I have four different sweaters. Every day I choose one of the four sweaters at random to wear. Let X be the number of different sweaters I wore during a 5-day week.

(For example, if my 5-day sweater sequence is 3,2,4,4,2 then X = 3 because that week I wore sweaters {2,3.4}.)


Find the mean of X.

Answer:

a) bell-shaped.              

b) mean, median, and mode all equivalent.      

d) symmetric around the mean.        

g) most of the data fall within 3 standard deviations from the mean.

Step-by-step explanation:

We have to describe a normal distribution.

a. bell-shaped.

This is true a normal distribution is a bell shaped distribution.

b. mean, median, and mode all equivalent.

This is true for a normal distribution.

Mean = Mode = Median

c. Bimodal

The is not true about the normal distribution. A normal distribution is unimodal and the mode is equal to the mean of the distribution.

d. symmetric around the mean.

This is true. The normal distribution is centered around the mean

e. skewed to the right.

This is not a property of normal distribution.

f. models discrete random variables.

Normal distribution is a continuous distribution.

g. most of the data fall within 3 standard deviations from the mean.

This is true. According to Empirical rule, almost all the data lies within three standard deviation of mean.

h. uniform-shaped

This is not true. A normal distribution is bell shaped.

The options that properly describe a normal distribution are;

A) Bell Shaped

B) Mean, median and mode are equivalent

D) Symmetric about the mean

G) most of the data fall within 3 standard deviations from the mean

Some of the properties of normal distribution are that;

Let us look at the options;

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Answer:

a) [tex]21.7-2.776\frac{2.718}{\sqrt{5}}=18.33[/tex]    

[tex]21.7+2.776\frac{2.718}{\sqrt{5}}=25.07[/tex]  

So on this case the 95% confidence interval would be given by (18.33;25.07)

b) [tex]20.52-2.776\frac{3.757}{\sqrt{5}}=18.84[/tex]    

[tex]20.52+2.776\frac{3.757}{\sqrt{5}}=22.20[/tex]    

So on this case the 95% confidence interval would be given by (18.84;22.20)

c) [tex]21.11-2.262\frac{3.154}{\sqrt{10}}=18.85[/tex]    

[tex]21.11+2.262\frac{3.154}{\sqrt{10}}=23.37[/tex]    

So on this case the 95% confidence interval would be given by (18.85;23.37)

And as we can see the confidence intervals are very similar for the 3 cases.

Step-by-step explanation:

Previous concepts  

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".  

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)  

s represent the sample standard deviation  

n represent the sample size  

Part a)  Build a 95% confidence interval for the mean time in the system using the first five averages collected.

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

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)  

Data: 25.2, 19.7, 23.6, 18.6, and 21.4

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)    

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)    

The mean calculated for this case is [tex]\bar X=21.7[/tex]  

The sample deviation calculated [tex]s=2.718[/tex]  

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:  

[tex]df=n-1=5-1=4[/tex]  

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that [tex]t_{\alpha/2}=2.776[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]21.7-2.776\frac{2.718}{\sqrt{5}}=18.33[/tex]    

[tex]21.7+2.776\frac{2.718}{\sqrt{5}}=25.07[/tex]  

So on this case the 95% confidence interval would be given by (18.33;25.07)

Part b: Build a 95% confidence interval for the mean time in the system using the second set of five averages collected.

Data: 22.1, 26.0, 20.2, 16.4, and 17.9

The mean calculated for this case is [tex]\bar X=20.52[/tex]  

The sample deviation calculated [tex]s=3.757[/tex]  

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:  

[tex]df=n-1=5-1=4[/tex]  

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that [tex]t_{\alpha/2}=2.776[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]20.52-2.776\frac{3.757}{\sqrt{5}}=18.84[/tex]    

[tex]20.52+2.776\frac{3.757}{\sqrt{5}}=22.20[/tex]    

So on this case the 95% confidence interval would be given by (18.84;22.20)

Part c: Build a 95% confidence interval for the mean time in the system using all ten averages collected.

Data: 25.2, 19.7, 23.6, 18.6, 21.4, 22.1, 26.0, 20.2, 16.4, and 17.9

The mean calculated for this case is [tex]\bar X=21.11[/tex]  

The sample deviation calculated [tex]s=3.154[/tex]  

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:  

[tex]df=n-1=10-1=9[/tex]  

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.262[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]21.11-2.262\frac{3.154}{\sqrt{10}}=18.85[/tex]    

[tex]21.11+2.262\frac{3.154}{\sqrt{10}}=23.37[/tex]    

So on this case the 95% confidence interval would be given by (18.85;23.37)

And as we can see the confidence intervals are very similar for the 3 cases.

Answer:

≈ 0.07

Step-by-step explanation:

To find the geometric mean over this three-year period, we plug in the values  for the yearly return rates into the equation for the geometric mean.

The geometric mean growth rate of sales over this three-year period is 7.36%. Also, the ending value of sales after this three-year period is $6,187,500.

Geometric Mean (GM) is the average value or mean which signifies the central tendency of the set of numbers by finding the product of their values. Basically, we multiply the numbers altogether and take the nth root of the multiplied numbers, where n is the total number of data values.

[tex](1+G)^3 = (1+0.1)(1-0.1)(1+0.25)\\\\(1+G)^3 = 1.2375\\\\(1+G) = \sqrt[3]{1.2375} = 1.0736\\\\G = 0.0736[/tex]

G  = 7.36 %

Also,

sales in year 0 = $5,000,000

annual Growth rate of year 1 = 10%

Sales in year 1 : $5,000,000 + 10% of $5,000,000 = [tex]5,000,000 + \frac{10}{100} *$5,000,000 = $5,500,000[/tex]

annual Growth rate of year 2 = -10%

Sales in year 2 : $5,500,000 - 10% of $5,500,000 = [tex]5,500,000 - \frac{10}{100} *$5,500,000 = $4,950,000[/tex]

annual Growth rate of year 3 = 25%

Sales in year 3 : $4,950,000 + 25% of $4,950,000 = [tex]4,950,000 + \frac{25}{100} *4,950,000 = $6,187,500[/tex]

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Final answer:

The weighted mean number of days to maturity for dollars invested in the 5 money market funds is approximately 11.35 days. This is calculated by taking the product of the days to maturity and the corresponding money value for each fund, summing these products, and then dividing by the total money value invested.

Explanation:

The question asks to calculate the weighted mean of days to maturity for dollars invested in several money market funds with varying maturities and dollar values. To compute this, we multiply each fund's days to maturity by its dollar value (in millions), sum these products, and then divide by the total of the dollar values. Here's the calculation:

(20 days * $20 million) + (12 days * $30 million) + (7 days * $10 million) + (5 days * $15 million) + (6 days * $10 million) = $400 million-days + $360 million-days + $70 million-days + $75 million-days + $60 million-days

Total million-days = $965 million-days

Total value of all funds = $85 million

Weighted mean days to maturity = Total million-days / Total value of all funds = $965 million-days / $85 million = 11.35 days

So, the weighted mean number of days to maturity for the dollars invested in these 5 money market funds is approximately 11.35 days.

Final answer:

Explaining the polar coordinates for given Cartesian coordinates.

Explanation:

Polar Coordinates of Points:

Point A(2, 2√3): Polar coordinates are (4, π/6).

Point B(-4, 7π/6): Wrong polar coordinates as it should be (4, 11π/6).

Point C(4, 13π/6): Wrong polar coordinates.

Answer:

1) P(E) = 1 - 0.14 = 0.86

2) P(E) = 1 - 0.92 = 0.08

3) P(E) = 1 - 17/35 = 18/35

4) P(E) = 1 - 61/100 = 39/100

Step-by-step explanation:

The probability that an event will happen equals one minus the probability that an event will not happen.

P(E) = 1 - P(E') ......1

Where;

P(E) is the probability that an event will happen.

P(E') is the probability that an event will not happen.

1. P(E') = 0.14

Applying equation 1

P(E) = 1 - 0.14 = 0.86

2. P(E') = 0.92

P(E) = 1 - 0.92 = 0.08

3. P(E') = 17/35

P(E) = 1 - 17/35 = 18/35

4. P(E') = 61/100

P(E) = 1 - 61/100 = 39/100

The problem involves computing z-scores and finding corresponding proportions in a normal distribution, then using these proportions to estimate rod discard rates and the amount of rods to manufacture. The mean and standard deviation of rod lengths provided are used for z-score calculations.

In this problem, we are dealing with a normal distribution scenario where the mean (μ) is given as 29 cm and the standard deviation (σ) is 0.07 cm.

To find the proportion, we can firstly calculate the z-score using the formula z = (x - μ)/σ, where x is the required value (28.9 cm). The z-score helps us express how far our data point is from the mean in terms of standard deviations. Using standard normal distribution tables or a calculator, we can then determine the proportion of rods with lengths less than 28.9 cm.

A similar strategy is used. Calculate the z-scores for 28.84 cm and 29.16 cm, then use these to find proportions. The proportion of discarded rods would be the sum of these two.

Using the proportion obtained in (b), multiply it by the total number of rods manufactured in a day (i.e., 5000) to get the expected number of rods to be discarded.

In this part, calculate the z-scores for 28.9 and 29.1 cm, then find the proportion of rods within these lengths using the tables or calculator. As this proportion represents the acceptable rods, divide the required amount of rods (i.e., 10,000) by this proportion to calculate the expected total rods to be manufactured to fulfill the order.

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To find the proportion of rods that have a length less than 28.9 cm, we use the standard normal distribution table. The proportion of rods that will be discarded is found by calculating the z-scores for the lower and upper bounds of the range. The expected number of discarded rods can be estimated by multiplying the proportion of discarded rods by the number of rods manufactured. The number of rods that should be manufactured within a specified range can be calculated using the proportion of rods that fall within that range.

To solve this problem, we will use the standard normal distribution table to find the proportion of rods that have a length less than 28.9 cm. We will also use the standard normal distribution table to find the proportion of rods that will be discarded when their length is outside the range of 28.84 cm to 29.16 cm. Finally, we will use the proportion of discarded rods to estimate the number of rods that the plant manager should expect to discard when manufacturing 5000 rods in a day. For part (d), we will assume that the lengths of the rods are normally distributed and calculate the number of rods that fall within the range of 28.9 cm to 29.1 cm when manufacturing 10,000 rods.

(a) To find the proportion of rods with a length less than 28.9 cm, we need to find the z-score corresponding to 28.9 cm. The z-score formula is given by z = (x - mean) / standard deviation. Plugging in the values, we get z = (28.9 - 29) / 0.07 = -1.43. Using the standard normal distribution table, we find that the proportion of rods with a length less than 28.9 cm is 0.0764 or 7.64%.

(b) To find the proportion of rods that will be discarded, we need to find the z-scores corresponding to 28.84 cm and 29.16 cm. The z-score for 28.84 cm is z = (28.84 - 29) / 0.07 = -2.29 and the z-score for 29.16 cm is z = (29.16 - 29) / 0.07 = 2.29. Using the standard normal distribution table, we find that the proportion of rods with a length shorter than 28.84 cm or longer than 29.16 cm is 0.0485 or 4.85%. Therefore, the proportion of rods that will be discarded is 4.85%.

(c) To estimate the number of rods that the plant manager should expect to discard when manufacturing 5000 rods in a day, we multiply the proportion of discarded rods by the number of rods manufactured. The expected number of discarded rods is 0.0485 x 5000 = 242.5 or approximately 243 rods.

(d) To calculate the number of rods that the plant manager should expect to manufacture when the order states that all rods must be between 28.9 cm and 29.1 cm, we need to find the proportion of rods that fall within this range. The z-scores for 28.9 cm and 29.1 cm are calculated as z1 = (28.9 - 29) / 0.07 = -1.43 and z2 = (29.1 - 29) / 0.07 = 1.43. Using the standard normal distribution table, we find that the proportion of rods that fall within this range is 0.8471 or 84.71%. Therefore, the number of rods that the plant manager should expect to manufacture when the order states that all rods must be between 28.9 cm and 29.1 cm is 0.8471 x 10000 = 8471 rods.

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The probability of drawing a 10 or Jack will be 8/52.

The chances of an event occurring are defined by probability. Probability has several uses in games, and in business to create probability-based forecasts.

The number of jack cards and 10 number cards from the pack of cards is 4,

The probability of drawing a 10 or jack is,

P = P(10) + P(J)

P = (4/52)+(4/52)

P= 8/52

Hence, the probability of drawing a 10 or Jack will be 8/52.

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Answer:

Part a

The probability that the event disk has high shock resistance is 0.86

Part b

The probability that a disk has high scratch resistance given that the disk has high shock resistance is 0.8140

Step-by-step explanation:

(a).

From the given information,

Let A denote the event that a disk has high shock resistance,

and let B denote the event that a disk has high scratch resistance.

                          SHOCK HIGH(A)     SHOCK LOW(A)          TOTAL

SCRATCH  

HIGH(B)               70                               9                                79

RESISTANCE       16                               5                                  21

LOW(B)

TOTAL                86                                 14                                100

Compute P(A).

Therefore, the probability value of the event A is 0.86.

Part a

The probability that the event disk has high shock resistance is 0.86

Explanation | Hint for next step

Based on the given information, the probability that the event disk has high shock resistance is 0.86. That means it is approximately equal to 86%.

Step 2 of 2

(b)

From the given information,

Let A denote the event that a disk has high shock resistance,

and let B denote the event that a disk has high scratch resistance.

                      SHOCK HIGH(A)     SHOCK LOW(A)            TOTAL

SCRATCH

HIGH(B)             70                               9                               79

SCRATCH

LOW(B)             16                                 5                               21

TOTAL             86                                15                               100

ComputeP(B/A) = P(A∩B) /P(A)  ;P(A) > 0

P(A∩B)  =70/100

​  =0.70

​  

From the part [a], the probability value of the event A is P(A) =0.86 .

Therefore,

P( {B/A}  = P(A∩B)  /P(A)

​=  0.70 /0.86

​  

=0.8140

​  

Part b

The probability that a disk has high scratch resistance given that the disk has high shock resistance is 0.8140

Explanation | Common mistakes

The disk with high scratch resistance is found to be 81.40% with the condition that the disk with high shock resistance is maintained.

Final answer:

Events A (high shock resistance) and B (high scratch resistance) in the context of polycarbonate plastic disks are not independent, as the calculated probabilities P(A ∩ B) and P(A)P(B) do not match.

Explanation:

Whether two events are independent can be determined by checking if the probability of one event occurring does not affect the probability of the other event. Mathematically, two events A and B are independent if and only if-

P(A ∩ B) = P(A)P(B)

Let's calculate this using the data provided from the analysis of 100 disks of polycarbonate plastic.

Total number of disks (n) = 100

Number of disks with high shock resistance (A) = 70 + 9 = 79

Number of disks with high scratch resistance (B) = 70 + 16 = 86

Number of disks with both high shock and scratch resistance (A ∩ B) = 70

The probability of A (P(A)) is thus 79/100, the probability of B (P(B)) is 86/100, and the probability of both A and B occurring (P(A ∩ B)) is 70/100. Now we multiply P(A) and P(B):

P(A)P(B) = (79/100) × (86/100) = 0.6794

However, P(A ∩ B) as observed is-

70/100 = 0.7

Since P(A ∩ B) does not equal P(A)P(B), the events A and B are not independent.

Answer with Step-by-step explanation:

We are given that a sample space

S={a,b,c,d,e}

P(a)=0.1

P(b)=0.1

P(c)=0.2

P(d)=0.4

P(e)=0.2

a.A={a,b,c}

P(A)=P(a)+P(b)+P(c)

P(A)=0.1+0.1+0.2=0.4

b.B={c,d,e}

P(B)=P(c)+P(d)+P(e)=0.2+0.4+0.2=0.8

c.A'=Sample space-A={a,b,c,d,e}-{a,b,c}={d,e}

P(A')=P(d)+P(e)=0.4+0.2=0.6

d.[tex]A\cup B[/tex]={a,b,c,d,e}

[tex]P(A\cup B)[/tex]=P(a)+P(b)+P(c)+P(d)+P(e)=0.1+0.1+0.2+0.4+0.2=1

e.[tex]A\cap B[/tex]={c}

[tex]P(A\cap B)=P(c)=0.2[/tex]

Answer:4000 bottle containers and 1000

Medical kits

Step-by-step explanation:

The total weights the plane can take per trip is 90000lb. It implies that if we multiply the weight of each bottle container by the number of bottle container and add it to the product of the number medical kit and the weight of each medical kit, we will obtain a total weight of 90000lb.

If x is the number of bottle and y is the number of medical kit

20×x+10×y=90000....i

Also

The total volume the plane can take per trip is 6000ft³ It implies that if we multiply the volume of each bottle container by the number of bottle container and add it to the product of the number medical kit and the volume of each medical kit, we will obtain a total volume of 6000ft³.

Recall, x is the number of bottle and y is the number of medical kit

1×x+2×y=6000....ii

Combining equation i and ii and solving simultaneously,

x=4000 units and y= 1000 units in each plane trip

Answer:

Bottled water = 4000.

Medical kits = 1000.

Step-by-step explanation:

Let x = bottled water

y = medical kits

For the mass (in pounds):

20x + 10y = 90000

For the cubic ft:

x + 2y = 6000

Solving equation i and ii simultaneously,

x = 4000.

y = 1000.

Bottled water = 4000.

Medical kits = 1000.

Using the empirical rule, approximately 84% of apple diameters are greater than 7.07 cm, since this value is within one standard deviation below the mean diameter of 7.46 cm.

The empirical rule (also known as the 68-95-99.7 rule) is a statistical rule which states that for a normally distributed set of data, approximately 68% of data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

In this particular problem, we know the mean (7.46cm) and standard deviation (0.39cm) of the apple diameters. We are asked to find the percentage of apple diameters that are greater than 7.07cm. Since 7.07cm is less than one standard deviation below the mean (7.46 - 0.39 = 7.07), from the normal distribution we can say that approximately 84% (50% of the data is above the mean and 34% is between the mean and one standard deviation below the mean) of the apple diameters are larger than 7.07cm according to the Empirical rule.

Thus, using Empirical rule and given statistical mean and standard deviation, we can estimate that about 84% of the apples from this species have a diameter greater than 7.07 cm.

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Using the Empirical Rule, approximately 84% of the apples have diameters greater than 7.07cm since 7.07cm is exactly one standard deviation below the mean of 7.46cm in a normal distribution.

To answer the student's question, we need to apply the Empirical Rule which is used in statistics to describe the distribution of data in a bell-shaped curve, also known as a normal distribution. This rule states that for a data set with a normal distribution, approximately 68% of the data falls within one standard deviation from the mean, 95% falls within two standard deviations, and over 99% falls within three standard deviations.

Given that the mean diameter of the apples is 7.46cm and the standard deviation is 0.39cm, first we need to determine how many standard deviations 7.07cm is from the mean. This is calculated as:

(7.07cm - 7.46cm) / 0.39cm = -1 standard deviation.

According to the Empirical Rule, 68% of apples are within one standard deviation above and below the mean. Since 7.07cm falls exactly at one standard deviation below the mean, we have:

Therefore, to find the percentage of apples that have diameters greater than 7.07cm, we would add these two percentages together:

50% + 34% = 84%

Thus, we conclude that approximately 84% of the apples have diameters that are greater than 7.07cm.

Answer:

The probability of a false alarm is 0.5116 .

Step-by-step explanation:

Let us indicate three events :

          Event A = Alarm system triggers

          Event B = Dangerous conditions exist

         Event B' = Normal conditions exist

Now we are given with P(A/B) which means probability of alarm getting triggered given the dangerous conditions exist , P(A/B') which means probability of alarm getting triggered given the normal conditions exist and P(B) which means the probability of dangerous conditions existing i.e.

P(A/B) = 0.95      P(A/B') = 0.005   P(B) = 0.005   P(B') = 1 - P(B) = 0.995

(a) The probability of a false alarm means given the alarm gets triggered probability that there was normal conditions i.e. P(B'/A)

  P(B'/A) = [tex]\frac{P(A\bigcap B')}{P(A)}[/tex]

Now P(A) = P(B) * P(A/B) + P(B') * P(A/B')  {This representing Probability of alarm getting triggered in both the conditions]

 P(A) = 0.005 * 0.95 + 0.995 * 0.005 = 9.725 x [tex]10^{-3}[/tex]

Since P(A/B') = 0.005

         [tex]\frac{P(A\bigcap B')}{P(B')}[/tex] = 0.005       So, [tex]P(A\bigcap B')[/tex] = 0.005 * 0.995 = 4.975 x  [tex]10^{-3}[/tex]

Therefore,  P(B'/A) = [tex]\frac{P(A\bigcap B')}{P(A)}[/tex]  = [tex]\frac{4.975*10^{-3} }{9.725*10^{-3} }[/tex] = 0.5116 .

Answer:

We conclude that its value change for 277.8$.

Step-by-step explanation:

We know that a perpetuity pays $50 per year and interest rates are 9 percent.  We calculate how much would its value change if interest rates decreased to 6 percent. We know that

9%=0.09

6%=0.06

We get

\frac{50}{0.09}=555.5

\frac{50}{0.06}=833.3

Therefore, we get 833.3-555.5=277.8

We conclude that its value change for 277.8$.

The value change for $277.8.

[tex]\frac{50}{0.09}=555.5\\\\\frac{50}{0.06}=833.3[/tex]

So,

=  $833.3  - $555.5

= $277.8

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Answer:

Here are Boolean expressions in Java syntax.

Step-by-step explanation:

For each part:

    a) (x > y && y > z)

    b) x == y && x < 0, or x < 0 && y < 0, or x == y && y < 0 (which is essentially the first example)

    c) (x == y && x >= 0), or (x >= 0 && y >= 0), or (x == y && y >= 0) (for the first and third, once the first condition establishes that x is equal to y, if either x or y is greater than or equal to 0, then they are both not less than 0)

    d) (x == y && x != z), or (x == y && y != z)

Answer:

0.9544

Step-by-step explanation:

We are given that mean=18 and standard deviation=1 and we have to find P(16<X<20).

P(16<X<20)=P(z1<Z<z2)

z1=(x1-mean)/standard deviation

z1=(16-18)/1=-2

z2=(x2-mean)/standard deviation

z2=(20-18)/1=2

P(16<X<20)=P(z1<Z<z2)=P(-2<Z<2)

P(16<X<20)=P(-2<Z<0)+P(0<Z<2)

P(16<X<20)=0.4772+0.4772=0.9544

The probability that the height of a tree is between 16 and 20 feet is 95.44%

Answer:

Bryant: 2059.5 (RIP)

Durant:2353.5

Step-by-step explanation:

To find how many points Kevin Durant and Kobe Bryant scored, set up equations based on the given information. Kevin Durant scored 2280 points and Kobe Bryant scored 2133 points during the 2012-2013 NBA regular season.

The question asks us to determine how many points Kevin Durant and Kobe Bryant scored individually during the 2012-2013 NBA regular season given that together they scored 4413 points and Kobe Bryant scored 147 fewer points than Kevin Durant.

To solve this, we can set up two equations based on the information provided:

We can substitute the second equation into the first to find Durant's score:

Now that we have Durant's score, we can use it to find Bryant's score:

Therefore, Kevin Durant scored 2280 points and Kobe Bryant scored 2133 points during the 2012-2013 NBA regular season.

Answer:

20 inches and 5 inches

Step-by-step explanation:

Let the diameter of the largest and the smallest pizza be Y and X respectively.

Then,

Y+X = 25 ........................... Equation 1

Y-X = 15 ............................ Equation 2

Solve equation 1 and equation 2 simultaneously.

Add equation 1 and equation 2

Y+Y = +X+(-X)+25+15

2Y = 40

Y = 40/2

Y = 20 inches.

Also,

Substitute the value of Y into equation 1

20+X=25

X = 25-20

X = 5 inches.

Hence the diameter of the largest and the smallest pizzas = 20 inches and 5 inches

Final answer:

The smallest pizza has a diameter of 5 inches, and the largest pizza has a diameter of 20 inches.

Explanation:

The diameters of the largest and smallest pizzas are 20 inches and 5 inches, respectively.

To find the diameters of the pizzas:

Let x be the diameter of the smallest pizza and y be the diameter of the largest pizza.

We have the system of equations x + y = 25 and y - x = 15.

Solving these equations simultaneously, we get y = 20 and x = 5.

Answer:

from my opinion second option is right

the last cow gives 202 pints of milk

Given :

There are 199 cows line up in the barn and he begins milking them.

The first cow gives 4 pints of milk. The next cow gives 5 pints of milk. The next gives 6 pints, and so on, increasing by 1 pint each cow.

pints of milk is increasing 1 by a sequence

4,5,6,7 ......... and so on

when n=1 , the cow given 4 pints of milk

The sequence is arithmetic

nth cow is 199. Lets find out pints of milk when n=199

apply nth formula of arithmetic

[tex]T_n = T_1 + (n-1) d\\[/tex]

n=199

difference d=1

T1=4

Substitute all the values

[tex]T_n = T_1 + (n-1) d\\\\\\T_{199}=4+(199-1)1\\T_{199}=4+198\\T_{199}=202[/tex]

So, the last cow gives 202 pints of milk

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Answer:

The required probability is 0.00144 or 0.144%.

Step-by-step explanation:

Consider the provided information.

Five cards are drawn from an ordinary deck of 52 playing cards.

A full house is a hand that consists of two of one kind and three of another kind.

The total number of ways to draw 5 cards are: [tex]^{52}C_5=\frac{52!}{5!47!}[/tex]

Now we want two of one kind and three of another.

Let the hand has the pattern AAABB, where A and B are from distinct kinds. The number of such hands are:

[tex]^{13}C_1\times^{4}C_3\times^{12}C_1\times^{4}C_2=\frac{13!}{12!}\times\frac{4!}{3!}\times\frac{12!}{11!}\times\frac{4!}{2!2!}[/tex]

Thus, the required probability is:

[tex]\frac{^{13}C_1\times^{4}C_3\times^{12}C_1\times^{4}C_2}{^{52}C_5}=\frac{\frac{13!}{12!}\times\frac{4!}{3!}\times\frac{12!}{11!}\times\frac{4!}{2!2!}}{\frac{52!}{5!47!}}[/tex]

[tex]=\frac{3744}{2598960}\\\\\approx0.00144[/tex]

Hence, the required probability is 0.00144 or 0.144%.

College Mathematics 10+5 pts

"Consider the following argument:

a. George and Mary are not both innocent.

b. If George is not lying, Mary must be innocent.

c. Therefore, if George is innocent, then he is lying.

Let ???? be the proposition "George is innocent", m be the proposition "Mary is innocent", and let ???? be the proposition "George is lying"."

1. Write a propositional formula F involving variables g, m, l such that the above argument is valid if and only if F is valid.

2. Is the above argument valid? If so, prove its validity by proving the validity of F. If not, give an interpretation under which F evaluates to false.

Answer

1. [tex]F = ((g\cap \sim m)\cup(\sim g\cap m)\cap(\sim l\to m))\to(g\to l) [/tex]

2. It is valid. See explanation for proof.

Step-by-step explanation

1. Statement a is an exclusive OR of [tex]g[/tex] and [tex]m[/tex]. This is because only one of them is innocent while the other is not. That is [tex]g[/tex] is true and [tex]m[/tex] is false or [tex]g[/tex] is false and [tex]m[/tex] is true.

Statement b is an implication that [tex]\sim l \to m[/tex].

Statement c is another implication that [tex]g\to l[/tex].

The presence of the word "therefore" in statement c means it is an implication from statement a AND statement b. So we have

[tex]F = ((g\cap \sim m)\cup(\sim g\cap m)\cap(\sim l\to m))\to(g\to l) [/tex]

2.

The validity will be proved using a truth table. [tex]F[/tex] is valid if the last column contains only true values i.e. truth values of T.

From the table (in the attached image), [tex]c[/tex] represents the statement a, [tex]d[/tex] represents statement b and [tex]f[/tex] represents statement c. The last column, which represents [tex]F[/tex], is valid as all its entries are T.

To translate the argument into logic notation, one could write it as ((¬g ∨ m) ∧ (¬l → m)) → ((g → ¬l). Upon considering the case where both George and Mary are innocent, and George is not lying, it can be observed that the argument is invalid.

1. The propositional formula F can be written as: ((¬g ∨ m) ∧ (¬l → m)) → ((g → ¬l). This translates the given argument into logical notation.

2. This argument is not valid. An easy way to prove this is by considering a scenario where both George and Mary are innocent, hence g and m are both true, and let l be the proposition 'George is lying', so l is false. In this case, the left-hand side of your main implication results true ((¬g ∨ m) ∧ (¬l → m)) = (false ∨ true) ∧ (true → true) = true ∧ true = true), but the right-hand side of your main implication results false (g → ¬l = true → false = false) hence the full implication results false, which makes the argument invalid.

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Answer: a) C = 0.126x + 1.5

b) C = 2.382 c) approximately 179 units

Step-by-step explanation: profit P(x), total revenue R(x) and total cost (C) are related by the formulae below.

Profit = total revenue - total cost

P(x) = R(x) - C

From the question, P(x) = 0.084x - 1.5, R(x) = 0.21x.

Question a)

C = R(x) - P(x)

C = 0.21x - {0.084x - 1.5}

C = 0.21x - 0.084x + 1.5

C = 0.126x + 1.5

Question b)

If C = 0.126x + 1.5, then C at x = 7 units is gotten below as

C = 0.126(7) + 1.5

C = 0.882 + 1.5

C = 2.382.

Question c)

The break even point is the point where total revenue R(x) equals total cost C.

R(x) = 0.21x and C = 0.126x + 1.5

0.21x = 0.126x + 1.5

0.21x - 0.126x = 1.5

0.084x = 15

x = 15/ 0.084

x = 178.57 which is approximately 179 units (since quantity of units can't be decimal)

To find the cost equation, subtract the profit equation from the revenue equation. The cost of producing 7 units is approximately 2.382 million dollars. The break-even point occurs at approximately 17.857 units.

a. To find the cost equation, we need to subtract the profit equation from the revenue equation. Since the revenue equation is R(x) = .21x and the profit equation is P(x) = .084x - 1.5, the cost equation is C(x) = R(x) - P(x). Substituting the given equations, we have C(x) = .21x - (.084x - 1.5).

Simplifying the equation, we get C(x) = .21x - .084x + 1.5.

Combining like terms, the cost equation is C(x) = .126x + 1.5.

b.

To find the cost of producing 7 units, we can substitute x = 7 into the cost equation. C(7) = .126(7) + 1.5 = .882 + 1.5 = 2.382 million dollars.

c.

The break-even point occurs when the revenue equals the cost. To find the break-even point, we set R(x) = C(x). Substituting the given equations, we have .21x = .126x + 1.5.

Solving for x, we get .084x = 1.5.

Dividing both sides by .084, we find x = 17.857 units.

Therefore, the break-even point is approximately 17.857 units.

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Answer:

[tex]T(t) = 42000 + 21t[/tex]

Step-by-step explanation:

There are two costs for the operation of the equipment, the purchase cost and the hourly cost. The total cost is the sum of these two costs:

So

[tex]T = F_{c} + H_{c}[/tex]

In which [tex]F_{c}[/tex] is the fixed cost and [tex]H_{c}[/tex] is the hourly cost.

Fixed cost

A roofing contractor purchases a shingle delivery truck with a shingle elevator for $42,000. This means that the fixed cost is 42000. So [tex]F_{c} = 42000[/tex]

Hourly cost

The vehicle requires an average expenditure of $9.50 per hour for fuel and maintenance, and the operator is paid $11.50 per hour. So for each hour, there are expenses of 9.5 + 11.5 = $21.

So the hour cost is

[tex]H_{c}(t) = 21t[/tex]

In which t is the number of hours

Total cost

[tex]T = F_{c} + H_{c}[/tex]

[tex]T(t) = 42000 + 21t[/tex]

Answer:

[tex]z=\frac{83.5-75}{6.5}=1.31[/tex]    

The rejection zone for this case would be:

[tex] z> 1.96 \cup Z<-1.96[/tex]

[tex]p_v =2*P(z>1.31)=0.1901[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly different from 75 at 5% of significance.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=83.5[/tex] represent the sample mean

[tex]\sigma=6.5[/tex] represent the population standard deviation for the sample  

[tex]\mu_o =75[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 75, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 75[/tex]  

Alternative hypothesis:[tex]\mu \neq 75[/tex]  

For this case we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\sigma}[/tex]  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

[tex]z=\frac{83.5-75}{6.5}=1.31[/tex]    

Cutoff for the rejection regon

Since our significance level is [tex] \alpha=0.05[/tex] and we are conducting a bilateral test we need to find a quantile in the standard normal distribution that accumulates 0.025 of the area on each tail.

And for this case those values are [tex] z_{crit}= \pm 1.96[/tex]

So the rejection zone for this case would be:

[tex] z> 1.96 \cup Z<-1.96[/tex]

Our calculated value is not on the rejection zone. So we fail to reject the null hypothesis.

P-value

Since is a two sided test the p value would be given by:  

[tex]p_v =2*P(z>1.31)=0.1901[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly different from 75 at 5% of significance.

Final answer:

To determine if the new polygraph test's accuracy significantly differs from the current one, we compare the calculated test statistic with the critical z-score of ±1.96 at an alpha level of 0.05. The process involves hypothesis testing, where rejecting or failing to reject the null hypothesis depends on whether the test statistic exceeds the critical value.

Explanation:

The question is about determining if the accuracy of a new type of polygraph test is statistically different from the current one using hypothesis testing. Given an alpha level (α) of 0.05 (5% level of significance), the critical z-score for a two-tailed test is ±1.96. This critical value defines the cutoff points for the rejection regions. To calculate the test statistic (Zobt), we use the formula:

Zobt = (X - μ) / (σ / √n),

where X is the sample mean (83.5%), μ is the population mean (75%), σ is the standard deviation (6.5%), and √n is the square root of the sample size. Given that only one participant's data is used, √n is 1 for this scenario, which simplifies our formula. Unfortunately, without the exact sample size for a more precise calculation or this being a theoretical scenario with one participant, we remark on the process rather than providing a numerical value for Zobt.

Based on the calculated Zobt, if it exceeds the critical z-score (1.96 or -1.96), we reject the null hypothesis indicating that there is a statistically significant difference between the new polygraph's accuracy and the current one. If Zobt does not exceed the critical value, we fail to reject the null hypothesis, indicating no significant difference.

Answer:

($3.61, $4.27) is the range in which at least 88.9% of the data will reside.

Step-by-step explanation:

Chebyshev's Theorem

[tex]\mu \pm 2\sigma\\\\1 - \dfrac{1}{(2)^2}\% = 75\%[/tex]

Atleast 75% of data lies within two standard deviation of the mean for a non-normal data.

[tex]\mu \pm 3\sigma\\\\1 - \dfrac{1}{(3)^2}\% = 88.9\%[/tex]

Thus, range of data for which 88.9% of data will reside is

[tex]\mu \pm 3\sigma\\= (\mu - 3\sigma , \mu + \3\sigma)[/tex]

Now,

Mean = $3.94

Standard Deviation = $0.11

Putting the values, we get,

Range =

[tex]3.94 \pm 30.11\\= (3.94 - 3(0.11) , 3.94 + 3(0.11))\\=(3.61,4.27)[/tex]

($3.61, $4.27) is the range in which at least 88.9% of the data will reside.

Answer:

Using Chebyshev's Theorem, the range in which at least 88.9% of the data reside is [$3.82, $4.06].

Step-by-step explanation:

Given:

The mean of the prices of a gallon of milk is, [tex]\mu=\$3.94[/tex]

The standard deviation of the prices of a gallon of milk is, [tex]\sigma=\$0.11[/tex]

The Chebyshev's Theorem states that for a random variable X with finite mean μ and finite standard deviation σ and for a positive constant k the provided inequality exists,

                                      [tex]P(|X-\mu|\geq k)\leq \frac{\sigma^{2}}{k^{2}}[/tex]

The value of [tex]\frac{\sigma^{2}}{k^{2}} = 0.889[/tex]

Then solve for k as follows:

[tex]\frac{\sigma^{2}}{k^{2}} = 0.889\\\frac{(0.11)^{2}}{k^{2}}=0.889\\k^{2}=\frac{(0.11)^{2}}{0.889}\\k=\sqrt{0.0136108} \\\approx0.1167[/tex]

The range in which at least 88.9% of the data will reside is:

[tex]P(|X-\mu|\geq k)\leq \frac{\sigma^{2}}{k^{2}}\\P(\mu-k\leq X\leq \mu+k)\leq 0.889\\P(3.94-0.1167\leq \leq X\leq 3.94+0.1167)\leq 0.889\\P(3.8233\leq X\leq 4.0567)\leq 0.889\\\approxP(3.82\leq X\leq 4.06)\leq 0.889[/tex]

Thus, the probability of prices of a gallon of milk between $3.82 and $4.06 is 0.889.

Answer:

A. Stratification

Step-by-step explanation:

Stratified random sampling is used when the researcher wants to highlight a specific subgroup within an entire population.

Stratification technique is mainly used to reduce the population differences and to increase the efficiency of the estimates. In this method the population is divided into a number of subgroups or strata.

Each strata should be so formed such that they are homogeneous as far as possible.

Final answer:

When examining the hypothesis about drug side effects in different populations, Randomization is the most appropriate statistical technique. It helps in reducing bias and ensures equal chances of group assignment, providing more reliable results in comparing the effects across populations.(Option D)

Explanation:

A researcher examining the hypothesis that a specific drug may exhibit a higher prevalence of side effects among the African American population compared to the Caucasian population could employ several statistical techniques to design the study. However, the most appropriate option provided is Randomization. Randomization helps in mitigating bias and ensures that each participant has an equal chance of being assigned to either the experimental or control group. This process decreases the likelihood of systematic differences between groups and allows any effects observed to be more confidently attributed to the drug under study rather than external factors.

Other options like Stratification, Crossover matching, and Matching could also play roles in different aspects of study design, but when testing a hypothesis about a drug's effects across different populations, randomization is crucial. It aligns with principles of experimental design that seek to control, to the extent possible, for variables that could influence the outcome, ensuring that the treatment group and control group are comparable at the beginning of the study.

Using it's concept, it is found that the proportion of Protestants that were calm for 2 out of the 3 days is of 0.0606.

In this problem:

Hence:

[tex]p = \frac{6}{99} = 0.0606[/tex]

The proportion of Protestants that were calm for 2 out of the 3 days is of 0.0606.

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Approximately 6.06% of Protestants were calm for 2 out of the 3 days, calculated by dividing the number of Protestants calm for 2 days (6) by the total surveyed (99).

To find the proportion of Protestants who were calm for 2 out of the 3 days, we need to look at the numbers given for Protestant individuals and calculate the ratio of those who were calm for exactly two days to the total number of Protestants surveyed. According to the data provided:

The total number of Protestants surveyed is the sum of those calm for 0, 1, 2, and 3 days, which is 50 + 27 + 6 + 16 = 99.
To find the proportion of Protestants calm for 2 days, we divide the number calm for 2 days by the total number of Protestants surveyed:

Proportion = Number calm for 2 days / Total number of Protestants
Proportion = 6 / 99
Proportion ≈ 0.0606

This means that approximately 6.06% of Protestants were calm for 2 out of the 3 days.

a) The probability that he or she has at least one food allergy and a history of severe reaction is 0.0287.

b) The probability that he or she is allergic to multiple foods is, 0.021

Given that;

A certain study found that 7% of children younger than 18 in the United States have at least one food allergy.

a. Since the probability that a child younger than 18 has at least one food allergy is given as 7%.

Among those with food allergies, the probability of having a history of severe reaction is 41%.

Hence for the probability that a child has both at least one food allergy and a history of severe reaction, multiply these probabilities together:

7% × 41% = 0.07 × 0.41

= 0.0287.

Therefore, the probability is 0.0287.

b) For the probability that a randomly selected child younger than 18 is allergic to multiple foods, consider the information given.

The probability of having at least one food allergy among children younger than 18 is 7%.

And among those with allergies, 30% are allergic to multiple foods.

Hence for the probability, multiply the probability of having at least one food allergy (7%) by the probability of being allergic to multiple foods (30% of those with allergies):

Probability = 7% × 30%

                  = 0.07 × 0.30

                  = 0.021.

Therefore, the probability that a randomly selected child younger than 18 is allergic to multiple foods is 0.021.

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The probability that a child younger than 18 has at least one food allergy and a history of severe reaction is approximately 0.029. The probability that a child younger than 18 is allergic to multiple foods is approximately 0.021.

To find the probability that a child younger than 18 has at least one food allergy and a history of severe reaction, we can use the information provided. We know that 7% of children younger than 18 have at least one food allergy and among those with food allergies, 41% had a history of severe reaction. To calculate the probability, we multiply these two probabilities together: 0.07 (the probability of having a food allergy) multiplied by 0.41 (the probability of having a severe reaction given a food allergy). So, the probability is 0.07 * 0.41 = 0.0287, which can be rounded to 0.0287 or approximately 0.029.

To find the probability that a child younger than 18 is allergic to multiple foods, we use the information that 30% of those with an allergy are allergic to multiple foods. So, the probability is 0.07 (the probability of having a food allergy) multiplied by 0.30 (the probability of being allergic to multiple foods given a food allergy). Hence, the probability is 0.07 * 0.30 = 0.021, which can be rounded to 0.021 or approximately 0.021.

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Answer:

a) 0.0989, Option a

Step-by-step explanation:

The concept of Poisson probability distribution is used as shown in the attached file.