"Suppose you draw a single card from a standard deck of 52 cards. How many ways arethere to draw either queen or a heart?"

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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Answer:

0.375 feet-lb

Step-by-step explanation:

We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.

We can represent our given information as:

[tex]6=\int\limits^2_0 {F(x)} \, dx[/tex]

We will use Hooke's Law to solve our given problem.

[tex]F(x)=kx[/tex]

Substituting this value in our integral, we will get:

[tex]6=\int\limits^2_0 {kx} \, dx[/tex]

Using power rule, we will get:

[tex]6=\left[ \frac{kx^2}{2} \right ]^2_0[/tex]

[tex]6=\frac{k(2)^2}{2}-\frac{k(0)^2}{2}[/tex]

[tex]6=\frac{4k}{2}-0\\\\k=3[/tex]

We know that 6 inches is equal to 0.5 feet.

Work needed to stretch it beyond 6 inches beyond its natural length would be [tex]\int\limits^{0.5}_0 {kx} \, dx =\int\limits^{0.5}_0 {3x} \, dx[/tex]

Using power rule, we will get:

[tex]\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0[/tex]

[tex]\frac{3(0.5)^2}{2}-\frac{3(0)^2}{2}\Rightarrow \frac{3(0.25)}{2}-0=\frac{0.75}{2}=0.375[/tex]

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.

Answer:

the helix intersects the sphere at t=4 and t=(-4)

Step-by-step explanation:

for the helix r(t) = [ sin(t) , cos(t) ,  t ] then x=sin(t) , y=cos(t) and z=t

thus the helix intersect the sphere  x² + y² + z² = 17 at

x² + y² + z² = 17

[sin(t)]²+[cos(t)]²+ t² = 17

1 + t² = 17

t² = 16

t = ±4

thus the helix intersects the sphere at t=4 and t=(-4)

The point wher the helix intersect the sphere is at t =  ±4

Given the coordinate of a helix expressed as;

If these coordinate intersects the sphere  x² + y² + z² = 17

Substitute the coordinate of the helix:

x² + y² + z² = 17

(sint)² + (cost)² + t² = 17

sin²t + cos²t + t² = 17

1 + t² = 17

t² = 17 - 1

t² = 16

t = ±√16

t = ±4

Hence the point wher the helix intersect the sphere is at t =  ±4

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Answer:

a) For this case we can see on the y axis the test scores and on the x axis the hours of homework. And as we can see we have a proportional relationship, because when the hours of homework increase the test scores increases too. If we fit a lineal model between y and x we will see an slope positive and a correlation coefficient positive based on the data observed.

b) [tex] y = 8x +42[/tex]

And we want to predict the test score for x = 3hr of homework we just need to replace the value of x=3 in the linear model and we got:

[tex] y = 8*3 +42= 24+42=66[/tex]

And that would be our predicted value for 3 h of homework

Step-by-step explanation:

For this case we consider the scatter plot attached to solve the problem.

Part a

For this case we can see on the y axis the test scores and on the x axis the hours of homework. And as we can see we have a proportional relationship, because when the hours of homework increase the test scores increases too. If we fit a lineal model between y and x we will see an slope positive and a correlation coefficient positive based on the data observed.

Part b

Assuming the following linear model for the situation:

[tex] y = 8x +42[/tex]

And we want to predict the test score for x = 3hr of homework we just need to replace the value of x=3 in the linear model and we got:

[tex] y = 8*3 +42= 24+42=66[/tex]

And that would be our predicted value for 3 h of homework

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:

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

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:

15 is the number of combination of 4 successes.

Step-by-step explanation:

We are given the following information:

We are given a binomial distribution, then probability of x succes is given by

[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 6 and x = 4

We have to evaluate the number of combination of success and failures.

It is given by:

[tex]\binom{n}{x} = \dfrac{n!}{x!(n-x)!}\\\\\binom{6}{4} = \dfrac{6!}{4!(6-4)!} = \dfrac{6!}{4!2!} = 15[/tex]

Thus, 15 is the number of combination of 4 successes.

The number of combinations of 4 successes out of 6 attempts, calculated using the binomial probability combination formula, is 15.

In binomial probability, the number of ways to get a specific number of successes is often calculated using the combination formulas. In this case, the number of combinations of X successes would be determined by the combination formula C(n, x), which stands for the number of combinations of n items taken x at a time. In your case, where n=6 and x=4, the required number of combinations (or ways to achieve 4 successes out of 6 trials) is C(6,4).

The formula for a combination is generally given by: n! / [x!(n - x)!]. Plugging the given numbers into the formula, we get: C(6,4) = 6! / [4!(6 - 4)!].  This simplifies to: 720 / (24*2), which equals to 15. So, there are 15 combinations of 4 successes out of 6 attempts.

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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:

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 and Step-by-step explanation:

a) probability of selling a medium, long sleeved printed shirt, P(M ∩LS ∩PR) = 0.07, directly from the table of probabilities

b) probability that the next shirt sold is a medium printed shirt, (M ∩Pr) = P(M,Pr,LS) + P(M,Pr,SS) = 0.07+0.10 = 0.17

c) probability that the next shirt is a short sleeved shirt, P(SS) = sum of 9 probabilities in Short Sleeved shirt table = 0.58

probability that the next shirt is a long sleeved shirt, P(LS) = 1 - 0.58 = 0.42 or add up the total probabilities in the long sleeved shirt table, P(LS) = sum of 9 probabilities in the long sleeved shirt table = 0.42

d) probability that the next shirt is a medium, P(M) = P(M,SS) + P(M,LS) = (0.07+0.10+0.12) + (0.06+0.07+0.07) = 0.49

probability that the next shirt sold is a print, P(Pr) = P(Pr,SS) + P(Pr,LS) = (0.02+0.10+0.07) + (0.02+0.07+0.02) = 0.40

e) probability that the shirt sold is a medium given that the shirt just sold was a short-sleeved plaid, P(M|SS,PL) = (P(M,SS,PL))/P(SS,PL) = 0.07/(0.04+0.07+0.03) = 0.5

f) probability that the shirt sold is short sleeved given that the shirt just sold was a medium plaid, P(SS|M,PL) = (P(M,SS,PL))/P(M,PL) = 0.07/(0.07+0.06) = 0.53846 = 0.54

probability that the shirt sold is long sleeved given that the shirt just sold was a medium plaid, P(LS|M,PL) = (P(M,LS,PL))/P(M,PL) = 0.06/(0.07+0.06) = 0.462 = 0.46

QED!

The following probabilities are as follow;

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.

(a) The next shirt sold is a medium, long-sleeved, print shirt.

Probability = 0.07

(b) The probability that the next shirt sold is a medium print shirt.

Probability = 0.7 + 0.10 = 0.17

(c) The probability that the next shirt sold is a long-sleeved shirt.

Probability = 0.42

(d) The probability that the size of the next shirt sold is medium. The probability that the pattern of the next shirt sold is a print.

Probability = 0.07 + 0.10 = 0.17

(e)The shirt just sold was a short-sleeved plaid, the probability that its size was medium.

Probability = 0.07

(f) The shirt just sold was a medium plaid, The probability that it was short-sleeved.

Probability = 0.07

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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.

Answer:

a) 0.0989, Option a

Step-by-step explanation:

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

Answer:

12%

Step-by-step explanation:

let x represent the actual number of members last year.

current members = 675 and with an increment of 13% over previous year

to find the increment

(675 - x) / x = 0.13

675 - x = 0.13x

collect the like terms

675 = 0.13x + x

675 = 1.13 x

x = approx 597 members were there last year

its hope to increase by the same number next year

percent increase = 675 - 597 / 675 = 78 / 675 = 0.12 = 12%

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:

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: There is a strong positive correlation between  number of games won by a minor league baseball team and the average attendance at their home games is analyzed.

Step-by-step explanation:

The Pearson's coefficient 'r' gives the correlation between the predicted values and the observed values .

Given :  A regression to predict the average attendance from the number of games won has an r = 0.73.

Since r=0.73 is positive and 0.70 <0.73 <1 , it means there is a strong positive correlation between number of games won by a minor league baseball team and the average attendance at their home games is analyzed.

Answer:

a. 15

b.

Sr.no Samples Sample mean

1           (79,64)         71.5

2          (79,84)          81.5

3          (79,82)          80.5

4          (79,92)          85.5

5          (79,77)           78

6         (64,84)            74

7          (64,82)            73

8          (64,92)            78

9          (64,77)            70.5

10        (84,82)            83

11         (84,92)            88

12        (84,77)             80.5

13        (82,92)            87

14        (82,77)            79.5

15        (92,77)            84.5

c.

mean of sample mean=population mean=79.67

Step-by-step explanation:

a.

The different samples of two test grade are nCr, where n=6 and r=2.

nCr=6C2=6!/2!(6-2)!=6*5*4!/2!4!=30/2=15.

Thus, there are 15 different samples of two test grade.

b.

All possible samples are listed below:

Sr.no Samples

1           (79,64)        

2          (79,84)          

3          (79,82)          

4          (79,92)        

5          (79,77)          

6         (64,84)            

7          (64,82)

8          (64,92)

9          (64,77)

10        (84,82)

11         (84,92)

12        (84,77)

13        (82,92)

14        (82,77)

15        (92,77)

The sample means for each sample can be calculated as

Sr.no Samples Sample mean

1           (79,64)         (79+64)/2=71.5

2          (79,84)          (79+84)/2=81.5

3          (79,82)          (79+82)/2=80.5

4          (79,92)          (79+92)/2=85.5

5          (79,77)           (79+77)/2=78

6         (64,84)            (64+84)/2=74

7          (64,82)            (64+82)/2=73

8          (64,92)            (64+92)/2=78

9          (64,77)            (64+77)/2=70.5

10        (84,82)            (84+82)/2=83

11         (84,92)            (84+92)/2=88

12        (84,77)             (84+77)/2=80.5

13        (82,92)            (82+92)/2=87

14        (82,77)            (82+77)/2=79.5

15        (92,77)            (92+77)/2=84.5

c.

The sample means of sample mean μxbar will calculated by taking average of sample means

μxbar=(71.5+ 81.5+ 80.5+ 85.5+ 78+ 74+ 73+ 78+ 70.5+ 83+ 88+ 80.5+ 87+ 79.5+ 84.5)/15

μxbar=1195/15=79.67

Population mean=μ=(79+64+84+82+92+77)/6

μ=478/6=79.67

Sample means of sample mean μxbar=Population mean μ.

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:

See below

Step-by-step explanation:

Essentially, we have to replace "quantifier words" like "All", "Every" by the universal quantifier ∀.

a) ∀ dinosaur x, x is extinct.

b) ∀ real number x, x is positive, negative, or zero.

c) ∀ irrational number x, x is not an integer.

d) ∀ logician x, x is not lazy.

e) ∀ integer x, x²≠ 2,147,581,953.

f)  ∀ real number x, x²≠ -1.

In a) and b) we replace the words without major changes. In the other statements, we modify the statement using negation. For example, "No irrational numbers are integers." is equivalent to "Every irrational number is not integer".

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:

A. 8 complete hotdogs were made.

B. The wiener.

C. The bun.

Step-by-step explanation:

1 complete hotdog = 1 wiener + 1 bun

You have, 3*10 bun and 8 wiener

= 30 bun and 8 wiener

A. Since, we have 30 bun and 8 wiener and for 1 complete hotdog = 1 wiener + 1 bun

Therefore, 8 hotdogs = 8 wiener + 8 buns

= 8 complete hotdogs are made

B.

Since 8 complete hotdogs are made, 8 wiener and 8 buns are used.

Therefore, the limiting ingredient is the wiener because it is the smallest number of ingredient and none of it was left after making the 8 hotdogs.

C. Since 8 complete hotdogs are made, 8 buns were used

So, the leftover buns = (30 - 8) buns

= 22 buns

The buns is the leftover ingredient

At the store, you buy four packages of eight wieners and three bags of 10 buns, you can make 8 hotdogs.

There are 8 hotdogs you can make then the limited ingredient is wieners.

The buns are the leftover ingredient.

You and your friends are making hot dogs.

A complete hot dog consists of a wiener and a bun.

At the store, you buy four packages of eight wieners and three bags of 10 buns.

Total number of buns = 30

Total number of wieners = 8

At the store, you buy four packages of eight wieners and three bags of 10 buns.

For every hot dog, 1 wiener one 1 buns are required. the calculation for the given question is given as follows;

1. The total number of hot dogs can you make is,

To make 1 hot dog 1 wiener and 1 bun

Then,

for 8 hot dogs, 8 wieners and 8 buns are required.

Therefore, you can make 8 hotdogs.

2. There are 8 hotdogs you can make then the limited ingredient is wieners.

3. After making 8 hot dogs only 8 buns are used,

Then,

Left buns = Total number of buns - used buns

Left buns = 30-8 = 22

The buns are the leftover ingredient.

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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%.

Answer:

[tex]45*10^{6}[/tex] orders of magnitude.

Step-by-step explanation:

We know that 900 km is equivalent to 20 cm, because we need to fit the United Kingdom length on a sheet of paper. So:

Therefore, if we divide 45 km by 0.00001 km (1 cm) we will have [tex]45*10^{6}[/tex] times to scale down.

I hope it helps you!

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 .

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.

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.

The total number of ways to choose the two groups, depending on the relationship between n and m, is:

M < N: C(100, n).

M ≥ N: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1.

Case 1: m < n (n tallest chosen first):

Choose the n tallest people from the 100. There are C(100, n) ways to do this.

The remaining 100 - n people are all shorter than the chosen n. Since m < n, we can simply choose all the remaining people (100 - n).

The total number of ways in this case is C(100, n).

Case 2: m ≥ n (not all n tallest need to be chosen):

This case requires considering all possible combinations of how many tall people to choose (from n down to 1).

For each value of k (n, n - 1, ..., 1), choose k tallest people from the 100. There are C(100, k) ways to do this.

The remaining 100 - k people are all shorter than the chosen k. Choose the remaining m - k people from this group. There are C(100 - k, m - k) ways to do this.

Sum the results for each value of k: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1.

Therefore, the total number of ways to choose the two groups, depending on the relationship between n and m, is:

M < N: C(100, n)

M ≥ N: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1

Complete question:

How many ways are there to pick a group of n people from 100 people, and then pick a second group of m people from the remaining 100 − n people, so that all the people you chose from the first group are taller than the people from the second group (assume that everyone in the group of 100 people has a different height.)

Note: it is implied in this question that 1 ≤ n ≤ 100 and 0 ≤ m ≤ 100 − n.

Answer:

(a) 0.970225.

(b) 0.999775.

(c) 2.25 x [tex]10^{-4}[/tex].

Step-by-step explanation:

Given the probability that either system will function satisfactorily during a flight is 0.985.

(a) To calculate the probability that both systems function satisfactorily we are given the probability of either system will function of 0.985 which means that both function have probability of functioning satisfactorily of 0.985 each i.e.,

= 0.985 x 0.985 = 0.970225.

(b) The probability of first function will not function satisfactorily is 1 - 0.985 because the probability of first system function satisfactorily is 0.985.

Similarly, probability of second function will not function satisfactorily is also 1 - 0.985 = 0.015

So Probability that during a given flight at least one system functions satisfactorily = 1 - both system will not function satisfactorily

                     = 1 - (0.015 x 0.015) = 0.999775.

(c) Probability that during a given flight both systems fail or not function  satisfactorily = (1 - 0.985) x (1 - 0.985) = 2.25 x [tex]10^{-4}[/tex]

Answer:

(a) P (Both systems functioning satisfactorily) = 0.970

(b) P (At least one system functions satisfactorily) = 0.999

(c) P (Both the systems failing) = 0.00023

Step-by-step explanation:

Let A = the system in use is working satisfactorily and B = the backup system is working satisfactorily.

The probability that either of the systems works satisfactorily is,

P (A) = P(B) = 0.985

(a)

Both the events A and B are independent, i.e. [tex]P(A\cap B)=P(A)\times P(B)[/tex]

Compute the probability that during a given flight both systems function satisfactorily as follows:

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

               [tex]=0.985\times0.985\\=0.970225\\\approx0.970[/tex]

Thus, the probability of both systems functioning satisfactorily is 0.970.

(b)

Compute the probability that during a given flight at least one system functions satisfactorily as follows:

P (At least one system functions satisfactorily) = 1 - P (None of the system functions satisfactorily)

                                                                             = [tex]1-[P(A^{c})\times P(B^{c})][/tex]

                                                                             [tex]=1-([1-P(A)]\times [1-P(B)])\\=1-([1-0.985]\times [1-0.985])\\= 1-0.000225\\=0.999775\\\approx0.999[/tex]

Thus, the probability that during a given flight at least one system functions satisfactorily is 0.999.

(c)

Compute the probability that during a given flight both the systems fail as follows:

P (Both the systems failing) = [tex]P(A^{c})\times P(B^{c})[/tex]

                                             [tex]=[1-P(A^{c})]\times [1-P(B^{c})]\\=(1-0.985)\times (1-0.985)\\=0.015\times 0.015\\=0.000225\\\approx0.00023[/tex]

Thus, the probability that during a given flight both the systems fail is 0.00023.

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.