Using your knowledge of exponential and logarithmic functions and properties, what is the intensity of a fire alarm that has a sound level of 120 decibels?



A.
1.0x10^-12 watts/m^2
B.
1.0x10^0 watts/m^2
C.
12 watts/m^2
D.
1.10x10^2 watts/m^2

Final answer:

The appropriate conclusion from the survey is that the true percentage of all people who regularly use sunscreen in the region is estimated to be between 18% and 20%, considering the margin of error of 1 percentage point.

Explanation:

Based on the estimate and the margin of error, an appropriate conclusion is that it can be said with a certain level of confidence, usually 95%, that the true percentage of all people living in that region who regularly use sunscreen when going outdoors is between 18% and 20%. The margin of error gives a range around the estimate to account for the sampling error, and it implies that the actual value in the population should fall within this range. This helps to understand the precision of the survey result and allows for the natural variability that comes from the fact that not every individual in the region was surveyed.

The variability also depends on the size of the sample, which relates to the confidence interval. For example, smaller samples tend to have larger margins of error, while larger samples tend to lead to smaller margins of error and thus, more precise estimates of the population parameter.

Answer:

option b

-x^2 + 3x - 11

Step-by-step explanation:

we have to add the X of the same order

x^2

-3x^2 + 2x^2 = -x^2

x^1

4x + ( -x ) = 3x

x^0

0 + (-11) = -11

finally we accommodate and get the final result

-x^2 + 3x - 11

Answer:-x^2 + 3x - 11

Step-by-step explanation:

I got it right on the test

Answer:

a. The angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(\sqrt{3}, 1\right)[/tex] and [tex]\mathbf{v}=\left(0, 5\right)[/tex] is [tex]\phi=\frac{\pi}{3}[/tex] or [tex]60\º[/tex].

b. The angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(0, 4, 4\right)[/tex] and [tex]\mathbf{v}=\left(3, -3, 0\right)[/tex] is [tex]\phi=\frac{2 \pi}{3}[/tex] or [tex]120\º[/tex].

Step-by-step explanation:

a. To calculate the angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(\sqrt{3}, 1\right)[/tex] and [tex]\mathbf{v}=\left(0, 5\right)[/tex] you must:

Step 1: Calculate the dot product.

The dot product is given as [tex]\displaystyle{\large{{\left({u}_{{x}},{u}_{{y}}\right)}\cdot{\left({v}_{{x}},{v}_{{y}}\right)}={u}_{{x}}\cdot{v}_{{x}}+{u}_{{y}}\cdot{v}_{{y}}}}[/tex].

So,

[tex]\left(\sqrt{3}, 1\right)\cdot\left(0, 5\right)=\left(\sqrt{3}\right)\cdot\left(0\right)+\left(1\right)\cdot\left(5\right)=5[/tex]

Step 2: Find the lengths of the vectors.

[tex]\left|\mathbf{u}\right|=\sqrt{\left(u_x\right)^2+\left(u_y\right)^2}=\sqrt{\left(\sqrt{3}\right)^2+\left(1\right)^2}=2[/tex]

[tex]\left|\mathbf{v}\right|=\sqrt{\left(v_x\right)^2+\left(v_y\right)^2}=\sqrt{\left(0\right)^2+\left(5\right)^2}=5[/tex]

Step 3: The angle is given by [tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}[/tex]

[tex]\frac{5}{2 \cdot 5}=\frac{1}{2}[/tex]

[tex]\phi=cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}=60^0[/tex]

b. To calculate the angle [tex]\phi[/tex] between the vectors [tex]\mathbf{u}=\left(0, 4, 4\right)[/tex] and [tex]\mathbf{v}=\left(3, -3, 0\right)[/tex] you must:

Step 1: Calculate the dot product.

[tex]\left(0, 4, 4\right)\cdot\left(3, -3, 0\right)=\left(0\right)\cdot\left(3\right)+\left(4\right)\cdot\left(-3\right)+\left(4\right)\cdot\left(0\right)=-12[/tex]

Step 2: Find the lengths of the vectors.

[tex]\left|\mathbf{u}\right|=\sqrt{\left(u_x\right)^2+\left(u_y\right)^2+\left(u_z\right)^2}=\sqrt{\left(0\right)^2+\left(4\right)^2+\left(4\right)^2}=4 \sqrt{2}[/tex]

[tex]\left|\mathbf{v}\right|=\sqrt{\left(v_x\right)^2+\left(v_y\right)^2+\left(v_z\right)^2}=\sqrt{\left(3\right)^2+\left(-3\right)^2+\left(0\right)^2}=3 \sqrt{2}[/tex]

Step 3: The angle is given by [tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}[/tex]

[tex]\cos\left(\phi\right)=\frac{\mathbf{u} \cdot \mathbf{v}}{\left|\mathbf{u}\right| \cdot \left|\mathbf{v}\right|}=\frac{-12}{4 \sqrt{2} \cdot 3 \sqrt{2}}=- \frac{1}{2}[/tex]

[tex]\phi=cos^{-1}\left(- \frac{1}{2}\right)=\frac{2 \pi}{3}=120^0[/tex]

To find the angle between two vectors, use the dot product formula. For the vectors (√3,1) and <0, 5 >, the angle is π/3 radians. For the vectors <0,4,4> and <3,-3,0>, the angle is 2π/3 radians.

To find the angle between two vectors, we can use the dot product formula. Let's start with the first set of vectors (√3,1) and <0, 5>. We can calculate the dot product by multiplying the corresponding components and adding them up: (√3)(0) + (1)(5) = 5. Next, we need to calculate the magnitudes of each vector. The magnitude of (√3,1) is √(√3)^2 + 1^2 = √4 = 2. The magnitude of <0, 5> is √0^2 + 5^2 = √25 = 5. Now, we can calculate the angle between the vectors using the formula: cosθ = dot product / (magnitude 1 * magnitude 2). Plugging in the values, cosθ = 5 / (2 * 5) = 1/2. Taking the inverse cosine of 1/2 gives us the angle in radians: θ = acos(1/2) = π/3 ≈ 1.047.

For the second set of vectors <0,4,4> and <3,-3,0>, we will follow the same steps. The dot product is (0)(3) + (4)(-3) + (4)(0) = -12. The magnitude of <0,4,4> is √0^2 + 4^2 + 4^2 = √32 = 4√2. The magnitude of <3,-3,0> is √3^2 + (-3)^2 + 0^2 = √18 = 3√2. Plugging in the values, cosθ = -12 / (4√2 * 3√2) = -1/2. Taking the inverse cosine of -1/2 gives us the angle in radians: θ = acos(-1/2) = 2π/3 ≈ 2.094.

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Answer:

d. field notes recorded during participant observation

Step-by-step explanation:

Qualitative data are data that can be gathered and described by observation, interviews and evaluation. Such data are not numerically based but are based on features of the phenomenon under study. It is also known as categorical data.

Methods of Qualitative data includes :

Focus groups, keeping records, direct interviews, case studies etc.

The area of a triangle is 1/2 x base x height

Area = 1/2 x 24 x 28 = 336 square inches.

Because the corners are rounded the are would actually be just under 336 but no radius was given to get the exact area.

Answer:

yes

Step-by-step explanation:

Answer:

The value of sensitivity obtained is equal to the one given in statement.

Step-by-step explanation:

Sensitivity is the ratio of number of positive test results to the number of individuals having disease. It is the ability of the test to detect all those with disease in the screened population.

Its formula is written as:

Sensitivity = [tex]\frac{number of true positives}{total with disease}[/tex]

So, in this case we have values

sensitivity = 14/16 = 0.875 So, the given statement is true.

Answer:

0.0202

Step-by-step explanation:

I got this because there is 1000 millimeters in a meter. if you divide 20.2 by 1000, you get 0.0202.

Have a good day :3

Answer:

Step-by-step explanation:

a) probability a person admitted to the hospital will suffer a treatment-caused injury due to negligence

P(injury) = 4%

P(negligence) = 1/4 = 0.25

We need to find probability (injury)(negligence)

P(injury) * P(negligence) = 0.04*0.25 = 0.01

b) probability a person admitted to the hospital will die from a treatment-caused injury

P(injury) = 4%

P(death) = 1/7

P(Injury) *P(death) = 0.04/7 = 0.00571

c) In the case of a negligent treatment-caused injury, what is the probability a malpractice claim will be paid

P(claim) = 1/7.5

P(payment) = 1/2

P(claim)*P(payment) = 1/7.5 * 1/2 = 0.06

Answer:

a.) 0.01

b.) 0.006

c.)0.00067

Step-by-step explanation:

Answer: m= 18g

Step-by-step explanation:

Let m= miles you can drive

g = g for gallons in tank

We know that there is directly proportional relation between the distance traveled by vehicle and the number of gallons of gas.

Then, rate of miles driven by you per gallon = [tex]\dfrac{m}{g}[/tex]

Since , rate of miles driven by you per gallon  = 18 miles per gallon (given)

Then,

[tex]\dfrac{m}{g}=18\\\\ m= 18g[/tex]

Hence, the equation to model this situation : m= 18g

Answer:

P(20≤x≤35) = 0.75 .

Step-by-step explanation:

We know that the probability distribution function of Uniform Distribution is represented as :

     If x follows Uniform(c,d) then,

          f(x) = [tex]\frac{1}{d-c}[/tex] where c < x < d

To fond the given probability it is better to first calculate the Cumulative Distribution Function(CDF) of Uniform Distribution.

The CDF of Uniform Distribution is P(X<=x) = [tex]\frac{x-c}{d-c}[/tex] where d > c .

Therefore, P(20<=x<=35) = P(x<=35) - P(x<20)

  P(x<=35) =  [tex]\frac{x-20}{40-20}[/tex]  because we are given c = 20 and d = 40.

                 = [tex]\frac{35-20}{40-20}[/tex] = 0.75

 P(x<20) =  [tex]\frac{x-20}{40-20}[/tex]  = [tex]\frac{20-20}{40-20}[/tex] = 0

Hence, P(20<=x<=35) = 0.75 - 0 = 0.75.

The probability of x being between 20 and 35 is 0.75 or 75%.

The uniform probability distribution is defined as:

P(x) = 1/(b-a) for a ≤ x ≤ b and 0 otherwise.

In this case, a=20 and b=40, so the probability P(20≤x≤35) can be calculated as:

P(20≤x≤35) = (35-20)/(40-20) = 15/20 = 0.75.

Therefore, the probability of x being between 20 and 35 is 0.75 or 75%.

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Answer:

7/12

Step-by-step explanation:

Step 1:  Find common denominators

The common denominator is 12

10/12 is already good

1/4 * 3/3 = 3/12

Step 2:  Subtract

10/12 - 3/12 = 7/12

Answer:  7/12

Final answer:

To subtract the fractions 10/12 and 1/4, find a common denominator, convert the fractions, and then subtract the numerators, resulting in the answer 7/12.

Explanation:

To subtract fractions with unlike denominators, you first need to find a common denominator. For the fractions 10/12 and 1/4, the common denominator is 12. You can convert 1/4 to 3/12, because 4 times 3 equals 12. Now you can subtract the fractions as follows:

10/12 - 3/12 = (10 - 3)/12 = 7/12

So, the answer to 10/12 - 1/4 is 7/12.

This process relies on the intuitive understanding that you can only subtract numerators if the denominators are the same. We never add or subtract the denominators themselves. When your fractions share the same denominator, the subtraction is straightforward as we are simply dealing with parts of the whole that are of the same size.

Answer:

1. 1/12 or 0.083

2. 5/36 or 0.1389

3. 7/12 or 0.5833

4. 5/12 or 0.4167

5. 11/12 or 0.9167

6. 1/9 or 0.1111

7. 1/3 or 0.3333

8. 5/12 or 0.4167

Step-by-step explanation:

This problem can easily be understood by making sample space first.

Outcomes in sample space=S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Total outcomes=36

The probability in the following scenario is calculated as

Probability=number of outcomes/Total outcomes

1.

The sum is 4

The sum is 4 ={(1,3),(2,2),(3,1)}

number of outcomes=3

P(The sum is 4)=3/36=1/12 or 0.083

2.

The sum is 6

The sum is 6 ={(1,5),(2,4),(3,3),(4,2),(5,1)}

number of outcomes=5

P(The sum is 6)=5/36 or 0.1389

3.

The sum is at least 7

The sum is at least 7=The sum is greater than or equal to 7

The sum is at least 7={(1,6),(2,5),(2,6),(3,4),(3,5),(3,6),(4,3),(4,4),(4,5),(4,6),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

number of outcomes=21

P(The sum is at least 7)=21/36

P(The sum is at least 7)=7/12 or 0.5833

4.

The sum is at least 8

The sum is at least 8=The sum is greater than or equal to 8

The sum is at least 8={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

number of outcomes=15

P(The sum is at least 8)=15/36

P(The sum is at least 8)=5/12 or 0.4167

5.

The sum is less than 11

The sum is less than 11={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(6,1),(6,2),(6,3),(6,4)}

number of outcomes=33

P(The sum is less than 11)=33/36

P(The sum is less than 11)=11/12 or 0.9167

6.

The sum is 2, 3, or 12

The sum is 2, 3, or 12={(1,1),(1,2),(2,1),(6,6)}

number of outcomes=4

P(The sum is 2, 3, or 12)=4/36

P(The sum is 2, 3, or 12)=1/9 or 0.1111

7.

The sum is odd and no more than 7

The sum is odd and no more than 7={(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(6,1)}

number of outcomes=12

P(The sum is odd and no more than 7)=12/36

P(The sum is odd and no more than 7)=1/3 or 0.3333

8.

The sum is odd or prime

The sum is odd or prime={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)}

number of outcomes=15

P(The sum is odd or prime)=15/36

P(The sum is odd or prime)=5/12 or 0.4167

Probabilities are,

1. The sum is 4 = 1/12

2. The sum is 6 = 5/36

3. The sum is at least 7 = 7/12

4. The sum is at least 8 = 5/12

5. The sum is less than 11 = 11/12

6. The sum is 2, 3, or 12 = 1/9

7. The sum is odd and no more than 7 = 1/3

8. The sum is odd or prime = 19/36

When six sided die toss two times, Then total possible Outcomes shown below

[tex](1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}[/tex]

Total number of  outcomes=36

Probability is calculated by dividing number of favourable outcomes by Total number of  outcomes

1. Favourable outcomes for the sum is 4 [tex]={(1,3),(2,2),(3,1)}[/tex]

number of outcomes=3

The probability of The sum is 4. = [tex]3/36=1/12 = 0.083[/tex]

2.Favourable outcomes for the sum is 6[tex]={(1,5),(2,4),(3,3),(4,2),(5,1)}[/tex]

number of outcomes=5

The probability of The sum is 6, =5/36  

3.The sum is at least 7,

={(1,6),(2,5),(2,6),(3,4),(3,5),(3,6),(4,3),(4,4),(4,5),(4,6),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

number of outcomes=21

Probability of the sum is at least 7 =21/36=7/12 = 0.5833

4.The sum is at least 8

The sum is at least 8={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

number of outcomes=15

The sum is at least 8=15/36 = 5/12 = 0.4167

5.The sum is less than 11

The sum is less than 11={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(6,1),(6,2),(6,3),(6,4)}

number of outcomes=33

The sum is less than 11=33/36 = 11/12

6.The sum is 2, 3, or 12={(1,1),(1,2),(2,1),(6,6)}

number of outcomes=4

The sum is 2, 3, or 12=4/36 = 1/9

7.The sum is odd and no more than 7={(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(6,1)}

number of outcomes=12

The sum is odd and no more than 7=12/36 = 1/3

8.The sum is odd or prime={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5), (3, 6), (6, 3), (4,5), (5, 4)

number of outcomes=19

The sum is odd or prime=19/36

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Answer:

skew lines

Step-by-step explanation:

Given are two lines in 3 dimension as

[tex]L1: x=23+6t, y=12+3t, z=19+5t\\L2: x=-9+7t, y=-7+5t, z=-12+8t[/tex]

To find out whether parallel or skew or intersect

If parallel direction ratios should be proportional

Direction ratios of I line are 6,3,5 and not proportional to that of II line (7,5,8)

So not parallel

If intersect we must have same point for the two lines

Let us change parameter for II line to s to avoid confusion.  If intersecting, then

[tex]23+6t = -9+7s\\12+3t = -7+5s\\19+5t =12+8s\\[/tex]

6t-7s=-32 and 3t-5s = -19

Solving

t=-3 and s =2

Check this with III equation

Left side = 19-15 =4 and right side = 12+16 =28

not equal

So there cannot be any point of intersection.

These two are skew lines

Answer:

See explanation

Step-by-step explanation:

In Saras class, [tex]\frac{2}{5}[/tex] of the students ride a bus, and [tex]\frac{1}{3}[/tex] ride a car to school.

The rest

[tex]1-\dfrac{2}{5}-\dfrac{1}{3}=\dfrac{15-6-5}{15}=\dfrac{4}{15}[/tex]

walk to school (simply subtract from one whole the fractions of students).

The fraction of students who ride a bus or car to school is

[tex]\dfrac{2}{5}+\dfrac{1}{3}=\dfrac{6+5}{15}=\dfrac{11}{15}.[/tex]

See attached diagram for pictorial representations (red diagram shows the fraction of students who ride a bus, blue - ride a car, green - walk to school)

The fraction of students who walk to school is [tex]$\boxed{\frac{4}{15}}$[/tex].

To find the fraction of students who walk to school, we first consider the fractions of students who ride a bus or car to school. According to the information given:

- The fraction of students who ride a bus is [tex]$\frac{2}{5}$[/tex].

- The fraction of students who ride a car is [tex]$\frac{1}{3}$[/tex].

To find the total fraction of students who use transportation (bus or car), we add these two fractions together. However, we must first find a common denominator to add them properly. The least common multiple (LCM) of 5 and 3 is 15, so we convert both fractions to have a denominator of 15:

[tex]- $\frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15}$ (for the bus riders)[/tex]

[tex]- $\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}$ (for the car riders)[/tex]

Now we can add these fractions:

[tex]- $\frac{6}{15} + \frac{5}{15} = \frac{11}{15}$[/tex]

This represents the fraction of students who either ride a bus or a car to school. To find the fraction of students who walk to school, we subtract this fraction from the whole, which is 1 (since the whole class is represented by 1, or [tex]$\frac{15}{15}$):[/tex]

[tex]- $1 - \frac{11}{15} = \frac{15}{15} - \frac{11}{15} = \frac{4}{15}$[/tex]

Therefore, the fraction of students who walk to school is [tex]$\frac{4}{15}$[/tex].

To visualize this with a diagram, we can imagine a whole circle representing the entire class. We divide this circle into 15 equal parts (since our denominators are 15). We shade 6 parts to represent the bus riders and another 5 parts to represent the car riders. The remaining unshaded parts, which are 4 in number, represent the students who walk to school.

The equation representing the total fraction of students who walk or use transportation is:

[tex]\[ \text{Bus riders} + \text{Car riders} + \text{Walkers} = \text{Whole class} \] \[ \frac{6}{15} + \frac{5}{15} + \text{Walkers} = 1 \][/tex]

Solving for Walkers:

[tex]\[ \text{Walkers} = 1 - \left( \frac{6}{15} + \frac{5}{15} \right) \] \[ \text{Walkers} = 1 - \frac{11}{15} \] \[ \text{Walkers} = \frac{15}{15} - \frac{11}{15} \] \[ \text{Walkers} = \frac{4}{15} \][/tex]

Thus, the fraction of students who walk to school is[tex]$\boxed{\frac{4}{15}}$.[/tex]

Answer:

The measure of dispersion which is likely to vary most between your first and second samples is the range.

Step-by-step explanation:

The range and standard deviation of a data are measures of dispersion, i.e. they measure the degree to which the data is dispersed.

The formula to compute the range is:

[tex]Range=X_{max}-X_{min}[/tex]

The formula to compute the sample standard deviation is:

[tex]s=\sqrt{\frac{1}{n-1}\sum (X-\bar X)^{2} }[/tex]

The sample size is: n = 50.

Thus, the measure of dispersion which is likely to vary most between your first and second samples is the range.

Final answer:

The range is more likely to vary between two samples from a population due to its sensitivity to extreme values, whereas the standard deviation is a more stable measure of dispersion that accounts for all values in the dataset.

Explanation:

The measure of dispersion most likely to vary between the first and second samples from the population is the range. The range is sensitive to the particular values in the dataset because it only takes into account the smallest and largest values. Since each sample may have a different set of extreme values, the range can vary greatly from sample to sample. In contrast, the standard deviation is less likely to vary significantly between samples, as it measures the amount of variation or dispersion from the mean of a dataset and takes into account all the values in the set.

When drawing multiple samples from a population, the distribution of the sample means will tend to form a normal distribution around the population mean according to the Central Limit Theorem. In the case of a normally distributed population with a known mean and standard deviation, the sampling distribution of the sample mean will have a standard deviation equal to the population standard deviation divided by the square root of the sample size (known as the standard error).

Answer:

a) P=0.0225

b) P=0.057375

Step-by-step explanation:

From exercise we have that  15​% of items produced are defective, we conclude that  probabiity:

P=15/100

P=0.15

a)  We calculate the probabiity  that two items are​ defective:

P= 0.15 · 0.15

P=0.0225

b)  We calculate the probabiity  that  two  of  three items are​ defective:

P={3}_C_{2} · 0.85 · 0.15 · 0.15

P=\frac{3!}{2!(3-2)!} · 0.019125

P=3 · 0.019125

P=0.057375

Using the binomial distribution, it is found that there is a:

a) 0.0225 = 2.25% probability that both are​ defective.

b) 0.0574 = 5.74% probability that two are​ defective.

For each item, there are only two possible outcomes, either it is defective, or it is not. The probability of an item being defective is independent of any other item, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

In this problem, 15​% of items produced are defective, hence [tex]p = 0.15[/tex].

Item a:

Two items, hence [tex]n = 2[/tex].

The probability is P(X = 2), then:

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

[tex]P(X = 2) = C_{2,2}.(0.15)^{2}.(0.85)^{0} = 0.0225[/tex]

0.0225 = 2.25% probability that both are​ defective.

Item b:

Three items, hence [tex]n = 3[/tex].

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

[tex]P(X = 2) = C_{3,2}.(0.15)^{2}.(0.85)^{1} = 0.0574[/tex]

0.0574 = 5.74% probability that two are​ defective.

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Answer:

a) 52.15

b) 58.53

c) 62.98        

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 56

Standard Deviation, σ = 3

We are given that the distribution of x is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) We have to find [tex]x_0[/tex] such that

P(X < x)  = 0.1

[tex]P( X < x) = P( z < \displaystyle\frac{x_0 - 56}{3})=0.1[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < -1.282) = 0.1[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = -1.282\\\\x_0 = 52.15[/tex]

b) We have to find [tex]x_0[/tex] such that

P(X < x)  = 0.8

[tex]P( X < x) = P( z < \displaystyle\frac{x_0 - 56}{3})=0.8[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < 0.842) = 0.8[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = 0.842\\\\x_0 = 58.53[/tex]

c) We have to find [tex]x_0[/tex] such that

P(X > x)  = 0.01

[tex]P( X > x) = P( z > \displaystyle\frac{x - 56}{3})=0.01[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{x - 56}{3})=0.01 [/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 56}{3})=0.99 [/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < 2.326) = 0.99[/tex]

[tex]\displaystyle\frac{x_0 - 56}{3} = 2.326\\\\x_0 = 62.98[/tex]

To find specific percentiles for a normally distributed variable x with mean 56 and standard deviation 3, you must use a z-score table or calculator to find the corresponding z-scores for the 10%, 80%, and top 1% percentages, then use the formula x0 = µ + z*σ to calculate x0.

To find the value x0 of the normally distributed random variable x with mean µ = 56 and standard deviation σ = 3, that corresponds to specific percentiles, one would typically use a z-score table (or a calculator with a normal distribution function). However, in this scenario, the values provided for the x variable and the associated z-scores in the question are incorrect and would not provide a reasonable answer. Instead, I will explain the steps needed to find x0 without using the incorrect values provided.

After finding each z-score, use the formula x0 = µ + z*σ to calculate the corresponding x0 for each case.

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Answer:

The number of rookies who scored 4 or better is 24.

Step-by-step explanation:

the frequency distribution of for the ranking of pro-football rookies is:

Ranking   Frequency

    1                  3

    2                 7

    3                16

    4                21

    5                  3

Compute the number of rookies who scored 4 or better as follows:

No. of rookies with rankings 4 or more = No. of rookies with rank 4 +

                                                                           No. of rookies with rank 5

                                                                 = 21 + 3

                                                                 = 24

Thus, the number of rookies who scored 4 or better is 24.

Answer:

Step-by-step explanation:

Given:

[tex]P(S\cap F)=0.44\\P(S\cap F^{c})=0.13\\P(S^{c}\cap F^{c}) = 0.32\\P(S^{c}\cap F) = 0.11[/tex]

The rule of total probability states that:

[tex]P(A) = P(A\cap B) + P(A\cap B^{c})[/tex]

Compute the individual probabilities as follows:

[tex]P(S) = P(S\cap F) + P(S\cap F^{c})\\=0.44+0.13\\0.57[/tex]

[tex]P(S^{c}) = 1 - P(S)\\=1-0.57\\=0.43[/tex]

[tex]P(F) = P(S\cap F) + P(S^{c}\cap F)\\=0.44+0.11\\=0.55[/tex]

[tex]P(F^{c})=1-P(F)\\=1-0.55\\=0.45[/tex]

Conditional probability of an event A given B is:

[tex]P(A|B)=\frac{P(A\cap B)}{P(B)}[/tex]

         [tex]P(S|F)=\frac{P(S\cap F)}{P(F)}\\=\frac{0.44}{0.55}\\=0.80[/tex]

        [tex]P(S|F^{c})=\frac{P(S\cap F^{c})}{P(F^{c})}\\=\frac{0.13}{0.45}\\=0.289[/tex]

        [tex]P(S^{c}|F)=\frac{P(S^{c}\cap F)}{P(F}\\=\frac{0.11}{0.55}\\=0.20[/tex]

       [tex]P(S^{c}|F^{c})=\frac{P(S^{c}\cap F^{c})}{P(F^{c})}\\=\frac{0.32}{0.45}\\=0.711[/tex]

The probability of successful product given favorable test market is 0.80, and the probability of successful product given unfavorable test market is 0.29. The probability of unsuccessful product given favorable test market is 0.20, and the probability of unsuccessful product given unfavorable test market is 0.71.

To solve this, we first need to determine the total probability of each market test outcome. The probability that the market test is favorable (P(favorable test market)) can be found by summing the probabilities of a successful product with a favorable test market and an unsuccessful product with a favorable test market. Therefore, P(favorable test market) = 0.44 + 0.11 = 0.55. Similarly, P(unfavorable test market) = 0.13 + 0.32 = 0.45.

To find the conditional probabilities, we use the formula P(A|B) = P(A and B) / P(B):

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Answer:

a) So 25% probability that exactly two heads occur, given that the first outcome was a tail.

b) 50% probability that exactly two heads occur, given that the first two outcomes were heads.

c) 0% probability that exactly two heads occur, given that the first two outcomes were tails.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

We have the following sample space, that is, the possible outcomes:

In which h is heads and t is tails

h - h - h

h - h - t

h - t - h

h - t - t

t - h - h

t - h - t

t - t - h

t - t - t

What is the probability that exactly two heads occur, given that

a. the first outcome was a tail

Four outcomes in which the first outcome was a tail. They are:

t - h - h

t - h - t

t - t - h

t - t - t

In only 1 one them, exactly two heads occur.

1/4 = 0.25

So 25% probability that exactly two heads occur, given that the first outcome was a tail.

b. the first two outcomes were heads

Two possibilities in which the first two outcomes were heads.

h - h - h

h - h - t

In 1 of them, we have exactly two heads.

1/2 = 0.5

So 50% probability that exactly two heads occur, given that the first two outcomes were heads.

c. the first two outcomes were tails

Two possibilities in which the first two outcomes were tails.

t - t - h

t - t - t

In none of them we have exactly 2 heads.

0% probability that exactly two heads occur, given that the first two outcomes were tails.

The probability of getting exactly two heads in three coin tosses is 5/8.

To find the probability of exactly two heads occurring in three coin tosses, we need to consider three different scenarios:

If the first outcome was a tail: In this case, we have two remaining tosses and we need both of them to be heads. The probability of getting a head on any single toss is 1/2, so the probability of getting two heads after a tail is (1/2) * (1/2) = 1/4.

If the first two outcomes were heads: In this case, we have one remaining toss and we need it to be a tail. The probability of getting a tail on the remaining toss is 1/2, so the probability of getting exactly two heads after two heads is (1/2) = 1/2.

If the first two outcomes were tails: In this case, we again have one remaining toss and we need it to be a head. The probability of getting a head on the remaining toss is 1/2, so the probability of getting exactly two heads after two tails is (1/2) = 1/2.

Adding up the probabilities of these three scenarios, the overall probability of getting exactly two heads in three coin tosses is 1/4 + 1/2 + 1/2 = 5/8.

Answer:

a) Selling price y= a + b (age x)

b)

a= 728.025

b= -38.217

c)

Selling price y = 728.025 - 38.217 age x

d)

SSE=8280.25

Step-by-step explanation:

a)

The regression model can be written as

y=a+bx

Here y=selling price and x is age.

So, the regression model will be

Selling price y= a + b (age x)

b)

We have to find the values of "a" and "b"

[tex]b=\frac{nsumxy-(sumx)(sumy)}{nsumx^{2} -(sumx)^2}[/tex]

sumx=5+10+12+14+15=56

sumy=500+400+300+200+100=1500

sumxy=5*500+10*400+12*300+14*200+15*100=14400

sumx²=5²+10²+12²+14²+15²=690

n=5

[tex]b=\frac{5(14400)-(56)(1500)}{5(690) -(56)^2}[/tex]

b=-12000/314

b=-38.217

ybar=a+bxbar

a=ybar-bxbar

ybar=sumy/n=1500/5=300

xbar=sumx/n=56/5=11.2

a=300-(-38.217)(11.2)

a=300+428.025

a=728.025

c)

Selling price y = a - b(age x)

Selling price y = 728.025 - 38.217 age x

d)

SSE known as sum of square of error can be calculated as

SSE=sum(y-yhat)²

y  500 400 300 200 100

x   5      10    12     14    15

yhat= 728.025 - 38.217 age x 536.940  345.855 269.421  192.987  154.770

y-yhat  -36.940  54.145  30.579 7.013  -54.770

(y-yhat)² 1364.56  2931.68  935.08  49.18  2999.75

SSE=sum(y-yhat)²

SSE=1364.56 +2931.68 +935.08 +49.18 +2999.75

SSE =8280.25

Answer:

[tex] A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196 [/tex]

Step-by-step explanation:

For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by [tex] V= 10 +t[/tex]

Since the initial volume is 10 L and the volume increase at a rate of 1L/min.

For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:

[tex] \frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}[/tex]

Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of [tex] \frac{A}{10+t}[/tex]

So then we can reorder the differential equation like this:

[tex] \frac{dA}{dt} +\frac{A}{10+t} =8[/tex]

We find the solution using the integration factor:

[tex] \mu = -\int \frac{1}{10+t} dt = -ln(10+t)[/tex]

And then the solution would be given by:

[tex] A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})[/tex]

And if we simplify this we got:

[tex] A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)[/tex]

And after do the integral we got:

[tex] A= \frac{c}{10+t} +4 (10 +t) [/tex]

And using the initial condition t=0 A= 20 we have this:

[tex] 20 = \frac{c}{10} +40[/tex]

[tex] c= -200[/tex]

So then we have this function for the solution of A:

[tex] A= \frac{-200}{10+t} +4 (10 +t) [/tex]

And now replacinf t= 40 we got:

[tex] A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196 [/tex]

The amount of salt in the tank after 40 minutes is determined by first calculating the total amount of salt added to the tank, then considering the amount of salt-water mixture that exited the tank during this period due to the outgoing flow.

This question relates to a typical problem within the field of differential equations. You are asked to find the amount of salt in the tank after 40 minutes. The initial amount of salt in the tank is 20g, and additional salt-water solution is being pumped in at a rate of 2L/min with a salt concentration of 4g/L, hence adding 8g of salt per minute. At the same time, the well-stirred mixture is flowing out at a rate of 1L/min.

Therefore, after 40 minutes:
Additional salt added = 8g/min * 40 min = 320g.
Total salt content now = initial salt content + added salt = 20g + 320g = 340g.

However, within these 40 minutes,40 L of the mixture has flown out. Since the mixture is well-stirred, it retains the same concentration of salt as the mixture in the tank. Hence, if x is the amount of salt left in the tank, it will be on average the same concentration as that which flowed out and can be represented as follows: x/50 = (340-x)/40.

Solving for x we get the amount of salt left in the tank after 40 minutes.

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Answer:

No solution.

Step-by-step explanation:

A system has only one solution if it ends at ax = b, in which a is different than 0.

If it ends at 0x = 0, the system has infinitely many solutions.

If it ends at a division by 0, or 0 = constant(different than 0), the system is inconsistent.

Our system is:

−20x + 30y = −3

8x − 12y = −3

I am going to multiply the top equations by 2 and the bottom equation by 5, and add them. So

-40x + 60y = -6

40x - 60y = -15

So

-40x + 40x + 60y - 60y = -6 - 15

0x + 0y = -21

We cannot divide 21 by 0, which means that this system of equations has no solution.

Answer:

a) [tex]P(F) = \frac{672}{1254}=\frac{112}{209}=0.536[/tex]

b) [tex]P(M) = \frac{582}{1254}= \frac{97}{209}=0.464[/tex]

c) [tex] P(A') = 1-P(A) = 1- \frac{124}{672}= \frac{137}{168}=0.815[/tex]

Step-by-step explanation:

For this case we have a total of 1254 people. 672 are women and 582 are female.

We know that 124 women wnat on to graduate school.

And 198 male want on to graduate school

We can define the following events:

F = The alumnus selected is female

M= The alumnus selected is male

A= Female and attend graduate school

And we can find the probabilities required using the empirical definition of probability like this:

Part a

[tex]P(F) = \frac{672}{1254}=\frac{112}{209}=0.536[/tex]

Part b

[tex]P(M) = \frac{582}{1254}= \frac{97}{209}=0.464[/tex]

Part c

For this case we find the probability for the event A: The student selected is female and did attend graduate school

[tex] P(A) =\frac{124}{672}=\frac{31}{168}=0.185[/tex]

And using the complement rule we find P(A') representing the probability that the female selected did not attend graduate school like this:

[tex] P(A') = 1-P(A) = 1- \frac{124}{672}= \frac{137}{168}=0.815[/tex]

Answer:

There were 28.75 large boxes and 96.25 small boxes.

Step-by-step explanation:

Create a system of equations to solve.

State your variables

let x be the number of small boxes

let y be the number of large boxes

35x + 55y = 4950               Equation for weight

x + y = 125                          Equation for number of boxes

Rearrange equation for number of boxes to isolate "x".

x = 125 - y

Substitute the new equation into the equation for weight

35x + 55y = 4950

35(125 - y) + 55y = 4950

Expand the brackets

4375 - 35y + 55y = 4950       Combine like terms

4375 + 20y = 4950

Start isolating "y"

4375 + 20y = 4950

4375 - 4375 + 20y = 4950 - 4375         subtract 4375 from both sides

20y = 575

20y/20 = 575/20                         divide both sides by 20

y = 28.75                   number of large boxes

Calculate "x". Rearrange the equation for number of boxes to isolate "y".

y = 125 - x                    Substitute this expression

35x + 55y = 4950                          equation for weight

35x + 55(125 - x) = 4950                 expand the brackets

35x + 6875 - 55x = 4950                     combine like terms

6875 - 20x = 4950                              start isolating "x"

6875 - 6875 - 20x = 4950 - 6875            subtract 6875 on both sides

-20x = -1925

-20x/-20 = -1925/-20                     divide both sides by -20

x = 96.25                    number of small boxes

There is a mistake with this question because you should not be able to have partial boxes. However, the answer is:

There were 28.75 large boxes and 96.25 small boxes.

Answer:

[tex]45^{\circ}[/tex]

Step-by-step explanation:

We are given that two lines equation

[tex]x+\sqrt 3y=1[/tex]...(1)

[tex](1-\sqrt 3)y+(1+\sqrt 3)y=8[/tex]

Compare with the equation of line

ax+by+c=0

[tex]a_1=1,b_1=\sqrt 3[/tex]

[tex]a_2=(1-\sqrt 3),b_2=(1+\sqrt 3)[/tex]

The angle between two lines =Angle between two vectors

The angle between two vector

[tex]a_1i+b_1j[/tex] and

[tex]a_2i+b_2j[/tex]

is given by

[tex]cos\theta=\frac{a_1a_2+b_1b_2}{\sqrt{a^2_1+b^2_1}\sqrt{a^2_2+b^2_2}}[/tex]

Using the formula

Therefore, the angle between two lines

[tex]cos\theta=\frac{1(1-\sqrt 3)+\sqrt 3(1+\sqrt 3)}{\sqrt{(1)+(\sqrt 3)^2}\times \sqrt{(1-\sqrt 3)^2+(1+\sqrt 3)^2}}[/tex]

[tex]cos\theta=\frac{1-\sqrt 3+\sqrt 3+3}{\sqrt{1+3}\times\sqrt{1+3-2\sqrt 3+1+3+2\sqrt 3}}[/tex]

[tex]cos\theta=\frac{4}{2\times\sqrt 8}=\frac{2}{2\sqrt 2}[/tex]

[tex]cos\theta=\frac{1}{\sqrt 2}[/tex]

[tex]cos\theta=cos45^{\circ}[/tex]

By using [tex]cos45^{\circ}=\frac{1}{\sqrt 2}[/tex]

[tex]\theta=45^{\circ}[/tex]

Hence, the angle between two lines =45 degree

Answer:

parameter tested=σ(population standard deviation)

two tailed

Step-by-step explanation:

We are given that null hypothesis that σ is equal to 88 and alternative hypothesis that σ is not equal to 88. The parameter that is tested here is σ.

σ denotes the population standard deviation. Thus, the population parameter σ is tested here.

The type of test depends on the alternative hypothesis we have taken. The alternative hypothesis states that σ is not equal to 88. It means that σ can either be greater than 88 or less than 88. Thus, the test can either be right tailed or left tailed. This type of test is called as two tailed test.

Answer:

60th percentile.

Step-by-step explanation:

When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.

In this problem, we have that:

300 observations, there are 180 points less than 81.2.

180/300 = 0.6

So 60% of the observation are lower than 81.2, which means that 81.2 is the 60th percentile.

Answer:

a) see attached

b) see attached

Step-by-step explanation:

There is existence of outlier in the Ni data and there is none is Cr data.

########################################

# You can try this out in R programming

cr = c(34, 1, 511, 2, 574, 496, 322, 424, 269, 140, 244, 252, 76, 108, 24,

38, 18, 34, 30, 191)

Ni = c(23, 22, 55, 39, 283, 34, 159, 37, 61, 34, 163, 140, 32,

23, 54, 837, 64, 354, 376, 471)

par(mfrow=c(1,2))

hist(cr, col='green')

hist(Ni, col='brown')

par(mfrow=c(1,2))

boxplot(cr, main = 'Boxplot of Cr')

boxplot(Ni, main = 'Boxplot of Ni')

boxplot(cr, Ni)

The question involves constructing histograms and boxplots for two sets of data on soil concentrations of certain metals. After construction, these should be analysed for differences in central tendency, spread, variability, skewness, or presence of outliers.

This question seems to be related to Mathematics, specifically Statistics and Data Analysis. Firstly, to construct a histogram, you need to group the data into 'bins' or 'intervals', count how many data points fall into each bin, and then plot these counts as bars in your histogram. It's similar for both Chromium (Cr) and Nickel (Ni) data.

To construct comparative boxplots, it's essential to: calculate the quartiles (Q1, Q2 aka median, and Q3), identify the minimum and maximum points, and define possible outliers. With these statistical measures, you can draw the boxplots for both sets and analyse them. The dataset Cr seems to have several values clustered in the low range while Ni seems to have more spread, which can be verified by the boxplot comparison.

The interpretation of the boxplots should reflect on any differences in central tendency and spread between the two sets of concentrations. The variability, skewness, or outliers could also be significantly different. Any such inferred details from the boxplots may hint towards different contamination levels or heterogeneity in each of the datasets.

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Answer:

[tex]1332.1 \:cm^3[/tex] is neither significantly low nor significantly high.

Step-by-step explanation:

From the information given, we know that:

The mean is [tex]\bar{x}=1105.1 \:cm^{3}[/tex] and the standard deviation is [tex]\sigma=123.5 \:cm^{3}[/tex].

The range rule of thumb says that:

[tex]minimum \:usual \:value=\bar{x}-2\sigma\\\\maximum \:usual \:value=\bar{x}+2\sigma[/tex]

Applying these definitions, we get that

[tex]minimum \:usual \:value=1105.1-2(123.5)=858.1 \:cm^3\\\\maximum \:usual \:value=1105.1+2(123.5)=1352.1 \:cm^3[/tex]

We note that [tex]1332.1 \:cm^3[/tex] is between [tex]858.1 \:cm^3[/tex] and [tex]1352.1 \:cm^3[/tex], which indicates  [tex]1332.1 \:cm^3[/tex] is neither significantly low nor significantly high.