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This is about CIRCLES AND POLYGONS!

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: 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:

There is a 0.57% probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").

Step-by-step explanation:

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

In this problem, we have that:

Desired outcomes:

The number of male nannies selected. 24 of the nannies placed were men. So the number of desired outcomes is 24.

Total outcomes:

The number of nannies selected. 4,176 nannies were placed in a job in a given year. So the number of total outcomes 4176.

Find the probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").

[tex]P = \frac{24}{4176} = 0.0057[/tex]

There is a 0.57% probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").

Final answer:

The probability that a randomly selected nanny is a male is 0.57%.

Explanation:

The question asks us to find the probability that a randomly selected nanny who was placed during the last year is a male nanny, also referred to humorously as a "mannie." We are given that a total of 4,176 nannies were placed, and of these, only 24 were men. To find the probability, we use the formula for probability, which is the number of favorable outcomes divided by the total number of possible outcomes.

Here, the number of favorable outcomes is the number of male nannies placed, which is 24. The total number of possible outcomes is the total number of nannies placed, which is 4,176.

Thus, the probability (P) that a selected nanny is male is:

P(male nanny) = Number of male nannies / Total number of nannies

P(male nanny) = 24 / 4,176

P(male nanny) = 0.0057

To express this as a percentage, we multiply by 100:

P(male nanny) = 0.0057 * 100%

P(male nanny) = 0.57%

Therefore, the probability that a randomly selected nanny is a male is 0.57%.

Answer:

The sample space is:

[tex]AB,\ AC,\ AD,\ AE,\\BC,\ BD,\ BE\\CD,\ CE\\DE[/tex]

Step-by-step explanation:

The sample space of an experiment is a set of all the possible values that satisfies that experiment.

The five supervisors are A, B, C, D and E.

Two are to selected.

The number of ways to select 2 supervisors  from 5 is: [tex]{5\choose 2}=\frac{5!}{2!(5-2)!} =\frac{5!}{2!\times 3!} = 10[/tex]

The sample space will consist of 10 combinations.

The sample space is:

[tex]AB,\ AC,\ AD,\ AE,\\BC,\ BD,\ BE\\CD,\ CE\\DE[/tex]

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:

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:

the union of the four sets have 1094 elements

Step-by-step explanation:

denoting as N as the number of elements, since

N(A U B) =   N(A) + N(B) - N(A ∩ B)

then

N(A U B U C) =  N(A U B) + N(C) - N(A U B ∩ C ) = N(A) + N(B) - N(A ∩ B) + N(C)  - [ N(A∩C) +  N(B ∩ C ) - N(A ∩ B ∩ C )]

then for the union of 4 sets , we have

N (A U B U C U D) = N(A) + N(B) + N(C) +N(D) - N(A ∩ B) - N(A ∩ C) - N(A ∩ D)- N(B ∩ C) - N(B ∩ D) - N(C ∩ D) + N(A ∩ B ∩ C) + N(A ∩ B ∩ D) + N(A ∩ C ∩ D) + N(B ∩ C ∩ D) - N(A ∩ B ∩ C ∩ D)

thus replacing values for the sets, union of 2 sets , union of 3 sets and union of 4 sets

N (A U B U C U D) = ( 4*400 ) - ( 6*115 ) + ( 4*53 ) - 28 =  1094 elements

then the union of the four sets have 1094 elements

To determine the Cartesian components of a three-dimensional vector, one typically needs a minimum of two angles. These include the angle in the XY direction and the 'altitude' angle from the XY plane to the vector.

To determine the Cartesian components of a three-dimensional vector, a minimum of "two angles" is required. Unlike two-dimensional vectors that need just one angle for direction, 3D vectors are more complex, existing within an x, y, and z Cartesian coordinate system. Therefore, to describe a vector fully in a 3D space, we need direction in the XY plane (measured from the positive x-axis counterclockwise), hence angle one, and the direction from the XY plane to the vector (its 'altitude'), hence angle two.

Let's consider a 3D vector A. The magnitude A and the direction angles can be used to find the components Ax, Ay, Az. Trigonometry helps us here: Ax=Acos(θ), Ay=Asin(φ), and Az=Asin(φ), where θ is the angle with the x-axis in the XY plane, and φ is the angle with the XY plane.

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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:

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) 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:

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:

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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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:

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:

A) Private colleges:

mean = 42.7, SD = 6.72

In thousand dollars:

mean =  $42700, SD= $6720

B) Public colleges:

mean = 22.3, SD = 4.53

In thousand dollars:

mean =  $22300, SD= $4530

Step-by-step explanation:

A) Private Colleges:

Mean:

Total no. of samples = n =10

Sample values in dollar = x= [53.8, 42.2, 44.0, 34.3, 44.0,31.6, 45.8, 38.8, 50.5, 42.0]

Sum of samples = ∑x= 427

[tex]sample\,\,mean = \bar{x} =\frac{\sum x}{n}\\\\\bar{x}=\frac{427}{10}\\\\\bar{x}=42.7\\[/tex]

Sample mean in thousand dollars is $ 42700.

Standard Deviation:

Formula for standard deviation of sample data is

[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_{i}-\bar{x})^2}\\\\\sum(x_i-\bar{x})^2=(53.8-42.7)^2+(42.2-42.7)^2+...+(42.0-42.7)^2\\\\\sum(x_i-\bar{x})^2=123.21+0.25+ 1.69+ 70.56+ 1.69+ 123.21+ 9.61+ 15.21+60.84+0.49\\\\\sum(x_i-\bar{x})^2=406.67\\\\\sigma=\sqrt{\frac{406.67}{9}}\\\\\sigma=6.72[/tex]

Standard deviation in thousand dollars is $ 6720.

B) Public Colleges:

Mean:

Total no. of samples = n =12

Sample values in dollar = x= [20.3, 22.0, 28.2, 15.6, 24.1, 28.5,22.8, 25.8, 18.5, 25.6, 14.4, 21.8]

Sum of samples = ∑x= 267.6

[tex]sample\,\,mean = \bar{x} =\frac{\sum x}{n}\\\\\bar{x}=\frac{267.6}{12}\\\\\bar{x}=22.3\\[/tex]

Sample mean in thousand dollars is $ 22300.

Standard Deviation:

[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_{i}-\bar{x})^2}\\\\\sum(x_i-\bar{x})^2=(20.3-22.3)^2+(22.0-22.3)^2+...+(21.8-22.3)^2\\\\\sum(x_i-\bar{x})^2=4+ 0.09+34.81+ 44.89+ 3.24+ 38.44+0.25+ 12.25+14.44+10.89+62.41+0.25\\\\\sum(x_i-\bar{x})^2=225.96\\\\\sigma=\sqrt{\frac{225.96}{11}}\\\\\sigma=4.53[/tex]

Standard deviation in thousand dollars is $ 4530.

Answer:

There is not enough evidence to support the claim that interested consumers would be willing, on average, to pay $20.00 for the product.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $20.00

Sample mean, [tex]\bar{x}[/tex] = $18.14

Sample size, n = 315

Population standard deviation, σ = $2.98

a) Reference value

[tex]\mu = 20[/tex]

b) First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 20.00\text{ dollars}\\H_A: \mu \neq 20.00\text{ dollars}[/tex]

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

Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{18.14 - 20.00}{\frac{2.98}{\sqrt{315}} } = -11.078[/tex]

c) Calculating the p-value at 5% significance level

P-Value < 0.00001

Thus,

p<0.001

d) Calculating the p-value at 1% level of significance

P-Value < 0.00001

Thus,

p<0.001

Thus, at both 5% and 1% level of significance, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it.

There is not enough evidence to support the claim that interested consumers would be willing, on average, to pay $20.00 for the product.

Answer:

a) There are 126 different ways are there to select four of the nine computers to be set up.

b) 31.75% probability that exactly three of the selected computers are desktops.

c) 35.71% probability that at least three desktops are selected.

Step-by-step explanation:

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

In this problem, there are no replacements. Which means that after one of the 9 computers is selected, there will be 8 computers.

Also, the order that the computers are selected is not important. For example, desktop A and desktop B is the same outcome as desktop B and desktop A.

These are the two reasons why the combinations formula is important to solve this problem.

Combinations formula:

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

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

(a) How many different ways are there to select four of the nine computers to be set up?

Four computers are selected from a set of 9.

So

[tex]T = C_{9,4} = \frac{9!}{4!(9-4)!} = 126[/tex]

There are 126 different ways are there to select four of the nine computers to be set up.

(b) What is the probability that exactly three of the selected computers are desktops?

Desired outcomes:

3 desktops, from a set of 5

One laptop, from a set of 4.

So

[tex]D = C_{5,3}*C_{4,1} = 40[/tex]

Total outcomes:

From a), 126

Probability:

[tex]P = \frac{40}{126} = 0.3175[/tex]

31.75% probability that exactly three of the selected computers are desktops.

(c) What is the probability that at least three desktops are selected?

Three or four

Three:

[tex]P = \frac{40}{126}[/tex]

Four:

Desired outcomes:

4 desktops, from a set of 5

Zero laptop, from a set of 4.

[tex]D = C_{5,4}*C_{4,0} = 5[/tex]

[tex]P = \frac{5}{126}[/tex]

Total(three or four) probability:

[tex]P = \frac{40}{126} + \frac{5}{126} = \frac{45}{126} = 0.3571[/tex]

35.71% probability that at least three desktops are selected.

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:

[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:

a) Observational study

b) Population of drivers in Louisiana who were involved in an accident as a result of talking on a cell phone while driving.

c) It is quantitative

d) It is discrete

e) The sample of the reports of all drivers who uses phone while driving and were involved in an accidents in the parish where the researcher lives only.

Step-by-step explanation:

a) The study is not done in a controlled environment. It happens randomly. As we must know not all drivers who use phone while driving have accidents. So, it is observational study.

b) We are told in the question the population of interest.

c) Since we are interested in the number of accidents that happens within the population of interest. It is a count data and therefore, it is quantitative.

d) It is count data, thus it is discrete. That is, it cannot take a decimal point. It must be whole number.

e) The kind of sample taken must be from the population of interest.

The study in question is an observational study examining the relationship between cell phone use while driving and accident incidence in Louisiana. It focuses on a quantitative, discrete variable - the count of accidents involving cell phone use - and may employ a convenience sample if only local reports are analyzed.

To answer the question of whether using a cell phone while driving increases the risk of an accident in Louisiana, the researcher is conducting an observational study since they are examining existing data and are not manipulating any variables or conducting a randomized experiment.

The population being studied in this research is all drivers in Louisiana, and specifically, those who have been involved in accidents. When examining the number of accidents where the driver was using a cell phone, we are dealing with a quantitative variable since it is a countable number of accidents.

This quantitative variable would be classified as a discrete variable because the number of accidents can only be expressed in whole numbers; a driver cannot be involved in a fraction of an accident.

If the researcher is only examining accident reports in the parish where they live, the sample taken is a convenience sample because it is not randomly selected and might not be representative of the entire population of Louisiana drivers.

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:

$7,000,000

Step-by-step explanation:

The taxable income is the amount of money (income earned or unearned) by an individual or an organization that creates a potential tax liability.

The formula is shown below:

Taxable Income Formula = Gross Total Income – Total Exemptions – Total Deductions

Note that $5,000,000 for 2015 is an operating loss which is a deduction.

Gross Total Income = $4000000+$4000000+$4000000 = $12,000,000

Total Exemption = 0

Total Deductions = $5,000,000

Taxable Income Formula = $12,000,000 - $5,000,000 = $7,000,000

The effective taxable income for 2017 after carrying forward the operating loss is -$1 million.

The question pertains to a business scenario and involves calculating the effective taxable income after carrying forward an operating loss. To calculate the effective taxable income for 2017, we need to consider the operating losses from previous years. In this case, your firm had a $5 million operating loss in 2015. This loss can be carried forward to offset taxable income in future years.

Therefore, the effective taxable income for 2017 after carrying forward the operating loss is -$1 million.

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Answer:

                                 Case a                               Case b

margin of error       0.0216                                   0.0231

Interval estimate   (0.7016 , 0.6795)                (0.5031 , 0.4569)

margin of error is not same in both cases.

Step-by-step explanation:

a

At 95% confidence interval the interval estimate of number of 20 year old drivers in year A can be computed as

  p'  ±  z  [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

= 0.68 ± 1.96 [tex]\sqrt{\frac{0.68(1-0.68)}{1800} }[/tex]

= 0.7016 , 0.6795

the margin of error can be written as

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

= 1.96 [tex]\sqrt{\frac{0.68(1-0.68)}{1800} }[/tex]

= 0.0216

b

At 95% confidence interval the interval estimate of number of 20 year old drivers in year B can be computed as

p'  ±  z  [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

= 0.48 ± 1.96 [tex]\sqrt{\frac{0.48(1-0.48)}{1800} }[/tex]

=  0.5031 , 0.4569

the margin of error can be written as

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

=  1.96 [tex]\sqrt{\frac{0.48(1-0.48)}{1800} }[/tex]

= 0.0231

c

Sample size is same in case A and B but proportion is different in both cases so margin of error is different in both cases

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:

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:

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:

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.

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:

2,929,687 pictures

Step-by-step explanation:

1 pixel requires 2 bytes of storage

A color digital photo uses 512 pixels (512×2 bytes = 1024 bytes) per row and 512 pixels (1024) bytes per column

Total storage used by 1 picture = 1024 + 1024 = 2048 bytes

Storage capacity my camera = 6GB = 6×10^9 bytes

Number of pictures I can have in my camera = 6×10^9/2048 = 2,929,687 pictures

By calculating the storage required for one 512 x 512 pixel photo and dividing the total storage capacity by this amount, we find that a camera with 6 GB of storage can hold approximately 12,288 such photos.

To calculate how many 512 x 512 pixel color digital photos can be stored in a camera with 6 GB of storage, where each pixel requires two bytes of storage, we follow these steps:

First, find the total number of pixels in one photo:

512 pixels/row × 512 pixels/column = 262,144 pixels/photo.

Next, calculate the storage required for one photo:

262,144 pixels/photo × 2 bytes/pixel = 524,288 bytes/photo.

Since there are 1,024 bytes in one kilobyte (KB) and 1,024 KB in one megabyte (MB), we convert the photo size to megabytes:

524,288 bytes/photo / 1,024 bytes/KB / 1,024 KB/MB
≈ 0.5 MB/photo.

Now, convert 6 GB of storage to MB:

6 GB × 1,024 MB/GB = 6,144 MB.

Finally, divide the total storage by the size of one photo to determine the number of photos:

6,144 MB / 0.5 MB/photo = 12,288 photos.

Therefore, you could store approximately 12,288 such pictures on a camera with 6 GB of storage.