A bacteria colony increases in size at a rate of 4.0565e1.3t bacteria per hour. If the initial population is 54 bacteria, find the population four hours later. (Round your answer to the nearest whole number.) bacteria

Answer: 2a + b - 3cd

Step-by-step explanation:

a 2 + b - cd 3

2a + b - 3cd

Answer:

P(A) = 3/20

Step-by-step explanation:

P(A)=P(blue)P(head)=(3/10)(1/2)=3/20

as there are 10 cards in total, out of which 3 are blue so the probability to get the blue card is, P(blue) = 3/10. and the probability of getting a head when a coin is tossed is P(head) = 1/2.

So in total

P(A) = P(blue)*p(head) = (3/10)*(1/2) = 3/20 = 0.15

Final answer:

P(A) = 0.15.

Explanation:

To find P(A), the probability of event A happening, we need to consider two separate events: picking a blue card and tossing a head on a coin.

First, the probability of picking a blue card from the special deck is 3 out of 10, because there are 3 blue cards among the 10 total cards.

Next, the probability of tossing a head on a coin is 1 out of 2, since a coin has two sides and both outcomes (heads and tails) are equally likely.

To find the combined probability P(A) of the two independent events, we multiply the probabilities:

[tex]P(A) = P(Blue Card) \(\times\) P(Head Coin Toss)[/tex]

Thus:

[tex]P(A) = (3/10) \(\times\) (1/2) = 3/20 = 0.15[/tex]

Rounded to two decimal places, P(A) = 0.15.

Answer:

Step-by-step explanation:

Given a rectangular tank with dimension (10ft by 20ft by 16ft)

Then the volume of the tank is

Volume =length × breadth ×height

Volume=10×20×16=3200ft³

V=3200ft³

Then,

∆F= weight density × volume

∆F= 62.4×3200

∆F= 199,680 lb

Then,

Let the 0-point on the x-axis be at the bottom of the tank, so the level of the water ranges from x = 0 to x = 16ft. (It would just as well to let 0 be ground level and let x range from x = −26ft to x − 0.) Then a slice of water at level x is raised (26− x)ft

Then ∆x=(26-x)ft

Work is given as

W= -∫F∆xdx. From 0 to 16

W= -∫199,680(26-x)dx From 0 to 16

W=-199,680∫26-x dx From 0 to 16

W=-199,680 (26x-x²/2). From 0 to 16

W=-199,680(26×16-0.5×16² -0-0)

W=-199,680(288)

W=-57,507,840J

The work done to empty the tank is 57,507,840J

Answer:

2 X 10∧6 ft.lb

Step-by-step explanation:

Volume of rectangular tank =  Base area X Height = (10 X 20 X 16)ft³ = 3200ft³

Mass of tank filled with water = 3200ft³ X 62.4 lb/ft³ = 199680lb or 90706.37kg

Hence, work done due to gravity,Δg = m * g * h, where, m = mass of tank, g = gravity = 9.8m/s² and h = height = 10ft or 3.05 meter(m)

∴ Δg = 90706.37 * 9.8 * 3.05 =  2.711 X10∧6 J or 2 X 10∧6 ft.lb

Answer:

The lowest 10 percent of the citizens would need at least 52.8 minutes to complete the form.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 40, \sigma = 10[/tex]

The slowest 10 percent of the citizens would need at least how many minutes to complete the form

This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.28 = \frac{X - 40}{10}[/tex]

[tex]X - 40 = 1.28*10[/tex]

[tex]X = 52.8[/tex]

The lowest 10 percent of the citizens would need at least 52.8 minutes to complete the form.

Final answer:

To find the time needed for the lowest 10 percent of citizens to complete the 2010 U.S. Census 'long' form, we use the z-score formula to calculate the value. The lowest 10 percent of citizens would need at least 26.8 minutes to complete the form.

Explanation:

To find how many minutes would be needed for the lowest 10 percent of citizens to complete the 2010 U.S. Census 'long' form, we need to use the z-score formula.

First, we calculate the z-score using the formula: z = (x - mean) / standard deviation, where x is the value we want to find the z-score for, mean is the mean of the distribution, and the standard deviation is the standard deviation of the distribution.

Next, we use a z-table or a calculator to find the z-score that corresponds to a cumulative probability of 0.1. The z-table gives us a z-score of -1.28. We can rearrange the z-score formula to solve for x: x = (z * standard deviation) + mean. Plugging in the values, we get: x = (-1.28 * 10) + 40 = 26.8. Therefore, the lowest 10 percent of citizens would need at least 26.8 minutes to complete the form.

Answer:

a) Parameter of interest [tex] p[/tex] representing the true proportion of the plates have blistered.

b) Null hypothesis:[tex]p\leq 0.1[/tex]  

Alternative hypothesis:[tex]p > 0.1[/tex]  

c) [tex]z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33[/tex]  

d) For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:

[tex] z_{critc}= 1.28[/tex]

e) For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1

f) [tex]p_v =P(z>1.33)=0.0917[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

Part a

Parameter of interest [tex] p[/tex] representing the true proportion of the plates have blistered.

X=14 represent the number of the plates have blistered.

[tex]\hat p=\frac{14}{100}=0.14[/tex] estimated proportion of the plates have blistered.

[tex]p_o=0.1[/tex] is the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Part b: Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that  more than 10% of all plates blister under such circumstances.:  

Null hypothesis:[tex]p\leq 0.1[/tex]  

Alternative hypothesis:[tex]p > 0.1[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Part c: Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.14 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=1.33[/tex]  

Part d: Rejection region

For this case we need to find a value in the normal standard distribution that accumulates 0.1 of the area in the right tail and for this case is:

[tex] z_{critc}= 1.28[/tex]

Part e

For this case since our calculated value is higher than the critical value 1.33>1.28 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion is significantly higher than 0.1

Part f

Since is a right taild test the p value would be:  

[tex]p_v =P(z>1.33)=0.0917[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the true proportion is higher than 0.1 or 10%

Answer:children at home prefer hot breakfast than hot school breakfast because at home you could put it in the refrigerator but at school you have to throw it in the trash before you go to trash.

Step-by-step explanation:

Parental care in this context is considered the independent variable because it influences children's school performance.

In this scenario, parental care is considered an independent variable. The reason for this is because it is the variable that influences or predicts the outcome, which in this case, is children's performance in school. The independent variable is the one that is manipulated or controlled in a study to observe its effects on the dependent variable (children's school performance here). Examples of parental care might include ensuring the child eats a good breakfast, aiding with schoolwork, or providing emotional support.

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Answer: C.$10,000

Step-by-step explanation:

Assuming the interest was compounded annually, then we would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

A = $31066

r = 6.5% = 6.5/100 = 0.065

n = 1 because it was compounded once in a year.

t = 18 years

Therefore,

31066 = P(1+ 0.065/1)^1 × 18

31066 = P(1.065)^18

31066 = 3.12P

P = 31066/3.12

P = $9957.1

Approximately $10000 to the nearest dollar

Answer:

[tex](-\infty,3)[/tex]

Step-by-step explanation:

We are given that

[tex](x-3)y''+4y=x[/tex]

[tex]y''+\frac{4}{x-3}y=\frac{x}{x-3}[/tex]

y(0)=0

y'(0)=1

By comparing with

[tex]y''+p(x)y'+q(x)y=g(x)[/tex]

We get

[tex]p(x)=\frac{4}{x-3}[/tex]

[tex]g(x)=\frac{x}{x-3}[/tex]

q(x)=0

p(x),q(x) and g(x) are continuous for all real values of x except 3.

Interval on which p(x),q(x) and g(x) are continuous

[tex](-\infty,3)[/tex]and (3,[tex]\infty)[/tex]

By unique existence theorem

Largest interval which contains 0=[tex](-\infty,3)[/tex]

Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=[tex](-\infty,3)[/tex]

The largest interval on which includes x=0 for which given initial-value problem has unique solution is [tex](-\infty, 3)[/tex]

The given parameters are:

(x − 3)y'' + 4y = x,

y(0) = 0

y'(0) = 1

Divide the equation (x − 3)y'' + 4y = x through by (x - 3)

[tex]y'' + \frac{4y}{x - 3} = \frac{x}{x - 3}[/tex]

Compare the above equation to the following equation

y" + p(x) y' +  q(x)y = g(x)

Then, we have:

[tex]p(x) = \frac{4y}{x - 3}[/tex]

[tex]q(x) = 0[/tex]

[tex]g(x) = \frac x{x - 3}[/tex]

The domains of functions p(x) and g(x) are all set of real values except 3

This is represented as:

[tex](-\infty, 3)\ u\ (3,\infty)[/tex]

Using the unique existence theorem, we have:

The largest interval that contains x = 0 is [tex](-\infty, 3)[/tex]

Hence, the largest interval on which includes x=0 for which given initial-value problem has unique solution is [tex](-\infty, 3)[/tex]

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Answer:

The probability distribution of [tex]\hat p[/tex] is, [tex]N(0.70, 0.032)[/tex].

Step-by-step explanation:

The random variable X is defined as the number of brand A external hard drives work in a satisfactory manner throughout the warranty period.

The sample selected is of size, n = 15.

The probability of selecting a hard drive that works in a satisfactory manner throughout is, p = 0.70.

Every hard drive works independently of the others.

The random variable X follows a Binomial distribution with parameters n = 15 and p = 0.70.

The probability mass function of X, the binomial random variable is:

[tex]P(X=x)={15\choose x}0.70^{x}(1-0.70)^{15-x};\ x=0,1,2,3...[/tex]

The probability distribution of X is shown below.

Now a statistic is defined as:

[tex]\hat p=\frac{X}{n}[/tex]

This statistic is known as the sample proportion.

A Normal approximation to Binomial can be used to approximate the distribution of sample proportion if the following conditions are satisfied:

Then the sampling distribution of [tex]\hat p[/tex] follows a Normal; distribution with mean [tex]p=0.70[/tex] and standard deviation [tex]\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.70(1-0.70)}{200}}=0.032[/tex].

Check the conditions:

[tex]np=15\times0.70=10.5\\n(1-p)=15\times (1-0.70)=4.5\approx5[/tex]

So, the probability distribution of [tex]\hat p[/tex] is, [tex]N(0.70, 0.032)[/tex].

The sampling distribution of X/n is approximately normally distributed with a mean of 0.7 and a standard deviation of 0.117.

To determine the sampling distribution of the statistic X/n, we need to consider the probability distribution of the number of successes (X) in the sample of n = 15 drives.

Since the probability of a success (a working hard drive) is p = 0.7 and the probability of a failure (a non-working hard drive) is q = 1 - p = 0.3, we can model X using the binomial distribution with parameters n = 15 and p = 0.7.

The probability of getting exactly x successes (working hard drives) in a sample of n = 15 drives is given by the binomial probability mass function (PMF):

P(X = x) = (nCx) * px * (1-p)^(n-x)

where:

nCx is the binomial coefficient, which represents the number of ways to choose x successes out of n trials.

px is the probability of success (working hard drive) on each trial.

(1-p)^(n-x) is the probability of failure (non-working hard drive) on each trial.

For x = 3 successes (working hard drives), we calculate the probability:

P(X = 3) = (15C3) * (0.7)^3 * (0.3)^12 ≈ 0.193

This means that the probability of observing 3 successes (working hard drives) in a sample of 15 drives is approximately 0.193.

The sample proportion (fraction) of successes, X/n, is a continuous random variable that can take values between 0 and 1. To determine the sampling distribution of X/n, we need to consider the distribution of X and how it transforms to X/n.

The distribution of X/n is approximately normal for large sample sizes (n ≥ 30), but for smaller sample sizes, it can be skewed. In this case, with n = 15, the distribution of X/n is slightly skewed, but it is still close to being normal.

The mean of the sampling distribution of X/n is equal to the population proportion, which is p = 0.7.

The standard deviation of the sampling distribution of X/n is given by:

σ(X/n) = √(p * q / n) ≈ √(0.7 * 0.3 / 15) ≈ 0.117

Therefore, the sampling distribution of X/n is approximately normally distributed with a mean of 0.7 and a standard deviation of 0.117.

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Answer:

0.7262

Step-by-step explanation:

Let random variable = X

Probability of getting a cookie with at least 55 chips:

P(55≤ X ≤78), where X1 = 55 and X2 = 78

Let mean, U = 66 and  standard deviation, S.D. = 10

Let Z = X - U/S.D; Z1 = X1 - U/S.D and Z2 = X2 - U/S.D

Z1 = 55 - 66/10 = - 1.1

Z2 = 78 - 66/10 = 1.2

P(X1≤ X ≤ X2) = P(X1 -U/S.D ≤ X - U/S.D ≤ X2 - U/S.D)

P(Z1 ≤ Z ≤ Z2) = P(0 ≤ Z ≤ Z1) + P(0 ≤ Z ≤ Z2)

Hence, (55 ≤ X ≤ 78) = P( - 1.1 ≤ Z ≤ 1.2)

∴ P(-1.1 ≤ Z ≤ 1.2) = P(0 ≤ Z ≤ 1.1) + (0 ≤ Z ≤ 1.2)

=   0.3413 + 0.3849 = 0.7262

Answer:

the cost function is Cost=7000 m*$ /R + 50.265 $/m² * R²

Step-by-step explanation:

then the cost function is

Cost= cost of side area+ cost of top + cost of bottom = 2*π*R*L * 5$/m² +

π*R² * 8$/m² +  π*R² * 8$/m²

since the volume V is

V=π*R²*L → V/(π*R²)=L

then

Cost=2*π*R*V/(π*R²) * 5$/m² +  π*R² * 8$/m² +  π*R² * 8$/m²

replacing values

Cost=2*700 m³ /R * 5$/m² +  π*R² * 16$/m² = 7000 m*$ /R + 50.265 $/m² * R²

thus the cost function is

Cost=7000 m*$ /R + 50.265 $/m² * R²

Answer: 50.24r² + 7000/r

Step-by-step explanation:

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Since the volume of the can is 700cm³, then

πr²h = 700

h = 700/πr²

The formula for determining the total surface area of a cylinder is expressed as

Total surface area = 2πr² + 2πrh

The surface area of the top and bottom of the can is 2πr².

Since top and bottom are made up of a material that costs $8 per square centimeter, then the cost is

2πr² × 8 = 16πr²

Since π = 3.14, the surface area of the top and bottom of the cylindrical can is

16 × 3.14 × r² = 50.24r²

The surface area of the side of the can is

2πrh = 2πr × 700/πr²

= 1400/r

Since the the sides are made of material that costs $5 per square centimeter, then the cost is

1400/r × 5 = 7000/r

The total cost of the material as a function of the radius, r of the cylinder is

50.24r² + 7000/r

Answer:

Requirements are computed below.

Step-by-step explanation:

For ABC analysis:

Step1: Calculate the total value of each item.

SKU Number of Items Value ($/item) Value of the product ($)

A 220 67 14740

B 225 3 675

C 245 21 5145

D 145 7 1015

E 230 26 5980

F 240 83 19920

Step2: Reorganize the table on the basis of total value in descending order, & calculate % share of total value of investment and cumulative % share of total value of investment.

SKU Number of Items Value ($/item) Value of the product ($) % of Total value Cumulative % of total value Segment

SKU Number of Items Value ($/item) Value of the product ($) % of Total value Cumulative % of total value Segment

F 240 83 19920 41.96 41.96 A

A 220 67 14740 31.05 73.01 A

E 230 26 5980 12.60 85.60 B

C 245 21 5145 10.84 96.44 B

D 145 7 1015    2.14   98.58 C

B 225 3 675         1.42         100.00 C

----------

% of total number of items in segment B = (230+245)/1305 = 36.4%

% of total value of inventory in segment B = (12.6 + 10.4)/100 = 23.43%

% of the total weight of inventory in segment B = (9660+14700)/67795 = 35.93%

SKU Number of Items Value ($/item) Weight (lb/item) Total weight

A 220 67 52 11440

B 225 3 59 13275

C 245 21 60 14700

D 145 7 48 6960

E 230 26 42 9660

F 240 83 49 11760

Answer:

0.436

Step-by-step explanation:

Given that a professor planned to give an examination in a large class on the Monday before Thanksgiving vacation

Some students asked whether he could change the date because so many of their classmates had at least one other exam on that date.

They speculated that at least 40% of the class had this problem.

The professor agreed to poll the class, and if there was convincing evidence that the proportion with at least one other exam on that date was greater than .40, he would change the date.

No of students in total = 250

Reported they had another exam = 109

proportion of the class reported that they had another exam on the date

= [tex]\frac{109}{250} =0.436[/tex]

Answer:

Step-by-step explanation:

Hello!

X: amount spent in a supermarket impulse buying in a 10 min unplanned shopping interval by one customer.

It is known that the mean of this variable is μ= $48 and its standard deviation is δ=$7

a.

The Central limit theorem states that if there is a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

X[bar]≈N(μ;σ²/n)

The mean of the sampling distribution is μ= $48

The standard deviation of the sampling distribution is σ/√n= $7/√60= $0.90

It is not necessary to make any assumption about the distribution of X since n=60 is considered large enough, you can directly approximate the sampling distribution to normal regardless of the distribution of X.

b.

To calculate this probability you have to use the approximation of the sampling distribution:

Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

μ= $48

σ/√n= $0.90

P(46≤X[bar]≤50)= P(X[bar]≤50) - P(X[bar]≤46)

P(Z≤(50-48)/0.90) - P(Z≤(46-48)/0.90)

P(Z≤2.22) - P(Z≤-2.22)= 0.987 - 0.013= 0.974

c.

If we assume that X has an approximately normal distribution, then you will use it's a mean and standard deviation to reach the asked probability.

Z= (X-μ)/δ≈N(0;1)

μ= $48

δ= $7

P(46≤X≤50)= P(X≤50) - P(X≤46)

P(Z≤(50-48)/7) - P(Z≤(46-48)/7)

P(Z≤0.29) - P(Z≤-0.29)= 0.61409 - 0.38591= 0.22818≅ 0.2282

I hope it helps!

The Central Limit Theorem ensures that the average amount spent by a sample of 60 customers due to impulse buying is normally distributed with a mean of $48 and a standard error of approximately $0.90.

Regarding part (a) of the question, by applying the Central Limit Theorem (CLT), we can state that for the sample of n = 60 customers, the sampling distribution of the sample mean x will be approximately normal due to the large sample size, even if the distribution of x is not normal. The mean of the distribution μx remains the same at $48, and the standard deviation (often referred to as the standard error, σx) can be calculated by dividing the population standard deviation by the square root of the sample size (n), which is σx = $7/√60 ≈ $0.90. Thus, the correct statement is: 'The sampling distribution of x is approximately normal with mean μx = $48 and standard error σx = $0.90.'

Answer:

Let X the random variable who represent the number of occurences in a period of time for the calls.

For this case we have the following parameter [tex] \lambda = 1 \frac{call}{2 minutes}[/tex]

And we are interested in the expected number of calls in one hour.

We know that 1 hr = 60 mins so then the expected number of calls that arrive in one hour are:

[tex] \lambda = \frac{1 call}{2 minutes} * \frac{60 minutes}{1 hour} = 30 calls per hour[/tex]

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]

And the parameter [tex]\lambda[/tex] represent the average ocurrence rate per unit of time.

Solution to the problem

Let X the random variable who represent the number of occurences in a period of time for the calls.

For this case we have the following parameter [tex] \lambda = 1 \frac{call}{2 minutes}[/tex]

And we are interested in the expected number of calls in one hour.

We know that 1 hr = 60 mins so then the expected number of calls that arrive in one hour are:

[tex] \lambda = \frac{1 call}{2 minutes} * \frac{60 minutes}{1 hour} = 30 calls per hour[/tex]

Answer:

Step-by-step explanation:

You should try going too jeeska

Answer:

Step-by-step explanation:

Elise’s garden is rectangular. The formula for determining the area of a rectangle is expressed as

Area = length × width

From the information given,

Length of garden = 6 feet

Width of garden = 12.6 feet

Area of garden = 6 × 12.6 = 75.6 square feet

Tomato plants require 2 square feet of space. This means that the number of tomato plants that Elise can fit in her garden is

75.6/2 = 37.8

Since the number of tomato plants must be whole number, then the number of tomato plants that Elise can fit in her garden is 37

Answer:

Therefore Rotten Tomato ratings explain about 75% of the variation in Metascore ratings is the correct answer here.

Step-by-step explanation:

The R2 value for regression or the coefficient of determination represents the proportion of variation in dependent variable that is explained by regression ( the rest of it is residual variation )

The given question is about the analysis of a linear regression model to examine the relationship between Rotten Tomatoes ratings and Metascore ratings for popular movies.

The given question pertains to linear regression analysis in statistics. It involves examining the relationship between Rotten Tomatoes ratings and Metascore ratings for a sample of 75 popular movies. The r-squared value of 0.75 indicates that approximately 75% of the variation in Metascore ratings can be explained by the linear relationship with Rotten Tomatoes ratings. Additionally, for every one-point increase in Rotten Tomatoes ratings, the regression model predicts a 0.75-point increase in Metascore ratings.

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Answer:

There is enough evidence to support the claim the mean nicotine content is actually higher than advertised.          

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 1.5

Sample mean, [tex]\bar{x}[/tex] = 1.53

Sample size, n = 100

Alpha, α = 0.05

Population standard deviation, σ = 0.17

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 1.5\\H_A: \mu > 1.5[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{1.53 - 1.5}{\frac{0.17}{\sqrt{100}} } = 1.7647[/tex]

Now, we calculate the p-value from the standard normal table.

P-value = 0.038807

Since the p-value is less than the significance level we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim the mean nicotine content is actually higher than advertised.

Answer:

Yes, the mean nicotine content is actually higher than advertised.

Step-by-step explanation:

We are given that the nicotine content in cigarettes of a certain brand is known to be right-skewed with mean (in milligrams) μ and a known standard deviation σ = 0.17.

Also, the brand advertises that the mean nicotine content of its cigarettes, [tex]\mu[/tex] is 1.5, but measurements on a random sample of 100 cigarettes of this brand give a sample mean of x bar = 1.53.

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 1.5

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 1.5

The test statistics used here will be;

     T.S. = [tex]\frac{xbar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

Here, n = sample size = 100

So, test statistics = [tex]\frac{1.53 - 1.5}{\frac{0.17}{\sqrt{100} } }[/tex] = 1.765

Now, at 0.05 level of significance, the standard z table gives critical value of 1.6449. Since our test statistics is more than the critical value as 1.765 > 1.6449 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region.

Therefore, we conclude that the mean nicotine content is actually higher than advertised.

Answer:

True, this is the meaning of the z-score. Z = 0.65 means that the raw score is 0.65 standard deviations above the mean.

Step-by-step explanation:

The Z-score measures how many standard deviations a raw score is from the mean.

For example, a z-score of -2 means that the raw score is 2 standard deviations below the mean.

Another example, a z-score of 2 means that the raw score is 2 standard deviations above the mean.

If z= 0.65, then the raw score is 0.65 standard deviations above the mean.

True, this is the meaning of the z-score. Z = 0.65 means that the raw score is 0.65 standard deviations above the mean.

Answer:

22.96% probability that a hotel room costs between $250.00 and $285.00

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 244, \sigma = 55[/tex]

What is the probability that a hotel room costs between $250.00 and $285.00?

This is the pvalue of Z when X = 285 subtracted by the pvalue of Z when X = 250. So

X = 285

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{285 - 244}{55}[/tex]

[tex]Z = 0.75[/tex]

[tex]Z = 0.75[/tex] has a pvalue of 0.7734

X = 250

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{250 - 244}{55}[/tex]

[tex]Z = 0.11[/tex]

[tex]Z = 0.11[/tex] has a pvalue of 0.5438

0.7734 - 0.5438 = 0.2296

22.96% probability that a hotel room costs between $250.00 and $285.00

Answer:

The probability that a child with a rash does not have a fever is 22%

Step-by-step explanation:

1. Probability of having fever:

[tex]P(fever)=0.70[/tex]

2. Probability of not having fever:

[tex]P(not fever)=1-P(fever)\\P(not fever)=1-0.70\\P(not fever)=0.30[/tex]

3. Probability of fave fevers and a rash:

[tex]P(fever and rash)=(0.70)(0.30)=0.21[/tex]

4. Probability of having a rash but not a fever:

[tex]P(rash and not fever)=(0.30)(0.20)=0.60[/tex]

5. Probability of having a rash:

[tex]P(rash)=P(rash and fever)+P(rash and no fever)\\P(rash)=0.21+0.06=0.27[/tex]

6. Probability a child with a rash does not have a fever

[tex]P=\frac{P(rash and not fever)}{P(rash)} =\frac{0.06}{0.27} =0.22[/tex]

22% of the child at the doctor's office with a rash does not have a fever.

Final answer:

To find the probability that a child with a rash does not have a fever, we analyze the given percentages, calculate how many children have a rash with and without fever, and then find the ratio of those without a fever to the total number with a rash, resulting in a probability of approximately 22.22%.

Explanation:

The question asks: What is the probability that a child at the doctor's office with a rash does not have a fever? To solve this, let's start by analyzing the given percentages.

70% of children who go to the doctor have fevers.

Of those with fevers, 30% also have a rash.

Of those without fevers, 20% have a rash.

To find the probability that a child with a rash does not have a fever, we need to calculate the proportion of children with a rash who are fever-free compared to all children with a rash.

Step-by-step Calculation:

Assuming 100 children visit the doctor: 70 will have fevers, and 30 will not.

Of the 70 with fevers, 21 (30% of 70) have a rash.

Of the 30 without fevers, 6 (20% of 30) have a rash.

In total, 27 children have a rash (21 with fever + 6 without).

The probability a child with a rash does not have a fever is the number of children with a rash but no fever divided by the total number of children with a rash: 6/27.

This calculation shows that the probability of a child having a rash but no fever is 6/27 or approximately 22.22%.

Answer:

interest income in amount of $4000 will be accrued

Step-by-step explanation:

given data

principal = $100,000

rate = 8 percent = 0.08  

time period = 6 month ( July - December ) = [tex]\frac{1}{2}[/tex] year

solution

we get here interest at December 31, 2015 that is express as

interest = principal × rate × time ..................1

put here value and we will get

interest =  $100,000 × 0.08 × [tex]\frac{1}{2}[/tex]    

interest = $100,000 × 0.08 × 0.5

interest = $4000

so interest income in amount of $4000 will be accrued

Answer:

A. No

City driving mean requires less miles than Highway driving mean.

B. No

City driving median requires less miles than Highway driving median.

C. No

City driving mode requires less miles than Highway driving mode

D.

The driving conditions in the City is better than the driving conditions in the Highway because all measures of centers are lesser than those of Highway.

Step-by-step explanation:

Answer:

99% confidence interval for the proportion of adults who professed to be baseball fans in November 2003 is (46.9%, 55.1%)

This interval means that the lower limit of the proportion of adults who professed to be baseball fans is 46.9% and the upper limit of the proportion is 55.1%

Step-by-step explanation:

Confidence interval = P' + or - t×sqrt[P'(1-P') ÷ n]

P' is sample proportion = 51% = 0.51

n = 1001

confidence level = 99%

t-value corresponding to 99% confidence interval and infinity degree of freedom is 2.576

t × sqrt[P'(1-P') ÷ n] = 2.576 × sqrt[0.51(1-0.51) ÷ 1001] = 2.576 × 0.0158 = 0.041

Lower limit = P' - 0.041 = 0.51 - 0.041 = 0.469 = 46.9%

Upper limit = P' + 0.041 = 0.51 + 0.041 = 0.551 = 55.1%

99% Confidence interval is (46.9%, 55.1%)

The interval means that the proportion of adults who professed to be baseball fans in November 2003 is between 46.9% and 55.1%

Answer:

a) {4,5,6,7,8,9,10}

b) {1,2,3,4,5}

c) {4,5,6}

Answer:

The interval is approximately a 98% confidence interval

Step-by-step explanation:

From the question :  Error, E= 3% = 0.03, Total population, n=1000,  number of people that are concerned that their taxes will be audited, p = 19% = 0.19

E^{2}=z_{\alpha/2}^{2}\cdot \frac{p(1-p)}{n}

0.03^{2}=z_{\alpha/2}^{2}\cdot \frac{0.19(1-0.19)}{1000}

z_{\alpha/2}^{2}=5.848

z_{\alpha/2}=2.418

Area right to 2.418 is 0.0078. So

\alpha/2=0.0078

Therefore \alpha=0.0156\approx 0.02

Thus, the interval is approximately a 98% confidence interval.

Answer:

98 percent

Step-by-step explanation:

plato

Answer:

(a) Initial concentration of salt = 0.07 kg/L

(b) [tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]

(c) [tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]

(d)Therefore the amount of salt after 5 hours is  =38.18 kg

(e) The concentration of salt in the solution in the tank as time approaches infinity is = 0.035 kg/L

Step-by-step explanation:

Given that,

A tank contains 70 kg of salt and 1000 L of water.

(a)

[tex]\textrm{Concentration of salt }=\frac{\textrm{mass of salt}}{\textrm{volume of water}}[/tex]

                                 [tex]=\frac{70}{1000} kg/L[/tex]

                                =0.07 kg/L

(b)

Let Y(t) be the amount of salt at any instant time t.

Therefore

[tex]Y'(t)= Y_{in}-Y_{out}[/tex]

[tex]Y_{in}=\textrm{concentration of salt} \times \textrm{rate of enter}[/tex]

     =(8×0.035) kg/min

     =0.28 kg/min

Since the rate of water in and out are same , the amount of solution remain constant.

[tex]Y_{out}= (\frac{y}{1000}\times 8) kg/min[/tex]

       [tex]=\frac{y}{125} kg/L[/tex]

[tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]

(c)

The above equation can be rewrite as

[tex]Y'(t) +\frac{y}{125} =0.28[/tex]

The coefficient of y is p(t) [tex]=\frac{1}{125}[/tex]

The integrating factor of the D.E is[tex]=e^{\int p(t) dt=[/tex] [tex]e^{\int \frac{1}{125} dt[/tex]  [tex]=e^{\frac{1}{125} t[/tex]

Multiplying the integrating factor of both sides of D.E

[tex]e^{\frac{1}{125} t} \ \frac{dY}{dt} +e^{\frac{1}{125} t} .\frac{1}{125} Y=0.28 \ e^{\frac{1}{125} t}[/tex]

Integrating both sides

[tex]\int e^{\frac{1}{125} t} \ \ dY+\int e^{\frac{1}{125} t} .\frac{1}{125} Y \ dt=\int0.28 \ e^{\frac{1}{125} t}\ dt[/tex]

[tex]\Rightarrow e^{\frac{1}{125} t} \ Y= \frac{0.28e^{\frac{1}{125}t} }{\frac{1}{125}} +C[/tex]

[tex]\Rightarrow Y=35+Ce^{-\frac{1}{125}t}[/tex]

At initial when t= 0, y =70

Therefore

[tex]70=35+Ce^0[/tex]

⇒C= 70-35

⇒C=35

Therefore

[tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]

(d)

When t= 5 hour = 300 min

To find the amount of salt after 5 hour , we need to put the value of t in the general solution of D.E

Therefore the amount of salt after 5 hours is =  [tex]Y(300)=35+35e^{-\frac{300}{125} }[/tex]

                                                                                          = 38.18 Kg

(e)

When t=∞

[tex]Y(\infty )= 35 +e^{-\infty}[/tex]

         = 35 Kg   [tex][e^{-\infty}=0][/tex]

Since the amount of water is remain same i.e 1000 L

Therefore the concentration of salt is [tex]=\frac{35}{1000}kg/L[/tex]

                                                                 =0.035 kg/L

This problem involves using differential equations to model the concentration of salt in a tank over time. The initial concentration is determined, an initial value problem is set up, the initial value problem is solved, the amount of salt after a specific time is found, and the approaching concentration as time goes to infinity is identified.

(a) The initial concentration of the solution can be found by dividing the amount of salt by the volume of water. So, 70 kg / 1000 L = 0.07 kg/L.

(b) Let y be the amount of salt in the tank at time t. Then, dy/dt = 0.035 kg/L * 8 L/min - 0.07 kg/L * 8 L/min since salt is being added at a rate of 0.035 kg/L *8 L/min and leaving at a concentration of 0.07 kg/L * 8 L/min. Thus, dy/dt = 0.28 kg/min - 0.56 kg/min = -0.28 kg/min. The initial condition is y(0) = 70 kg.

(c) To solve this differential equation, we use the method of separation of variables. We get y(t) = 70 - 0.28t in kg.

(d) After 5 hours, or 300 minutes, the amount of salt in the tank is y(300) = 70 - 0.28*300 = 16 kg.

(e) As time approaches infinity, the concentration of salt will approach the incoming salt concentration, which is 0.035 kg/L since the solution is mixed and leaves at that rate.

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The area of the cross section of the column is [tex]18 \pi \ m^2[/tex]

Explanation:

Given that a building engineer analyzes a concrete column with a circular cross section.

Also, given that the circumference of the column is [tex]18 \pi[/tex] meters.

We need to determine the area of the cross section of the column.

The area of the cross section of the column can be determined using the formula,

[tex]Area= \pi r^2[/tex]

First, we shall determine the value of the radius r.

Since, given that circumference is [tex]18 \pi[/tex] meters.

We have,

[tex]2 \pi r=18 \pi[/tex]

   [tex]r=9[/tex]

Thus, the radius is [tex]r=9[/tex]

Now, substituting the value [tex]r=9[/tex] in the formula [tex]Area= \pi r^2[/tex], we get,

[tex]Area = \pi (9)^2[/tex]

[tex]Area = 81 \pi[/tex]

Thus, the area of the cross section of the column is [tex]18 \pi \ m^2[/tex]

Answer:

[tex]\frac{x-2}{-1} =\frac{y-4}{1} =\frac{z-4}{-4} =t[/tex]

Step-by-step explanation:

Given that a line passes through P(2,4,4)

Also the line is perpendicular to the plane

[tex]-1x+1y-4z=1.[/tex]

From the equation of the plane we can say that normal to the plane has direction ratios as (-1,1,-4)

Since the required line is also perpendicular to the plane, the direction ratios of the required line is

(-1,1,4)

It passes through (2,4,4)

If Q(x,y,z) are general points on the line then

Direction ratios of PQ are = (x-2, y-4, z-4)

These are proportional to (-1,1,4)

So parametric form of the line is

[tex]\frac{x-2}{-1} =\frac{y-4}{1} =\frac{z-4}{-4} =t[/tex]

Whem t=0 we get the point P.

Step-by-step explanation:

sec² x − 2 = 0

sec² x = 2

cos² x = ½

cos x = ±√½

x = π/4, 3π/4, 5π/4, 7π/4