million m³ is approximately 0.1922.

a)The mean water volume in the lake at the end of the month can be calculated using the formula given below:

Mean water volume = Starting water volume + Total rainfall + Total flow - Total evaporation - Water drawn from the lake

Given:

Starting water volume = 20 million m³

Total rainfall = random variable with mean = 1 million m³ and standard deviation = 0.5 million m³

Total flow = random variable with mean = 8 million m³ and standard deviation = 2 million m³

Total evaporation = 3 million m³

Water drawn from the lake = 10 million m³

Now, let's calculate the mean water volume at the end of the month.

Mean water volume = Starting water volume + Total rainfall + Total flow - Total evaporation - Water drawn from the lake= 20 + 1 + 8 - 3 - 10= 16 million m³

Therefore, the mean water volume at the end of the month is 16 million m³.

The standard deviation of the water volume in the lake at the end of the month can be calculated using the formula given below:

σ = √{σr² + σf² + σe²}

σr = standard deviation of rainfall = 0.5 million m³

σf = standard deviation of flow = 2 million m³

σe = standard deviation of evaporation = 1 million m³σ = √{σr² + σf² + σe²}σ = √{0.5² + 2² + 1²}= √{5.25}≈ 2.29 million m³

Therefore, the standard deviation of the water volume in the lake at the end of the month is approximately 2.29 million m³.b)Given that all the random variables in the problem are normally distributed, we can find the probability that the end-of-month volume will remain greater than 18 million m³ using the z-score formula.

z = (x - μ) / σ

Where,

z = z-scorex = 18 μ = 16σ = 2.29

Now, let's calculate the z-score.

z = (x - μ) / σ= (18 - 16) / 2.29= 0.87

Using the z-table, we can find that the probability of z being less than 0.87 is 0.8078.

Therefore, the probability of the end-of-month volume being greater than 18 million m³ is:

1 - 0.8078 = 0.1922 (rounded to 4 decimal places)

Hence, the probability that the end-of-month volume will remain greater than 18 million m³ is approximately 0.1922.

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Answer:

The guy above me is completely wrong hahaha.

The correct answer should have a 63%.

It's probably D. The terminology is a little confusing.

The equation of the output looks like:

-7.611 + .186(x) = y

x --> Arm span

y --> Foot span

The linear relationship is formed with the armspan.

Answer: It’s D

Step-by-step explanation:

I took the test

Answer:

2.2

Step-by-step explanation:

Answer:

11/5

Step-by-step explanation:

Answer:

x=-2

Step-by-step explanation:

Divide both sides by the numeric factor on the left side, then solve.

Answer:

C

Step-by-step explanation:

Answer:

232cm

Step-by-step explanation:

A cheese usually has the shape of a cuboid.

Characteristics of a Cuboid

The amount of paper that would be needed to cover the block of cheese can be determined by calculating the perimeter of the block of cheese which is shaped as a cuboid

Perimeter of a cuboid = 4 x (length + breadth + height)

4 x (30 + 20 + 8) = 232cm

9514 1404 393

Answer:

  (x -y -1)(x +y)

Step-by-step explanation:

The expression can be factored by grouping.

  x^2 -y^2 -x -y = (x^2 -y^2) -(x +y)

  = (x -y)(x +y) -1(x +y)

  = (x -y -1)(x +y)

_____

It is useful to know that a difference of squares is factored as ...

  a^2 -b^2 = (a -b)(a +b)

i had this but your numbers aren't the same i thought i could help you..

Answer:

2.75 x (c) = 500

Step-by-step explanation:

i think this is it

Answer:

Option 3 Or number 5

Step-by-step explanation:

Answer:

8

Step-by-step explanation:

Eight people are at least 169.5 cm tall because the frequency of people that height is five. The frequency of people taller than 169.5 (because it said 'at least') is three. 5 + 3 = 8

[tex]\huge\text{Hey there!}[/tex]

[tex]\mathsf{7\times 10^{-4}}[/tex]

[tex]\rightarrow\mathsf{7\times\dfrac{1}{10\times10\times10\times10}}[/tex]

[tex]\mathsf{= \dfrac{7}{10\times10\times10\times10}}[/tex]

[tex]\mathsf{10\times10\times10\times10=100\times100=10,000=\bf 10^4}[/tex]

[tex]\mathsf{=\dfrac{7}{10^4}}[/tex]

[tex]\mathsf{=\dfrac{7}{10,000}}[/tex]

[tex]\boxed{\boxed{\large\textsf{Possible ANSWERS: }\mathsf{\bf {\dfrac{7}{10^4}\ or \ \dfrac{7}{10,000}}}}}\huge\checkmark[/tex]

[tex]\text{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]