An instructor is administering a final examination. She tells her class that she with give an A grade to the 10% of the students who earns the highest marks. Past experience with the same examination has yielded grades that are normally distributed with a mean of 70 and a standard deviation of 10. If present class runs true to form, what numerical score would a student need to earn an A grade?

The list that gives the sample space for pulling two slips of paper out of the bucket with replacement is List 4.

Probability is a branch of mathematics that deals with the study of randomness and uncertainty in events. It is the measure of the likelihood or chance that an event will occur. Probability is expressed as a number between 0 and 1, where 0 indicates that the event will not occur and 1 indicates that the event will occur with certainty.

The sample space for pulling two slips of paper out of the bucket with replacement means that after drawing the first slip of paper, it is returned to the bucket before drawing the second slip. This means that each draw is independent of the previous draw, and the possible outcomes of each draw remain the same for all subsequent draws.

The possible outcomes for each draw are {Red, Blue, Yellow}, and there are three possible outcomes for each draw.

Since we are drawing with replacement, each draw has the same possible outcomes and the same number of outcomes. So, the total number of possible outcomes for two draws is simply the number of outcomes for a single draw raised to the power of 2:

[tex]3^2 = 9[/tex]

Therefore, the list that gives the sample space for pulling two slips of paper out of the bucket with replacement is List 4.

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Answer: Imagine a rectangle with its centre (point of intersection of diagonals) situated at the origin of the XY plane and it's each of the sides being aligned with the co-ordinate axis.

Answer:

expand the square

2(x-6) (x-6)

2(x(x-6)-6(x-6)

2(x²-6x-6(x-6)

2(x²-6x-6x+36)

2(x²-12x+36)

2x²-24x+72

Step-by-step explanation:

hope this helps

Answer: To find the standard deviation of a probability distribution, we need to first calculate the mean or expected value of the distribution, which is given by:

E(X) = Σ [xi * P(xi)]

where xi is the ith outcome and P(xi) is its probability.

So, for the given distribution:

E(X) = (0 * 0.1) + (1 * 0.05) + (2 * 0.1) + (3 * 0.75) = 2.4

Next, we need to calculate the variance of the distribution, which is given by:

Var(X) = Σ [(xi - E(X))^2 * P(xi)]

So, for the given distribution:

Var(X) = (0 - 2.4)^2 * 0.1 + (1 - 2.4)^2 * 0.05 + (2 - 2.4)^2 * 0.1 + (3 - 2.4)^2 * 0.75 = 0.69

Finally, the standard deviation of the distribution is the square root of the variance:

SD(X) = sqrt(Var(X)) = sqrt(0.69) ≈ 0.83

Therefore, the standard deviation of this probability distribution is approximately 0.83, rounded to 2 decimal places.

Step-by-step explanation:

Answer:

Se depositan $ 8.000 en un banco que reconoce una tasa de interés del 36% anual, capitalizable mensualmente. ¿Cuál será el monto acumulado en cuatro años?

Step-by-step explanation:

Para resolver este problema, podemos utilizar la fórmula del interés compuesto:

A = P*(1 + r/n)^(n*t)

Donde:

A: el monto acumulado después de t años

P: el capital inicial

r: la tasa de interés anual

n: el número de veces que se capitaliza el interés por año

t: el tiempo en años

En este caso, tenemos:

P = $8.000

r = 36% = 0.36

n = 12 (ya que la tasa de interés se capitaliza mensualmente)

t = 4 años

Sustituyendo estos valores en la fórmula, obtenemos:

A = $8.000*(1 + 0.36/12)^(124)

A = $8.000(1 + 0.03)^48

A = $8.000*(1.03)^48

A = $16.751,83

Por lo tanto, el monto acumulado en cuatro años será de $16.751,83.

Answer:

x = 7

Step-by-step explanation:

The two angles are equal so the opposite sides are equal.

5x-2 =33

Add two to each side.

5x-2+2 = 33+2

5x=35

Divide by 5

5x/5 =35/5

x = 7

Answer: I really don't know i'm just trying to get a quest done sorry

Option C) The student calculated the marginal relative frequency for 10th grade students in geometry. The correct value of the joint relative frequency of 10th grade students in geometry is 20.7%.

Relative frequency is a measure of the proportion or percentage of times an event occurs in a given sample. It is calculated by dividing the frequency of an event by the total number of events in the sample. The formula for relative frequency is given as follows:

Relative frequency = Frequency of an event / Total number of events in the sample

For the joint relative frequency of 10th grade students in geometry, the parameters are given as follows:

Hence the joint relative frequency is obtained as follows:

20.7/100 x 100%= 20.7%.

Hence option C is correct.

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Step-by-step explanation:

"increase" means to take the original 100% and put an additional 3.5% of these 100% on top of it.

so, we have to calculate

100% + 3.5% of £16470.45

100% of £16470.45 = £16470.45 × 100/100

3.5% of £16470.45 = £16470.45 × 3.5/100

the sum is therefore

£16470.45 × (100/100 + 3.5/100) =

= £16470.45 × (1 + 0.035) = £16470.45 × 1.035 =

= £17,046.91575 ≈ £17,046.92

Answer: 42

Step-by-step explanation:

1. 1/3=14

2. 14x3=42

Some of the tips which is used to conduct a survey are:

Conducting a survey is an important tool for gathering information from a large and diverse group of people. Its allows allow researchers to obtain data from a representative sample of the population, which can then be used to make informed decisions, identify trends, and measure changes over time.

Also important, its can provide insight into people's attitudes, beliefs, and behaviors, which can be valuable in developing marketing strategies, designing programs and policies, and making important decisions.

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Answer:

s = 4g    ..........................................................                      

The derivative of the function z= x^2y+xy^2, x = 3t y = t^2 using the chain rule is given by dz/dt = 36t^3 + 15t^4.

Expressions are equals to,

z= x^2y+xy^2

x = 3t

y = t^2

Using the chain rule calculate dz/dt,

which states that if z is a function of x and y,

And x and y are both functions of t, then,

dz/dt = (dz/dx)(dx/dt) + (dz/dy)(dy/dt)

Using these expressions, calculate the value of dz/dt using the chain rule,

z= x^2y+xy^2

This implies,

dz/dx = 2xy + y^2

dz/dy = x^2 + 2xy

x = 3t

⇒ dx/dt = 3

y = t^2

⇒ dy/dt = 2t

Substituting these values into the chain rule formula, we get,

dz/dt = (2xy + y^2)(3) + (x^2 + 2xy)(2t)

        = [2(3t)(t^2 ) + (t^2)^2 ]3 + [(3t)^2 + 2(3t)(t^2)](2t )

        = [ 6t^3 + t^4 ]3 + [ 9t^2 + 6t^3 ]2t

        = 18t^3 + 3t^4 + 18t^3 + 12t^4

        = 36t^3 + 15t^4

Substituting the given expressions for x and y into z, we get,

z = (3t)^2(t^2) + (3t)(t^2)^2

   = 9t^4 + 3t^5

here also,

dz/dt = 36t^3 + 15t^4

Therefore, the value of the function using the chain rule dz/dt is equals to  36t^3 + 15t^4.

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Answer:

Step-by-step explanation:

[tex]y=2x-4[/tex]    (slope-intercept form is [tex]y=mx+b[/tex] where m=gradient

                     and b is where line intercepts y-axis)

The actual intake of carbohydrates is 138% as compare to recommended intake.

Recommended intake of carbohydrates or any other nutrient are,

Based on the information provided,

Consumed 159 grams of carbohydrates,

Recommended intake is between 115 and 166 grams.

Calculate the percentage of actual intake compared to the recommended intake, use the following formula,

Percentage = (Actual Intake / Recommended Intake) x 100%

Substituting the values in the formula we have,

⇒Percentage = (159 / 115) x 100%

⇒Percentage ≈ 138.3%

Therefore,  the actual intake of carbohydrates is about 138% of the recommended intake, indicating that consumption of more carbohydrates than recommended.

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[tex]\begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ \stackrel{ \textit{we'll use this one} }{log_a a^x = x}\qquad \qquad a^{log_a (x)}=x \end{array} \\\\[-0.35em] ~\dotfill\\\\ \log_2\left( \cfrac{1}{16} \right)+4\implies \log_2\left( \cfrac{1}{2^4} \right)+4\implies \log_2(2^{-4})+4\implies -4+4\implies \text{\LARGE 0}[/tex]

[tex]\rule{34em}{0.25pt}\\\\ \textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b\qquad\qquad \\\\[-0.35em] ~\dotfill\\\\ \log_2\left( \cfrac{1}{16} \right)=y\implies 2^y=\cfrac{1}{16}\implies 2^y=2^{-4}\implies y=-4[/tex]

The total number of elements in either Set A or Set B is 156.

A set is a collection of unique items, but a subset is a set that contains only components that belong to another set, termed the superset. In other words, set A is a subset of set B if all of its items are also found in set B.

Given that, Set A contains 98 elements and Set B contains 93 elements.

Total number of elements in either Set A or Set B = number of elements in Set A + number of elements in Set B - number of common elements

Total number of elements in either Set A or Set B = 98 + 93 - 35

Total number of elements in either Set A or Set B = 156

Hence, the total number of elements in either Set A or Set B is 156.

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Answer:

V=lwh

=2×2×2=8

720÷8=90

90 cubes

Answer:

False.

Step-by-step explanation:

A concave polygon can never be a regular polygon as it can never be equiangular. Each side of a regular polygon must be the same length, and all interior angles must also be equal.

Here using the correct order of operations that is PEMDAS, we get the value to be 13.

PEDMAS can be summed up as a mathematical acronym that lists the various arithmetic operations in order of greatest to least practical use.

The letters stand for:

P stands for parentheses.

Exponents are shown as E.

D stands for division.

The letter M stands for multiplication.

A stand for addition.

S stands for subtraction.

Now in the given question,

15 - (4-3)2

First the parentheses

15 - (1)2

Next is exponents.

15 - 2

At last, after subtracting, we get:

13

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The sample mean is 2.1, the sample variance is 212.5% and the standard deviation is 14.57%

a. The sample mean can be computed as the average of the quarterly percent total returns:

[tex](11.2 - 20.5 + 13.2 + 12.6 + 9.5 - 5.8 - 17.7 + 14.3) / 8 = 2.1[/tex]

So the sample mean is 2.1%, which can be interpreted as the average quarterly percent total return for the stock over the sample period.

b. The sample variance can be computed using the formula:

[tex]s^2 = sum((x - mean)^2) / (n - 1)[/tex]

where x is each quarterly percent total return, mean is the sample mean, and n is the sample size. Plugging in the values, we get:

[tex]s^2 = (11.2 - 2.1)^2 + (-20.5 - 2.1)^2 + (13.2 - 2.1)^2 + (12.6 - 2.1)^2 + (9.5 - 2.1)^2 + (-5.8 - 2.1)^2 + (-17.7 - 2.1)^2 + (14.3 - 2.1)^2 / (8 - 1) = 212.15[/tex]

So the sample variance is 212.15%. The sample standard deviation can be computed as the square root of the sample variance:

[tex]s = \sqrt(s^2) = \sqrt(212.15) = 14.57[/tex]

So the sample standard deviation is 14.57%.

c. To construct a 95% confidence interval for the population variance, we can use the chi-square distribution with degrees of freedom n - 1 = 7. The upper and lower bounds of the confidence interval can be found using the chi-square distribution table or calculator, as follows:

upper bound = (n - 1) * s^2 / chi-square(0.025, n - 1) = 306.05

lower bound = (n - 1) * s^2 / chi-square(0.975, n - 1) = 91.91

So the 95% confidence interval for the population variance is (91.91, 306.05).

d. To construct a 95% confidence interval for the standard deviation (as percent), we can use the formula:

lower bound = s * √((n - 1) / chi-square(0.975, n - 1))

upper bound = s * √((n - 1) / chi-square(0.025, n - 1))

Plugging in the values, we get:

lower bound = 6.4685%

upper bound = 20.1422%

So the 95% confidence interval for the standard deviation (as percent) is (6.4685%, 20.1422%).

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The given statement " when the population variance is not known (i.e., must be estimated from data), we use a z-statistic instead of a t-statistic for our hypothesis tests. " is false. Because in distribution of sample means, population variance is unknown.

When the population variance is not known and must be estimated from the data, we use a t-statistic instead of a z-statistic for our hypothesis tests.

This is because the distribution of the sample means follows a t-distribution when the population variance is unknown, whereas it follows a standard normal distribution (z-distribution) when the population variance is known.

The t-distribution has fatter tails than the z-distribution to account for the extra uncertainty introduced by estimating the population variance from the sample.

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The experimental probability of winning the contest based on the data of all 3 games is 0.432.

What is experimental probability ?

Experimental probability is a measure of the likelihood of an event occurring based on the results of an experiment or observation. It is determined by dividing the number of times the event occurred by the total number of trials or observations. The more trials or observations conducted, the more accurate the experimental probability will be. Experimental probability is often used in situations where the probability of an event cannot be determined theoretically or where the theoretical probability is difficult to calculate. It is also commonly used in scientific experiments, market research, and other fields where the results of an experiment or observation can be used to make predictions or inform decisions.

Finding the experimental probability of winning the contest :
In this case, the event is winning the contest by choosing a marble from a bag, and the trials are the three games played by Hal.

Total number of players in all three games = 123 + 155 + 172 = 450

Total number of winners in all three games = 52 + 63 + 65 = 180

Experimental probability of winning the contest = Number of winners / Total number of players = 180/450 = 0.432

Therefore, the experimental probability of winning the contest based on the data of all 3 games is 0.432.

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a. The number of shares purchased can be calculated as:

Number of Shares = (Investment Amount / Face Value)

= 25,059/ 10 = 2,505.9 = 2,506 (to the nearest whole number)

b. The amount of brokerage paid can be calculated as:

Brokerage = (Investment Amount x Brokerage Rate)

= 2,50,590 x 0.002 = Rs. 501.18

c. The GST paid for the trading can be calculated as:

GST = (Brokerage Amount x GST Rate)

= 501.18 x 0.18 = Rs. 90.22

The equation of the tangent line to the curve sin( x y) = 8x- 8y on the factor( π, π) is y = (7/9) x-( 2π/ 9).

To discover the equation of the tangent line to the curve sin( x y) = 8x- 8y on the point( π, π), we want to apply implicit differentiation to discover the pitch of the tangent line at that point.

We begin through differencing both sides of the equation with reference to xcos( x y)( 1 dy/ dx) = eight- 8dy/ dx

After, we can simplify the expression by isolating the terms beholding dy/ dx on one aspect

cos( x y) cos( x y) dy/ dx = 8- 8dy/ dx

8 cos( x y)) dy/ dx = 8- cos( x y)

dy/ dx = ( 8- cos( x y))( 8 cos( x y))

Now we're suitable to discover the pitch of the tangent line at the factor( π, π) by plugging in x = π and y = π into the expression we simply derived

dy/ dx = ( 8- cos( 2π))( 8 cos( 2π))

dy/ dx = ( 8- 1)/( 8 1)

dy/ dx = 7/ nine

Thus, the pitch of the tangent line to the curve sin( x y) = 8x- 8y at the factor( π, π) is7/9.

To find the equation of the tangent line, we can use the point- slope form of the equation

y- y1 = m( x- x1)

In which m is the pitch we simply set up, and( x1, y1) is the point( π, π). Plugging in the values, we get

y- π = ( 7/ nine)( x- π)

Simplifying, we get

y = ( 7/ nine) x-( 2π/ nine)

Thus, the equation of the tangent line to the curve sin( x y) = 8x- 8y on the factor( π, π) is y = (7/9) x-( 2π/ 9).

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The population distribution of scores has a similar average as the distribution of means.

In a sample distribution of means, the distribution will roughly resemble a normal distribution, the sampling distribution's mean will be equal to the population's mean, and the sampling distribution's standard deviation is based on the Central Limit Theorem.

The central limit theorem (CLT) of probability theory states that the distribution of a sample variable tends towards a normal distribution (i.e., a "bell curve") as the sample size increases, under the premise that all samples are of equal size and regardless of the population's actual distribution shape.

In other terms, the central limit theorem (CLT) is a statistical presumption that the mean of all sampled variables from the same population will be substantially equal to the mean of the entire population, given a large enough sample size from a population with little volatility. These samples likewise resemble a normal distribution in accordance with the law of large numbers, with their variances nearly matching the variance of the population as the sample size rises.

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the actual question is :

When comparing the size of the standard error of the mean with the size of the standard deviation of the underlying distribution of individual scores:

a. the standard error of the mean is always larger

b. the standard error of the mean is always smaller

c. the standard error of the mean is sometimes larger and sometimes smaller, depending on the sample size

d. none of these

The expected value of playing this game is -$0.251. This means that, on average, you can expect to lose 25.1 cents every time you play this game.

The term "outcome" generally refers to the result or effect of an action, process, or event. It refers to what happens as a consequence of a particular decision, activity, or circumstance.

To calculate the expected value, we multiply the possible outcomes by their probabilities and then sum them up.

Let's start with the possible outcomes:

If you win, you make a profit of $749, so the outcome is $749.

If you lose, you lose $1, so the outcome is -$1.

The probabilities are:

Probability of winning is 0.001.

Now we can calculate the expected value:

Expected value = (0.001 * $749) + (0.999 * -$1)

Expected value = $0.748 - $0.999

Expected value = -$0.251

Therefore, the expected value of playing this game is -$0.251. This means that, on average, you can expect to lose 25.1 cents every time you play this game.

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The equation that models the position of the weight at time t seconds is y(t) = 2.5 sin (πt) + 2.5

The equation that models the position of the weight at time t seconds is

y(t) = 2.5 sin (πt) + 2.5

The position of the weight at time t seconds can be modeled by a sinusoidal function of the form

y(t) = A sin (ωt + φ) + C

where

A is the amplitude of the oscillation (half the distance between the highest and lowest points), which is 2.5 in.

ω is the angular frequency of the oscillation, which is 2π divided by the period (the time for one complete cycle), which is 2 seconds. So, ω = 2π/2 = π radians/second.

φ is the phase angle, which depends on the initial position of the weight. We can assume that the weight starts at the highest point (the crest), so φ = 0.

C is the vertical shift, which is the midpoint of the oscillation (half the distance between the highest and lowest points added to the starting point). So, C = 2.5 + 0 = 2.5 in.

Putting it all together, we get

y(t) = 2.5 sin (πt) + 2.5

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Answer:

Greg bought a pack of X jumbo stickers during the back - to - school sale at Crafty's craft store. He ( uses 4  ( -4 ), ( you didn't give options ). There were 8 stickers left in the pack.

Hope this helps!

Step-by-step explanation: