A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable guessing is 0.30.3. At least 99 correct answers are required to pass. If the student makes knowledgeable guesses, what is the probability that he will pass? Round your answer to four decimal places.

Answer:

Range = 29

Variance=   (X₁- U) ² / N= 973/10 = 97.3

Standard Deviation= √variance= √97.3=  9.864

Step-by-step explanation:

Range = Difference between the highest and lowest value = 97-68= 29

Variance

                  X₁                X₁-U               (X₁- U) ²              

                 80               0                     zero

                 68                -12                   144

                 71                 -9                     81

                 72                -8                     64

                 95                 15                    225                  

                89                  9                     81

                97                   17                   289

                72                    -8                 64

                75                     -5                 25

                81                     1                   1

∑              800                ZERO            973  

u=  ∑X₁ /10=800/10=80

Variance=   (X₁- U) ² / N= 973/10 = 97.3

Standard Deviation= √variance= √97.3=  9.864

(b) The important feature of the data is not revealed through the different measures of​ variation is that the variability of two or more than two sets of data cannot be compared unless a relative measure of dispersion is used .

Answer:

0.9500, 0.4407, 0.3904

Step-by-step explanation:

(a) P(x ≤ 60).  We need to find the area under the standard normal curve to the left of x = 60.  The appropriate command when using a TI-83 Plus calculator with statistical functions is normcdf(-1000, 60, 49, 6.7).  This comes out to 0.9500.  P(x ≤ 60) = 0.9500

(b) P(x ≥ 50) would be normcdf(50, 1000,49, 6.7), or 0.4407

(c) P(50 ≤ x ≤ 60) would be normcdf(50,60,49,6.70, or 0.3904

The question involves finding probabilities for a normal distribution given mean μ and standard deviation σ. By converting x values to z-scores and using the standard normal distribution, we can calculate the desired probabilities.

The student is asking about probabilities related to a normally distributed random variable with a given mean (μ) and standard deviation (σ). To find these probabilities, we convert the x values to z-scores and use the standard normal distribution.

Calculations here are based on the normal distribution parameters provided and the standard normal distribution. The z-score is the key to converting any normal distribution to the standard normal distribution, enabling the use of standard tables or software functions for probability calculations.

Answer:

26% probability that they win both contracts.

Step-by-step explanation:

These following probabilities are important to solve this question:

0.4 = 40% probability of winning the first contract.

0.65 = 65% probability of winning the second contract if the first contract is won.

What is the probability that they win both contracts?

[tex]P = 0.4*0.65 = 0.26[/tex]

26% probability that they win both contracts.

Answer:(22,0)

Step-by-step explanation: you have to find when y equals 0

Answer:

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex]  the deviation

We know that the z score is given by:

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

And by properties the value that separate the half  area on one side and half is on the other is z=0, since we have this:

[tex] P(Z<0) =0.5[/tex]

[tex] P(Z>0)=0.5[/tex]

So then the correct answer for this case would be z =0.00

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex]  the deviation

We know that the z score is given by:

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

And by properties the value that separate the half  area on one side and half is on the other is z=0, since we have this:

[tex] P(Z<0) =0.5[/tex]

[tex] P(Z>0)=0.5[/tex]

So then the correct answer for this case would be z =0.00

The z-score that divides the standard normal distribution into two equal halves is 0.00 as the mean of this distribution is also 0. Z-scores are standardized values expressing how many standard deviations a value is above or below the mean and can be determined using a Z-Table of Standard Normal Distribution.

The value of z that divides the standard normal distribution so that half the area is on one side and half is on the other is 0.00. The standard normal distribution is a symmetrical distribution where half of the values are less than the mean and half of the values are greater than the mean. In a standard normal distribution, the mean is 0. Hence, the z-score that will divide the distribution into half is also 0.

A z-score is a standardized value, expressing how many standard deviations a value is above or below the mean. These scores are particularly useful when comparing values from different data sets that have different means and standard deviations, such as exam scores from different classes or schools. They help us understand whether a particular score is common, or exceptionally high or low.

To find this value, you can use a Z-Table of Standard Normal Distribution, which shows the cumulative probability of a random variable from a standard normal distribution (with mean 0 and standard deviation 1), as being less than a certain value.

https://brainly.com/question/30390016

#SPJ12

Answer:

Maximize Z = 6x1 + 3x2

other answers are as follows in the explanation

Step-by-step explanation:

Employee Glass Needed per product(sq feet) Glass available per                                                                                                  production

                            Product      

Wood framed glass Aluminium framed glass    

doug 6                         0                                            36  

linda 0                        8                                             32  

Bob         6                        8                                            48  

profit  $300              $150  

per batch

Z = 6x1 + 3x2,

with the constraint

6x1 ≤ 36    8x2 ≤ 32      6x1 + 8x2 ≤ 48  

and x1 ≥ 0, x2 ≥ 0

Maximize Z = 6x1 + 3x2

to get the points of the boundary on the graph we say

when 6x1= 36

x1=6

when

8x2= 32

x2=4

to get the line of intersect , we go to  

6x1 + 8x2 ≤ 48  

so,  6x1 + 8x2 = 48  

When X1=0

8x2=48

x2=6

when x2=0

x1=8

the optimal point can be seen on the graph as attached

In this exercise we have to write the maximum function of a company, in this way we find that:

A)[tex]M(Z) = 6x_1+ 3x_2[/tex]

B)[tex]X_1= 8 \ and \ x_2 = 6 \ or \ 0[/tex]

A)So to calculate the maximum equation we have:

[tex]Z = 6x_1 + 3x_2\\6x_1 \leq 36 \\ 8x_2 \leq 32 \\ 6x_1 + 8x_2 \leq 48[/tex]

B) To calculate the limits of the graph we have to do:

[tex]6x_1= 36\\x_1=6\\8x_2= 32\\x_2=4\\6x_1 + 8x_2 = 48\\X_1=0\\8x_2=48\\x_2=6\\x_2=0\\x_1=8[/tex]

See more about graphs at brainly.com/question/14375099

Answer:

$18 Millions

Step-by-step explanation:

Decline in total paid in capital=Total value of reacquisition - Decline in retained earnings

$19-$1=$18 millions

Answer:

$18 000 000

Step-by-step explanation:

The first step is to calculate the par value of shares

$1*35000000=35000000

=$35000000

Issued shares of the the par

amount over par

=$16-$1=$15

Then paid capital in excess of par

$15*1000000

=$525000000

The first reacquire remember there were issued at $16

so for first reacquire

$16-$14=$2

$2*35000000=$70 000000

so Agee total paid-in capital will decline by

$15+$2+$1=$18 per share

Answer:

[tex]P(5<X<7)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{5-6}{0.5}<Z<\frac{7-6}{0.5})=P(-2<z<2)[/tex]

And we can find this probability with thie difference:

[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(6,0.5)[/tex]  

Where [tex]\mu=6[/tex] and [tex]\sigma=0.5[/tex]

We are interested on this probability

[tex]P(5<X<7)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(5<X<7)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{5-6}{0.5}<Z<\frac{7-6}{0.5})=P(-2<z<2)[/tex]

And we can find this probability with thie difference:

[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/tex]

Final answer:

The probability that a 44 inch television screen will have a service life between 5 and 7 years, given a normal distribution with a mean of 6 years and a standard deviation of 0.5 years, is approximately 95.45%.

Explanation:

To calculate the probability that the service life of a 44 inch television screen is between 5 and 7 years, we use the properties of the normal distribution where the mean (μ) is 6 years and the standard deviation (σ) is 0.5 years. We need to find the z-scores for 5 and 7 years and then use the standard normal distribution table or a calculator to find the probability that the service life falls between these two z-scores.

First, calculate the z-score for 5 years:
z = (X - μ) / σ
z = (5 - 6) / 0.5
z = -2.0

Next, calculate the z-score for 7 years:
z = (7 - 6) / 0.5
z = 2.0

Once we have the z-scores, we look up the corresponding probabilities in the normal distribution table or use a calculator with normal distribution functions. The probability of z being between -2.0 and 2.0 in a standard normal distribution is approximately 0.9545.

Therefore, the probability that a television will last between 5 and 7 years is approximately 95.45%.

Final answer:

The probability of selecting the color of the first ball is 3/10. The probability of selecting the same color for the second ball depends on the remaining number of balls of that color divided by the total remaining number of balls.

Explanation:

The probability of selecting the color of the first ball is determined by the number of balls of that color divided by the total number of balls. In this case, there are 3 red balls out of 10, so the probability is 3/10.

After selecting the first ball without replacement, the probability of selecting the same color for the second ball depends on the remaining number of balls of that color divided by the total remaining number of balls. If the first ball was red, there are 2 red balls left out of 9, resulting in a probability of 2/9.

The probability of selecting a ball of the same color as the first ball drawn, without replacement, is as follows:

For red:[tex]\boxed{0.22}[/tex]

For white:[tex]\boxed{0.33}[/tex]

For blue:[tex]\boxed{0.22}[/tex]

Let's think step by step.To solve this problem, we need to calculate the probability of selecting a ball of the same color as the first ball drawn, without replacement. We will consider the two events separately: first, the probability of selecting a ball of the same color as the first ball when there are 10 balls in total, and second, the probability of selecting a ball of the same color as the first ball when there are only 9 balls left in the bucket (since the first ball is not replaced).

First Ball Selection:

When the first ball is picked, there are 10 balls in total, with 3 red, 4 white, and 3 blue balls. The probability of picking a ball of any specific color is the number of balls of that color divided by the total number of balls.

 For example, if the first ball drawn is red, the probability of drawing a red ball is:

[tex]\[ P(\text{first red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{10} \][/tex]

 Similarly, if the first ball is white, the probability is:

[tex]\[ P(\text{first white}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{10} \][/tex]

 And if the first ball is blue, the probability is:

[tex]\[ P(\text{first blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{3}{10} \][/tex]

 Second Ball Selection:

After the first ball is drawn and not replaced, there are 9 balls left in the bucket.

 If the first ball drawn was red, there are now 2 red balls left out of 9 total balls. The probability of drawing another red ball is:

[tex]\[ P(\text{second red} | \text{first red}) = \frac{\text{Number of red balls left}}{\text{Total number of balls left}} = \frac{2}{9} \][/tex]

If the first ball was white, there are now 3 white balls left out of 9 total balls. The probability of drawing another white ball is:

[tex]\[ P(\text{second white} | \text{first white}) = \frac{\text{Number of white balls left}}{\text{Total number of balls left}} = \frac{3}{9} = \frac{1}{3} \][/tex]

If the first ball was blue, there are now 2 blue balls left out of 9 total balls. The probability of drawing another blue ball is:

[tex]\[ P(\text{second blue} | \text{first blue}) = \frac{\text{Number of blue balls left}}{\text{Total number of balls left}} = \frac{2}{9} \][/tex]

Now, let's calculate the probabilities for each color and round them to two decimal places:

 For red:

[tex]\[ P(\text{second red} | \text{first red}) = \frac{2}{9} \approx 0.22 \][/tex]

 For white:

[tex]\[ P(\text{second white} | \text{first white}) = \frac{1}{3} \approx 0.33 \][/tex]

 For blue:

[tex]\[ P(\text{second blue} | \text{first blue}) = \frac{2}{9} \approx 0.22 \][/tex]

Final

The probability of selecting a ball of the same color as the first ball drawn, without replacement, is as follows:

For red:[tex]\boxed{0.22}[/tex]

For white:[tex]\boxed{0.33}[/tex]

For blue:[tex]\boxed{0.22}[/tex]

 These probabilities are rounded to two decimal places.

 The answer is: 0.22.

Answer:

28 inches

Step-by-step explanation:

∠A + ∠B + ∠C = 180°

∠A + 65° +93° = 180°

∠A + 158° = 180°

∠A= 180°-158° = 22°

Using Law of Sines

a/sinA= b/sinB

a/sin22= 68 inches/sin65

a/sin22 = 68/0.9063 = 75.03

a = 75.03 x sin 22

a = 75.03 x 0.3746 = 28.106238≈28 inches

Answer:

The probability that the card will be a heart or a face card is P=0.4231.

Step-by-step explanation:

We have a standard deck of 52 cards.

In this deck, we have 13 cards that are a heart.

We also have a total of 12 face cards (4 per each suit). So there are 3 face cards that are also a heart.

To calculate the probability that a card be a heart or a face card, we sum the probability of a card being a heart and the probability of it being a face card, and substract the probability of being a heart AND a face card.

We can express that as:

[tex]P(H\,or\,F)=P(H)+P(F)-P(H\&F)\\\\P(H\,or\,F)=13/52+12/52-3/52=22/52=0.4231[/tex]

Answer: units = 6198

Step-by-step explanation:

expected consumer demand is 5000, the units that must be planned given shrinkage percentage levels in different stages of the supply chain we have to use the trial and error method.

lets try 6200 units

6200 x 95% = 5890, 5890 x 96% = 5654.4, 5654.40 x 97% = 5484.768, 5484.768 x 96% = 5265.37728, 56265 x 95% = 5002.108416 ≈ 5002

6198 units

(6198  x 95% = 5888.10) (5888.10 x 96% = 5652.576) (5652.576 x 97% = 5482.99872) (5482.99872 x 96% = 5263.6787712) (5263.6787712 x 95% = 5000.4948326) ≈ 5000

using the same procedure for 6197 units the answer will be 4999.688041

units that should be produced to cover the demand of 5000 = 6198

To meet the 5,000 unit consumer demand, the materials plan should account for approximately 6,195 units, considering the cumulative loss percentages at each stage of the supply chain.

To meet the expected consumer demand of 5,000 units while accounting for shrinkage at various stages of the supply chain, we need to calculate the cumulative effect of theft, damage, and defects on the number of units. We need to work backwards from the consumer to the materials supplier to determine the initial quantity needed.

The materials plan should account for approximately 6,195 units to meet the expected consumer demand, accounting for expected losses due to theft, damage, and defects throughout the supply chain stages.

The 99% confidence interval for the mean peak power after training is approximately [tex](292.61, 337.39)[/tex] watts.

Identify the given data:

Mean peak power after training: [tex]\bar{x} = 315[/tex] watts

Standard deviation: [tex]s = 16[/tex] watts

Sample size: [tex]n = 7[/tex]

Confidence level: 99%

Find the critical value:

Since the sample size is small (n < 30) and the population standard deviation is not known, we use the t-distribution. For a 99% confidence interval with [tex](n-1) = 6[/tex] degrees of freedom, the critical value (t-value) can be found from the t-table. Using the t-table, [tex]t_{\frac{\alpha}{2},6} = 3.707[/tex].

Calculate the standard error (SE):
[tex]SE = \frac{s}{\sqrt{n}} = \frac{16}{\sqrt{7}} = \frac{16}{2.6458} \approx 6.04[/tex] watts

Compute the margin of error (ME):
[tex]ME = t_{\frac{\alpha}{2}} \times SE = 3.707 \times 6.04 \approx 22.39[/tex] watts

Construct the confidence interval:

Lower bound: [tex]\bar{x} - ME = 315 - 22.39 \approx 292.61[/tex] watts

Upper bound: [tex]\bar{x} + ME = 315 + 22.39 \approx 337.39[/tex] watts

Answer:

[tex]\displaystyle cos\theta=\frac{36}{164}=\frac{9}{41}[/tex]

[tex]\displaystyle tan\theta=\frac{160}{36}=\frac{40}{9}[/tex]

[tex]\displaystyle csc\theta=\frac{164}{160}=\frac{41}{40}[/tex]

[tex]\displaystyle sec\theta=\frac{164}{36}=\frac{41}{9}[/tex]

[tex]\displaystyle cot\theta=\frac{36}{160}=\frac{9}{40}[/tex]

Step-by-step explanation:

Trigonometric ratios in a Right Triangle

Let ABC a right triangle with the right angle (90°) in A. The longest length is called the hypotenuse and is the side opposite to A. The other sides are called legs and are shorter than the hypotenuse.

Some trigonometric relations are defined in a right triangle. Being [tex]\theta[/tex] one of the angles other than the right angle, h the hypotenuse, x the side opposite to [tex]\theta[/tex] and y the side adjacent to [tex]\theta[/tex], then

[tex]\displaystyle sin\theta=\frac{x}{h}[/tex]

[tex]\displaystyle cos\theta=\frac{y}{h}[/tex]

[tex]\displaystyle tan\theta=\frac{x}{y}[/tex]

[tex]\displaystyle csc\theta=\frac{h}{x}[/tex]

[tex]\displaystyle sec\theta=\frac{h}{y}[/tex]

[tex]\displaystyle cot\theta=\frac{y}{x}[/tex]

We are given the values of h=164 and x=160, let's find y

[tex]y=\sqrt{164^2-160^2}=36[/tex]

Now we compute the rest of the ratios

[tex]\displaystyle cos\theta=\frac{36}{164}=\frac{9}{41}[/tex]

[tex]\displaystyle tan\theta=\frac{160}{36}=\frac{40}{9}[/tex]

[tex]\displaystyle csc\theta=\frac{164}{160}=\frac{41}{40}[/tex]

[tex]\displaystyle sec\theta=\frac{164}{36}=\frac{41}{9}[/tex]

[tex]\displaystyle cot\theta=\frac{36}{160}=\frac{9}{40}[/tex]

Answer:

(1) The probability that an individual who tests negative does not carry the disease is 0.9709.

(2) The specificity of the test is 98%.

Step-by-step explanation:

Denote the events as follows:

X = a person carries the disease

Y = the test detected the disease.

Given:

[tex]P(X) = 0.01\\P(Y|X)=0.93\\P(Y|X^{c})=0.02[/tex]

The probability of a person not carrying the disease is:

[tex]P(X^{c})=1-P(X)=1-0.01=0.99[/tex]

The probability that the test does not detects the disease when the person is carrying it is:

[tex]P(Y^{c}|X)=1-P(Y|X)=1-0.93=0.07[/tex]

The probability that the test does detects the disease when the person is not carrying it is:

[tex]P(Y^{c}|X^{c})=1-P(Y|X^{c})=1-0.02=0.98[/tex]

(1)

Compute the probability that an individual who tests negative does not carry the disease as follows:

[tex]P(X^{c}|Y^{c})=\frac{P(Y^{c}|X^{c})P(X^{c})}{P(Y^{c}|X^{c})P(X^{c})+P(Y^{c}|X)P(X)} \\=\frac{(0.98\times 0.99)}{(0.98\times 0.99)+(0.07\times 0.01)} \\=0.9709[/tex]

Thus, the probability that an individual who tests negative does not carry the disease is 0.9709.

(2)

By specificity it implies that how accurate the test is.

Compute the probability of negative result when the person is not a carrier as follows:

[tex]P(Y^{c}|X^{c})=1-P(Y|X^{c})=1-0.02=0.98[/tex]

Thus, the specificity of the test is 98%.

Answer:

[tex](a) = \frac{144}{133225} \\\\(b) = \frac{1}{365}[/tex]

Step-by-step explanation:

Part (a) the probability that two people have a birthday on the 9th of any month.

Neglecting leap year, there are 365 days in a year.

There are 12 possible 9th in months that make a year calendar.

If two people have birthday on 9th; P(1st person) and P(2nd person).

[tex]=\frac{12}{365} X\frac{12}{365} = \frac{144}{133225}[/tex]

Part (b) the probability that two people have a birthday on the same day of the same month

P(2 people selected have birthday on the same day of same month) + P(2 people selected not having birthday on  same day of same month) = 1

P(2 people selected not having birthday on  same day of same month):

[tex]= \frac{365}{365} X \frac{364}{365} =\frac{364}{365}[/tex]

P(2 people selected have birthday on the same day of same month) [tex]= 1-\frac{364}{365} \\\\= \frac{1}{365}[/tex]

Final answer:

The probability that two people have a birthday on the 9th of any month is 1/133,225. The probability that two people have a birthday on the same day of the same month is also 1/133,225.

Explanation:

To calculate the probability that two people have a birthday on the 9th of any month, we need to consider the number of possible outcomes and the number of favorable outcomes. There are 12 months in a year, so the number of possible outcomes is 12. The probability of each person having a birthday on the 9th is 1/365. Therefore, the probability that two people have a birthday on the 9th of any month is (1/365) x (1/365) = 1/133,225.

To calculate the probability that two people have a birthday on the same day of the same month, we need to consider the number of possible outcomes and the number of favorable outcomes. There are 12 months in a year and each month has 30 or 31 days. So the number of possible outcomes is 12 x 31 = 372. The probability of each person having a birthday on a specific day is 1/365. Therefore, the probability that two people have a birthday on the same day of the same month is (1/365) x (1/365) = 1/133,225.

Answer:

1.83% probability there are no car accidents on that stretch on Monday

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays.

This means that [tex]\mu = 4[/tex]

What is the probability there are no car accidents on that stretch on Monday?

This is P(X = 0).

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

[tex]P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.0183[/tex]

1.83% probability there are no car accidents on that stretch on Monday

The probability of there being no car accidents on Monday on a certain section of I-40 can be calculated using the Poisson distribution. In this case, with an average of 4 accidents per weekday, the probability is approximately 1.83%.

To calculate the probability of there being no car accidents on Monday, we can use the Poisson distribution. In this case, the average number of accidents per weekday is given as 4. The Poisson distribution can be used to calculate the probability of a specific number of events occurring in a given time period.

The formula for calculating the probability of x events occurring in a Poisson distribution is:

P(x) = (e^-λ * λ^x) / x!

Where λ is the average number of events, and x is the number of events we want to calculate the probability for.

In this case, the average number of accidents per weekday is 4, so λ = 4. And we want to calculate the probability of there being no accidents, so x = 0.

Using the formula, we can calculate:

Therefore, the probability of there being no car accidents on Monday is approximately 0.0183, or 1.83%.

Answer:

The answer to your question is distance = 3500000 km

Step-by-step explanation:

Data

Scale  1 : 500000

7 cm

Process

1.- To solve this problem use direct proportions or rule of three.

                          1 : 500 000 :: 7 : x

2.- Multiply the middle numbers and the result divide it by the edge.

                x = (7 x 500000) / 1

Simplification

                x = 3500000 km                            

Answer:

Null Hypothesis: No first order autocorrelation

Alternative hypothesis: first order correlation exists

The assumptions to run this test are:

1) Errors are normally distributed with mean 0

2) Errors follows an stationary process

3) Independence condition between the erros

The statistic is defined as:

[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]

And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.

Step-by-step explanation:

The Durbin Watson test is a way to check autocorrelation in residuals for a time seeries or a regression.

We need to remember that the autocorrelation is the similarity of the time series in successive intervals. When we conduct this type of test we are checking if the time series can be modeled with and AR(1) process autoregressive.

The system of hypothesis on this case are:

Null Hypothesis: No first order autocorrelation

Alternative hypothesis: First order correlation exists

The assumptions to run this test are:

1) Errors are normally distributed with mean 0

2) Errors follows an stationary process

3) Independence condition between the errors

The statistic is defined as:

[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]

And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.

Answer:

The total number of possible complete staffs you could hire is 350.

Step-by-step explanation:

The order that the software engineers are is not important. For example, hiring John and Laura as the software engineers is the same as hiring Laura and John. The same applies for the computer engineers. So we use the combinations formula to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

What is the total number of possible complete staffs you could hire?

2 software engineers from a set of 5

3 computer engineers from a set of 7

So

[tex]T = C_{5,2}*C_{7,3} = \frac{5!}{2!3!}*\frac{7!}{3!4!} = 10*35 = 350[/tex]

The total number of possible complete staffs you could hire is 350.

Answer:

[tex]= - 2 x^{3} + 4x -8[/tex]

Step-by-step explanation:

The first step is to open the parenthesis,

Since there is a negative sign before the second parenthesis, so the sign of all the values in second parenthesis will be changed and the equation will look something like this

[tex]= 2x^{3} + 4x -2 - 4 x^{3} + 6[/tex]

The second step is to re arrange the equation

[tex]= 2x^{3} - 4 x^{3} + 4x - 2 + 6[/tex]

The last and final step is to solve the equation

[tex]= - 2 x^{3} + 4x -8[/tex]

This is our answer

Answer:

The difference is -2x³ + 4x + 4.

Step-by-step explanation:

Subtract the two expression as follows:

[tex](2x^{3}+4x-2)-(4x^{3}-6)=2x^{3}+4x-2-4x^{3}+6\\[/tex]

Combine the like terms together:

                                         [tex]=2x^{3}-4x^{3}+4x-2+6[/tex]

Simplify as follows:

                                         [tex]=-2x^{3}+4x+4[/tex]

Thus, the difference is -2x³ + 4x + 4.

Answer:

Monthly household expenditures on groceries

Step-by-step explanation:

The response variable is the one for which measurements are desired and that depends on other variables.

In this case, family size, household income, and household neighborhood are independent variables, while the response variable is the monthly household expenditures on groceries.

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

       T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = [tex]\frac{738.44-750}{\frac{38.2}{\sqrt{50} } }[/tex] ~ [tex]t_4_9[/tex]

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

QUESTION IS INCOMPLETE.

Nevertheless, I will explain how to find, without solving, the longest interval in which an initial value problem is certain to have a unique twice differentiable solution.

Step-by-step explanation:

Consider the Existence and uniqueness theorem:

Let p(t) , q(t) and r(t) be continuous on an interval a ≤ t ≤ b, then the differential equation given by:

y''+ p(t) y' +q(t) y = r(t) ;

y(t_0) = y_0, y'(t_0) = y'_0

has a unique solution defined for all t in the stated interval.

Example:

Consider the differential equation

ty'' + 9y = t

y(1) = y_0,

y'(1) = v_0

ty'' + 9y = t .................................(1)

First, write the differential equation (1) in the form:

y'' + p(t)y' + q(t)y = r(t) ..................(2)

by dividing (1) by t

So

y''+ (9/t)y = 1 ....................................(3)

Comparing (3) with (2)

p(t) = 0

q(t) = 9/t

r(t) = 1

For t = 0, p(t) and r(t) are continuous, but q(t) is undefined.

q(t) is continuous everywhere apart from the point t = 0.

We say (-∞, 0) and (0,∞) are the points where p(t), q(t) and r(t) are continuous.

But t = 1, which is contained in the initial conditions y(1) = y_0 and y'(1) = v_0 is found in (0,∞).

So, we conclude that this interval is the longest interval in which the initial value problem has a unique twice differentiable solution.

Answer:

[tex]0.322 - 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.283[/tex]

[tex]0.322 + 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.361[/tex]

The 94% confidence interval would be given by (0.283;0.361)

Step-by-step explanation:

Notation and definitions

[tex]n=500[/tex] random sample taken

[tex]\hat p=0.322[/tex] estimated proportion of people between 18 to 34 who live with their parents

[tex]p[/tex] true population proportion of people between 18 to 34 who live with their parents  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 94% of confidence, our significance level would be given by [tex]\alpha=1-0.94=0.06[/tex] and [tex]\alpha/2 =0.03[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.88, z_{1-\alpha/2}=1.88[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.322 - 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.283[/tex]

[tex]0.322 + 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.361[/tex]

The 94% confidence interval would be given by (0.283;0.361)

Answer:

a. 11

b. 3

c. homicide mortality rate

d. 11:3

Step-by-step explanation:

a.) If 15,555 homicide cases were recorded among males of 139,813,000 population, then in every 100,000 males, the number of homicides cases will become:

= [15,555/139,813,000] * 100,000

= 11.13 aproximately 11 homicide cases in every 100,000 males.

b.) If 4,753 homicide cases were recorded among Females of 144,984,000 population, then in every 100,000 Females, the number of homicides cases will become:

= [4753/144,984,000] * 100,000

= 3.28 approximately 3 homicide cases in every 100,000 females

C.) Homicide-mortality rate

d.) ratio of male to female homicide rate = 11 : 3

e.) What those rates means is that in every 100,000

Males in 2001, 11 of them were Victims of homicide and in every 100,000 females, 3 of them are victims of homicide.

Answer:

No

Step-by-step explanation:

A- full early flight atlanta to chicago

P(a)=0,8

B- full late night flightt

P(b)=0,7

A&b- both are full

P(a&b) probability that both flights are full

Suppose that they are independet, then we have:

P(a&b)=p(a)*p(b)=0,8*0,7=0,56.

So if they are independet then p(a&b)=0,56, and that is not true.

Answer:

In a sample of 1000 test the expected number of false negatives is 8.6.

Step-by-step explanation:

The table provided is:

                             Positive      Negative    TOTAL

Hepatitis C             335                  10            345

No Hepatitis C           2               1153           1155

TOTAL                     337               1163          1500

A false negative test result implies that, the person's test result for Hepatitis C was negative but actually he/she had Hepatitis C.

Compute the probability of false negative as follows:

[tex]P(Negative|No\ Hepapatits\ C)=\frac{n(Negative\cap No\ Hepapatits\ C)}{n(No\ Hepapatits\ C)} \\=\frac{10}{1163}\\ =0.0086[/tex]

Compute the expected number of false negatives in a sample is n = 1000 tests as follows:

E (False negative) = n × P (Negative|No Hepatitis C)

                              [tex]=1000\times0.0086\\=8.6[/tex]

Thus, in a sample of 1000 test the expected number of false negatives is 8.6.

÷Answer:

Standard Deviation  =  176.5

Step-by-step explanation:

To calculate the standard deviation, calculate the mean score for the 4 standard deviation scores:

mean, m = Σx ÷ n

where Σx represents summation of each value = 162 + 153 + 317 + 413

= 1045

n = number of samples to be considered = 4

mean, m = 1045 ÷4

= 261.25

To calculate the standard deviation, use the formula below

SD    =   [tex]\sqrt{\frac{Σ(x-m)}{n} ^{2} }[/tex]

where x =  each value from the week lead time

m = mean = 261.25

n = the size = 4

The Standard deviation formula can be simplified further

when x = 162

[tex]\sqrt{\frac{(x1-m)}{n} ^{2} }[/tex] = 49.625

when x = 153

[tex]\sqrt{\frac{(x2-m)}{n} ^{2} }[/tex] = 23.125

when x = 317

[tex]\sqrt{\frac{(x3-m)}{n} ^{2} }[/tex]=  27.875

when x = 413

[tex]\sqrt{\frac{(x4-m)}{n} ^{2} }[/tex]= 75.875

Note that the above 4 equations can be lumped up into one giant equation by applying a big square root function instead of breaking it down

SD = 49.625 + 23.125 + 27.875 + 75.875

SD = 176.5