Samâs Auto Shop services and repairs a particular brand of foreign automobile. Sam uses oil filters throughout the year. The shop operates fifty-two weeks per year and weekly demand is 150 filters. Sam estimates that it costs $20 to place an order and his annual holding cost rate is $3 per oil filter. Currently, Sam orders in quantities of 650 filters. Calculate the total annual costs associated with Samâs current ordering policy

Total annual costs = $

Answer:

has only one solution: [tex]x=8, y=3[/tex]

Step-by-step explanation:

[tex]\left \{ {{\frac{5x}{4} -\frac{2y}{3}=8(equation 1)} \atop {\frac{x}{4} +\frac{5y}{3} =7(equation 2)}} \right\\[/tex]

substract  [tex]\frac{1}{5}[/tex]×(equation 1) from equation2

[tex]\left \{ {{\frac{5x}{4} -\frac{2y}{3}=8(equation 1)} \atop {0x +\frac{9y}{5} =\frac{27}{5}(equation 2)}} \right\\[/tex]

from equation 2 we obtain

[tex]y=\frac{27}{9} =3[/tex]

and we replace 3 in equation 1

[tex]\frac{5x}{4} -\frac{2(3)}{3} =8\\\\\frac{5x}{4} =10\\\\x=\frac{40}{5} =8[/tex]

Answer: 0.7264

Explanation:

The number of independent questions (n)  = 100

Probability of answering a question (p) = 0.80

Let X be the no. of questions that need to be answered.

[tex]\therefore[/tex] random variable X follows binomial distribution

The probability function of a binomial distribution is given as

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

Now , we nee to find P(74 ≤ X ≤ 84)

[tex]\therefore P(74\leq X\leq 84) = P(X=74) + P(X=75).........+ P(X=84)[/tex]

P(74 ≤ X ≤ 84) = [tex]\sum_{74}^{84}\binom{100}{x}\times (0.80)^{x}(0.20)^{100-x}[/tex]

P(74 ≤ X ≤ 84) = 0.7264

Answer: 0.1052

Step-by-step explanation:

Given : Mean :[tex]\mu= 120[/tex]

Standard deviation : [tex]\sigma= 5.6[/tex]

We assume the variable is normally distributed.

The formula for z-score is given by :-

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

For x=120.4

[tex]z=\dfrac{120.4-120}{5.6}=0.0714285714286\approx0.07[/tex]

For x=121.9

[tex]z=\dfrac{121.9-120}{5.6}=0.339285714286\approx0.34[/tex]

The p-value =[tex]P(0.07<z<0.34)=P(0.34)-P(0.07)[/tex]

[tex]=0.6330717-0.5279031=0.1051686\approx0.1052[/tex]

The probability that the individual's pressure will be between 120.4 and 121.9 mm Hg = 0.1052

Answer:

C (0, 9)

Step-by-step explanation:

First, X is reflected across the line x=2.

X(4, -7) becomes X'(0, -7).

Next, X' is reflected across the line y=1.

X'(0, -7) becomes X"(0, 9).

Answer: The mean number of checks written per​ day = 0.3260

The standard deviation = 0.5710

The variance = 0.3260

Step-by-step explanation:

Given : The number of checks written by the person = 119

We assume that the year is not a leap year.

Thus, the number of days in the year must be 365.

Now, the mean number of checks written per​ day is given by :-

[tex]\lambda=\dfrac{119}{365}=0.3260273972\approx0.3260[/tex]

Also, in Poisson distribution , the variance is also equals to the mean value .

[tex]\text{Thus , Variance }=\sigma^2= 0.3260[/tex]

Then , [tex]\sigma= \sqrt{0.3260}=0.570964096945\approx0.5710[/tex]

Thus,  Standard deviation = 0.5710

Answer:

The commission would be $ 920.

Step-by-step explanation:

Given,

Sales in 1997 ​= $ 9500,

Since, there is a commision of 8 % on the first $ 7500 of​ sales, 16% on the next $ 7500 of​ sales, and 20% on sales over $ 15,000,

If x represents the total sales,

Commission for x ≤ 7500 = 8% of x = 0.08x,

Commission for 7500 < x ≤ 15000 = 8% of 7500 + 16% of (x-7500)

= 600 + 0.16(x-7500),

Commission for 7500 < x ≤ 15000 = 8% of 7500 + 16% of 7500 + 20% of (x-15000)

= 600 + 1200 + 0.20(x-15000)

= 1800 + 0.20(x-15000)

Thus, the function that shows the given situation,

[tex]f(x)=\left\{\begin{matrix}0.08x & x \leq 7500\\ 600+0.16(x-7500) & 7500 < x \leq 15000\\ 1800+0.20(x-15000) & 15000 < x\end{matrix}\right.[/tex]

Since, 9500 lies between 7500 and 15000,

Therefore, the commission would be,

f(9500)= 600 + 0.16(9500 - 7500) = 600 + 320 = $ 920

Shelby's gross commission for July 1997 on sales of $9500 is calculated as  $920.

Kelly Price sells college textbooks on commission, and the structure of her commission is a tiered rate: 8% on the first $7500 of sales, 16% on the next $7500 of sales, and 20% on sales over $15,000.

For Shelby’s sales totaling $9500 in July 1997, we must calculate her gross commission for that month.

Here is the step-by-step calculation:

Therefore, Shelby’s gross commission for July 1997 is $920.

Recall that for [tex]|z|<1[/tex], we have

[tex]\dfrac1{1-z}=\displaystyle\sum_{n\ge0}z^n[/tex]

Then

[tex]\dfrac1{1-z}=-\dfrac1z\dfrac1{1-\frac1z}=-\dfrac1z\displaystyle\sum_{n\ge0}z^{-n}=-\sum_{n\ge0}z^{-n-1}[/tex]

valid for [tex]|z|>1[/tex], so that

[tex]\dfrac1{1-z^2}=-\dfrac1{z^2}\dfrac1{1-\frac1{z^2}}=-\dfrac1{z^2}\displaystyle\sum_{n\ge0}z^{-2n}=-\sum_{n\ge0}z^{-2n-2}[/tex]

also valid for [tex]|z|>1[/tex].

Answer:

The answer is Parallel-forms reliability.

Parallel-forms reliability describes a measure used to compare two different tests with the same group of participants to see how closely correlated the two sets of scores are with each other.

Test-retest reliability is a measure used to compare the results of two instances of the same test taken by the same group of participants to establish how closely correlated the two sets of scores are.

The measure used to compare two different tests with the same group of participants to see how closely correlated the two sets of scores are with each other is called test-retest reliability.

In essence, test-retest reliability measures the consistency of results when a test is administered more than once. The higher the correlation between the two sets of scores, the higher the test’s reliability. For example, if a group of students were to take a vocabulary test one week, and then the same test a week later, a high correlation between the results would suggest that the test is reliable.

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[tex]x^2-6x + 8=x^2-6x+9-1=(x-3)^2-1[/tex]

Answer:

(x - 3)^2 - 1.

Step-by-step explanation:

x2 – 6x + 8.

We divide the coefficient of x by 2:

-6 / 2 = -3 so the contents of the parentheses is x - 3:

x^2 - 6x + 8

= (x - 3)^2 - (-3)^2 + 8

= (x - 3)^2 - 1.

Answer:

Length of B is 7.4833

Step-by-step explanation:

The vector sum of A and B vectors in 2D is

[tex]C=A+B=(a_1+b_1,a_2+b_2)[/tex]

And its magnitude is:

[tex]C=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2} =9[/tex]

Where

[tex]a_1=Asinx[/tex]

[tex]a_2=Acosx[/tex]

[tex]b_1=Bsin(x+90)[/tex]

[tex]b_2=Bcos(x+90)[/tex]

Using the properties of the sum of two angles in the sin and cosine:

[tex]b_1=Bsin(x+90)=B(sinx*cos90+sin90*cosx)=Bcosx[/tex]

[tex]b_2=Bcos(x+90)=B(cosx*cos90-sinx*sin90)=-Bsinx[/tex]

Sustituying in the magnitud of the sum

[tex]C=\sqrt{(Asinx+Bcosx)^2+(Acosx-Bsinx)^2} =9[/tex]

[tex]C=\sqrt{A^2sin^2x+2ABsinxcosx+B^2cos^2x+A^2cos^2x-2ABsinxcosx+B^2sin^2x} =9[/tex]

[tex]C=\sqrt{A^2(sin^2x+cos^2x)+B^2(cos^2x+sin^2x)}[/tex]

[tex]C=\sqrt{A^2+B^2} =9[/tex]

Solving for B

[tex]A^2+B^2 =9^2[/tex]

[tex]B^2 =9^2-A^2[/tex]

Sustituying the value of the magnitud of A

[tex]B^2=81-5^2=81-25=56[/tex]

[tex]B= 7. 4833[/tex]

The calculation reveals that vector B should be approximately 7.48 in length when magnitude of A s 5.00 and magnitude of their vector sum is 9.00.

The subject question asks for the length of vector B given that it is at right angles to vector A, whose magnitude is 5.00, to ensure that the magnitude of their vector sum is 9.00. To solve this problem, we can use the Pythagorean theorem because the vectors are at right angles to each other.

Let's designate the magnitude of vector B as |B|. According to the Pythagorean theorem for right-angled triangles, we have:
|A|² + |B|² = |A + B|²
So, if |A| = 5.00 and the resultant |A + B| = 9.00, we can substitute these values to find |B|:
5.00² + |B|² = 9.00²
|B|² = 9.00² - 5.00²
|B|² = 81.00 - 25.00
|B|² = 56.00

Therefore, |B| = √(56.00)

|B| = 7.48.

So the length of B should be approximately 7.48 to get a resultant magnitude of 9.00 when added to A.

To find maximum and minimum values using Lagrange multipliers, set up a Lagrangian function, apply the technique of setting partial derivatives to zero, and solve the resulting system of equations. This technique is applied to both given functions to solve for the variables and the multiplier.

The subject of this question relates to the application of Lagrange multipliers in mathematics, specifically to find the maximum and minimum values of functions with constraints.

Starting with the first function, we set up the Lagrangian function L = f(x, y) - λ(g(x, y) - c) where λ is the Lagrange multiplier, and g(x, y) is the constraint. So, we get L = 81x^2 + y^2 - λ(4x^2 + y^2 -9). By setting the partial derivatives of L to zero and solving that system of equations, we can find x, y and λ.

For the second function, we apply the same process, the Lagrangian function will be L = y^2 - 10z - λ(x^2 + y^2 + z^2 - 36). Setting partial derivatives equal to zero and solving the resulting system will provide x, y, z, and λ.

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The maximum for the first problem is 182.25 and the minimum is 9, while the maximum for the second problem is 61.

To solve these problems using Lagrange multipliers, we need to find the maximum and minimum values of the given functions subject to the provided constraints.

(i) Function f(x,y) = 81x^2 + y² with Constraint 4x² + y² = 9

→ Define the Lagrangian function: L(x, y, λ) = 81x² + y² + λ(9 - 4x² - y²).

Compute the partial derivatives and set them to zero:

→ [tex]L_x = 162x - 8x\lambda = 0[/tex]

→ [tex]L_y = 2y - 2y\lambda = 0[/tex]

→ [tex]L\lambda = 9 - 4x^2 - y^2 = 0[/tex]

Solve the system of equations:

→ 162x = 8xλ

⇒ λ = 20.25

→ 2y = 2yλ

⇒ λ = 1 (y ≠ 0)

But we need a consistent value of λ; thus, solve considering different cases (y = 0):

→ If y = 0, constraint becomes 4x² = 9

⇒ x = ±√(9/4)

      = ±3/2

→ If x = 0, constraint becomes y² = 9

⇒ y = ±3

Evaluate f(x, y) at these points:

→ f(3/2, 0) = 81 (3/2)²

                = 81 * 2.25

                = 182.25 (Maximum)

→ f(0, 3) = 3²

            = 9

(ii) Function f(x, y, z) = y² - 10z with Constraint x² + y² + z² = 36

Define the Lagrangian function: L(x, y, z, λ) = y² - 10z + λ(36 - x² - y² - z²).

Compute the partial derivatives and set them to zero:

→ [tex]L_x = -2x\lambda = 0[/tex]

⇒ x = 0

→ [tex]L_y = 2y - 2y\lambda = 0[/tex]

⇒ y = 0 or λ = 1

→ [tex]L_z = -10 - 2z\lambda = 0[/tex]

⇒ λ = -5/z

→ [tex]L\lambda = 36 - x^2 - y^2 - z^2 = 0[/tex]

Solve the system of equations considering constraints:

Substituting values, x = 0 and λ = 1, we need to solve for y and z:

→ 36 = y² + z²

→ λ = -5/z;

Thus z = -5

Solving y² + 25 = 36, y² = 11

⇒ y = ±√11

Evaluate f(x, y, z) at these points:

→ f(0, √11, -5) = (√11)² - 10(-5)

                     = 11 + 50              

                     = 61 (Maximum)

→ f(0, -√11, -5) = 61

Answer: 0.5

Step-by-step explanation:

Binomial probability formula :-

[tex]P(x)=^nC_x\ p^x(q)^{n-x}[/tex], where P(x) is the probability of getting success in x trials , n is the total trials and p is the probability of getting success in each trial.

Given : The probability that the adults follow more than one game = 0.30

Then , q= 1-p = 1-0.30=0.70

The number of adults surveyed : n= 15

Let X be represents the adults who follow more than one sport.

Then , the probability that fewer than 4 of them will say that football is their favorite sport,

[tex]P(X\leq4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)\\\\=^{15}C_{0}\ (0.30)^0(0.70)^{15}+^{15}C_{1}\ (0.30)^1(0.70)^{14}+^{15}C_{2}\ (0.30)^2(0.70)^{13}+^{15}C_{3}\ (0.30)^3(0.70)^{12}+^{15}C_{4}\ (0.30)^4(0.70)^{11}\\\\=(0.30)^0(0.70)^{15}+15(0.30)^1(0.70)^{14}+105(0.30)^2(0.70)^{13}+455(0.30)^3(0.70)^{12}+1365(0.30)^4(0.70)^{11}\\\\=0.515491059227\approx0.5[/tex]

Hence, the probability rounded to the nearest tenth that fewer than 4 of them will say that football is their favorite sport =0.5

one rupee coin be z

50 paisa coins=2z

25 paisa coins=4z

50/25=2 2z*2=4z

z+2x+4x=175

7z=175

7z/7=175/7

z=25

one rupee coin=25 value

50 paisa rupee coins=50 value

25 rupee coin=100 value

50/25=2 50*2=100

25+50+100=175

Answer is b) Rs 175. I think this is the answer.

Answer:

discount= $94.5

Proceeds=$2605.5

Step-by-step explanation:

amount Dawin has to pay= $2700

in 6 months time

at a discount rate of 7%

here future value = $2700

D= discount before m months is given by

[tex]D= \frac{FV\times r\times m}{1200}[/tex]

m= 6 months, r= 7% and FV= 2700

putting values

[tex]D= \frac{2700\times 7\times 6}{1200}[/tex]

solving we get

D= $94.5

now present value = future value - discount

=2700-94.5= $ 2605.5

here PV(present valve )= Proceeds= $2605.5

Parameterize [tex]S[/tex] by

[tex]\vec r(u,v)=\dfrac23u\cos v\,\vec\imath+\dfrac23u\sin v\,\vec\jmath+u\,\vec k[/tex]

with [tex]4\le u\le5[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal vector to [tex]S[/tex] to be

[tex]\vec r_u\times\vec r_v=-\dfrac23u\cos v\,\vec\imath-\dfrac23u\sin v\,\vec\jmath+\dfrac49u\,\vec k[/tex]

(orientation doesn't matter here)

Then the area of [tex]S[/tex] is

[tex]\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\frac{2\sqrt{13}}9\int_0^{2\pi}\int_4^5u\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\frac{4\sqrt{13}\,\pi}9\int_4^5u\,\mathrm du=\boxed{2\sqrt{13}\,\pi}[/tex]

Answer:  0.9612

Step-by-step explanation:

The binomial distribution formula is given by :-

[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of x successes out of n trials, p is the probability of success on a particular trial.

Given : The probability of a baby being a boy= 0.522.

Then the probability of a girl : [tex]p=1-0.522=0.478[/tex]

Number of trials  : n= 5

Now, the required probability will be :

[tex]P(x \geq1)=1-P(0)\\\\=1-[^{5}C_0(0.478)^{0}(1-0.478)^{5-0}]\\\\=1-[(1)(0.522)^{5}]=0.961242789206\approx0.9612[/tex]

Thus,  the probability that at least one of them is a girl​ = 0.9612

The probability that at least one of them is a girl​ is approximately 0.9565.

Probability of At Least One Girl in Five Births

To find the probability that at least one of the next five randomly selected births in the country will be a girl, we can use the complement rule. The complement rule states that the probability of at least one girl is equal to 1 minus the probability of no girls (i.e., all boys).

Given:

→ Probability of a boy (P(B)) = 0.522

→ Probability of a girl (P(G)) = 1 - P(B)

                                            = 1 - 0.522

                                            = 0.478

Step-by-step calculation:

1. Calculate the probability of all five births being boys: (P(B))⁵

2. Substitute the values: (0.522)⁵

3. Calculate: (0.522)⁵ ≈ 0.0435

Therefore, the probability of all five births being boys is 0.0435.

Using the complement rule:

→ Probability of at least one girl = 1 - Probability of all boys

                                                    = 1 - 0.0435

                                                    ≈ 0.9565

Thus, the probability that at least one of the next five births is a girl is approximately 0.9565.

Answer:

The number of adult tickets sold was 121

Step-by-step explanation:

The general equation is: S + A = T    ;where S means student tickets, A means adult tickets and T means total tickets. We know that S is two times A and T is 363. So, S will be 2x and A will be x. Then, the new equation is 2x + x = 363.

S + A = T  

2x + x = 363  

3x = 363  

x = 363 / 3

x = 121

Adult tickets: A = x = 121

Student tickets: S = 2x = 2 * 121 = 242

Final answer:

To solve for the number of adult tickets sold, set the number of adult tickets as x, the student tickets as 2x, and use the total of 363 tickets to create the equation 3x = 363. Solving for x, you find that 121 adult tickets were sold.

Explanation:

The numerical problem given is about determining the number of adult tickets and student tickets sold for a school play, with the total number of tickets sold being 363, and the number of student tickets being twice the number of adult tickets sold.

Let the number of adult tickets be x. Then the number of student tickets is 2x. The total number of tickets sold is the sum of the adult tickets and student tickets, which is x + 2x = 3x. Since we know that 363 tickets were sold in total, we can write the equation:

3x = 363

Now, solve for x:

Therefore, 121 adult tickets were sold.

Answer:

Find the slope of the line that passes through the points shown in the table.

The slope of the line that passes through the points in the table is

.

Step-by-step explanation:

Answer: 77.45 %

Step-by-step explanation:

We assume that the measurements are normally distributed.

Given : Mean : [tex]\mu=39[/tex]

Standard deviation : [tex]\sigma=4.0[/tex]

Let x be the randomly selected measurement.

Now we calculate z score for the normal distribution as :-

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

For x = 35, we have

[tex]z=\dfrac{35-39}{4}=-1[/tex]

For x = 45, we have

[tex]z=\dfrac{45-39}{4}=1.5[/tex]

Now, the p-value = [tex]P(35<x<45)=P(-1<z<1.5)[/tex]

[tex]=P(z<1.5)-P(z<-1)\\\\=0.9331927-0.1586553=0.7745374\approx0.7745[/tex]

In percent , [tex]0.7745\times100=77.45\%[/tex]

Hence, the percent of measurements should fall between 35 and 45 = 77.45 %

Final answer:

Approximately 68% of the hardness measurements on the Rockwell C scale should fall between 35 and 45, according to the empirical rule and the provided mean and standard deviation.

Explanation:

To determine the percentage of hardness measurements that fall between 35 and 45 on the Rockwell C scale, we utilize the concept of standard deviation and the empirical rule.

Given the average (or mean) Rockwell C value is 39 and the standard deviation is 4.0, the range from 35 to 45 encompasses from one standard deviation below to one standard deviation above the mean.

According to the empirical (68-95-99.7) rule, about 68% of data within a normal distribution falls within one standard deviation of the mean.

Since the range from 35 to 45 represents this bracket around the mean, we can conclude that approximately 68% of the hardness measurements should fall within this range, assuming a normal distribution of the measurements.

Answer: True, hope this helps! :)

Answer:  

Sample Response: Determine what should go on each axis. Grades will go on the x-axis, and the number of students will be plotted on the y-axis. Find the number of different scores and range. Choose an interval – say, 10 percent. Find the frequency within each interval. Choose a scale – say, 0–3. Draw bars the height of the frequency for each interval.

Step-by-step explanation:

(Sample Response)

Creating a histogram is a way to visually represent the distribution of a dataset. Here's how you can create a histogram for today’s math quiz scores step by step:
### Step 1: Collect the Data
First, gather all the scores you want to include in the histogram. In this case, the scores are 75, 95, 60, 75, 95, and 80.
### Step 2: Determine the Number of Bins
Choose an appropriate number of bins (groups or class intervals) for your histogram. The number of bins can significantly affect the appearance of the histogram. There are different rules of thumb for this, such as using the square root of the number of data points. However, you often have to decide based on the range of your data and the level of granularity you want.
### Step 3: Determine the Bin Intervals
The bins are ranges of scores that your data will be split into. You need to ensure the bins collectively cover the entire range of your data. To do this, find the minimum and maximum scores. Then, decide how wide each bin should be. This can be done either by dividing the entire range by the number of bins or by choosing a bin width that makes sense practically.
### Step 4: Tally Scores in Each Bin
Go through your data and count how many scores fall into each bin range. For instance, if your first bin includes scores from 60 to 70, count all the scores that are within this range.
### Step 5: Draw the Histogram
On graph paper or using graphing software:
- Draw a horizontal axis (the x-axis) and a vertical axis (the y-axis).
- Label the horizontal axis with the score ranges for each bin.
- Label the vertical axis with the frequency (the number of scores in each bin).
- For each bin, draw a bar that reaches up to the frequency of scores in that bin. The width of each bar should correspond to the bin width, and there should be no space between bars if the data is continuous.
### Step 6: Label Your Histogram
Finally, provide a title for your histogram and label the axes to make it clear what is being represented.
Keep in mind that histograms should give a clear picture of how the data is distributed. If you have bins that are too large, you'll lose detail. If they're too small, the histogram may be too choppy to identify any trends. Adjust the number of bins and bin width accordingly if your first histogram doesn’t seem to represent your data clearly.

Answer:  0.317

Step-by-step explanation:

Let A be the event that the selected household is prosperous and B the event that it is educated.

Given : P(A)=0.138,   P(B)=0.261

P(A and B)=0.082

We know that for any events M and N ,

[tex]\text{P(M or N)=P(M)+P(N)-P(M or N)}[/tex]

Thus , [tex]\text{P(A or B)=P(A)+P(B)-P(A or B)}[/tex]

[tex]\text{P(A or B)}=0.138+0.261-0.082\\\\\Rightarrow\text{ P(A or B)}=0.317[/tex]

Hence, the probability P(A or B) that the household selected is either prosperous or educated = 0.317

Final answer:

The probability that a randomly selected American household is either prosperous or educated is 0.317.

Explanation:

We are asked to find the probability P(A or B) that a randomly selected American household is either prosperous (income over $100,000) or educated (householder completed college). To calculate the probability of the union of two events A and B, we use the formula:

P(A or B) = P(A) + P(B) - P(A and B).

According to the given data:

P(A) is the probability that a household is prosperous, which is 0.138.

P(B) is the probability that a household is educated, which is 0.261.

P(A and B) is the probability that a household is both prosperous and educated, which is 0.082.

Now we apply the given probabilities to the formula:

P(A or B) = 0.138 + 0.261 - 0.082 = 0.317.

Therefore, the probability that a selected household is either prosperous or educated is 0.317.

Answer:

a) Velocity of the galaxy=573.756 km/s

b) Distance to the galaxy=716 Mpc

Step-by-step explanation:

a) [tex]v=H_0 D\\Where\ v=Velocity\ of\ the\ galaxy\\H_0= Hubble\ constant=69.8\pm 1.9\ as\ of\ 16\ July\ 2019\\D= Proper\ distance=8.22\ Mpc\\\Rightarrow v=69.8\times 8.22\\\therefore v=573.756\ km/s\\[/tex]

b)[tex]v=H_0 D\\v=50000\ km/s\\H_0= Hubble\ constant=69.8\pm 1.9\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{50000}{69.8}\\\therefore D=716\ Mpc[/tex]

Answer:

95%

Step-by-step explanation:

We have a normal distribution with a mean of 6 students and standard deviation of 26.

To solve this problem, we're going to need the help of a calculator.

P(16<z<120) = 0.9545 ≈ 95.45%

Rounding, we can say that 95% of the students scored between 16 and 120.

Answer:

Yes! There are 3 different explanations I have. Pick your favorite.

Step-by-step explanation:

So we are asked to see if the following equation holds:

[tex]\frac{4 \text{ ft }}{6 \text{ ft}}=\frac{12 \text{ sec }}{18 \text{sec}}[/tex]

The units cancel out ft/ft=1 and sec/sec=1.

So we are really just trying to see if 4/6 is equal to 12/18

Or you could cross multiply and see if the products are the same on both sides:

[tex]\frac{4}{6}=\frac{12}{18}[/tex]

Cross multiply:

[tex]4(18)=6(12)[/tex]

[tex]72=72[/tex]

Since you have the same thing on both sides then the ratios given were proportional.

OR!

Put 4/6 and 12/18 in your calculator.  They both come out to have the same decimal expansion of .66666666666666666(repeating) so the ratios gives are proportional.

OR!

Reduce 12/18 and reduce 4/6 and see if the reduced fractions are same.

12/18=2/3  (I divided top and bottom by 6)

4/6=2/3  (I divided top and bottom by 2)

They are equal to the same reduced fraction so they are proportional.

Answer:

The list of bijections from A into B are shown below.

Step-by-step explanation:

A function f is called one-to-one or injective, if and only if

[tex]f(x)=f(y)\Rightarrow x = y[/tex]

for all x and y in the domain of f.

A function f from X to Y is called onto or surjective, if and only if

for every element y∈Y there is an element x∈X with f(x)=y.

If a function is one-one and onto, then it is called bijective.

Part (a):

A={q,r,s} and B={2,3,4}

We need to find all the bijections from A into B.

(1) [tex]A\rightarrow B=\{(q,2),(r,3),(s,4)\}[/tex]

(2) [tex]A\rightarrow B=\{(q,2),(r,4),(s,3)\}[/tex]

(3) [tex]A\rightarrow B=\{(q,3),(r,2),(s,4)\}[/tex]

(4) [tex]A\rightarrow B=\{(q,3),(r,4),(s,2)\}[/tex]

(5) [tex]A\rightarrow B=\{(q,4),(r,2),(s,3)\}[/tex]

(6) [tex]A\rightarrow B=\{(q,4),(r,3),(s,2)\}[/tex]

Part (b):

A={1,2,3,4} and B={5,6,7,8}

We need to find all the bijections from A into B.

(1) [tex]A\rightarrow B=\{(1,5),(2,6),(3,7),(4,8)\}[/tex]

(2) [tex]A\rightarrow B=\{(1,5),(2,6),(3,8),(4,7)\}[/tex]

(3) [tex]A\rightarrow B=\{(1,5),(2,7),(3,6),(4,8)\}[/tex]

(4) [tex]A\rightarrow B=\{(1,5),(2,7),(3,8),(4,6)\}[/tex]

(5) [tex]A\rightarrow B=\{(1,5),(2,8),(3,6),(4,7)\}[/tex]

(6) [tex]A\rightarrow B=\{(1,5),(2,8),(3,7),(4,6)\}[/tex]

(7) [tex]A\rightarrow B=\{(1,6),(2,5),(3,7),(4,8)\}[/tex]

(8) [tex]A\rightarrow B=\{(1,6),(2,5),(3,8),(4,7)\}[/tex]

(9) [tex]A\rightarrow B=\{(1,6),(2,7),(3,5),(4,8)\}[/tex]

(10) [tex]A\rightarrow B=\{(1,6),(2,7),(3,8),(4,5)\}[/tex]

(11) [tex]A\rightarrow B=\{(1,6),(2,8),(3,5),(4,7)\}[/tex]

(12) [tex]A\rightarrow B=\{(1,6),(2,8),(3,7),(4,5)\}[/tex]

(13) [tex]A\rightarrow B=\{(1,7),(2,6),(3,5),(4,8)\}[/tex]

(14) [tex]A\rightarrow B=\{(1,7),(2,6),(3,8),(4,5)\}[/tex]

(15) [tex]A\rightarrow B=\{(1,7),(2,5),(3,6),(4,8)\}[/tex]

(16) [tex]A\rightarrow B=\{(1,7),(2,5),(3,8),(4,6)\}[/tex]

(17) [tex]A\rightarrow B=\{(1,7),(2,8),(3,6),(4,5)\}[/tex]

(18) [tex]A\rightarrow B=\{(1,7),(2,8),(3,5),(4,6)\}[/tex]

(19) [tex]A\rightarrow B=\{(1,8),(2,6),(3,7),(4,5)\}[/tex]

(20) [tex]A\rightarrow B=\{(1,8),(2,6),(3,5),(4,7)\}[/tex]

(21) [tex]A\rightarrow B=\{(1,8),(2,7),(3,6),(4,5)\}[/tex]

(22) [tex]A\rightarrow B=\{(1,8),(2,7),(3,5),(4,6)\}[/tex]

(23) [tex]A\rightarrow B=\{(1,8),(2,5),(3,6),(4,7)\}[/tex]

(24) [tex]A\rightarrow B=\{(1,8),(2,5),(3,7),(4,6)\}[/tex]

Answer:

Given

[tex]f(m+n)=f(m)f(n)[/tex]

If we assume

[tex]f(x)=ae^{x}\\\\f(m+n)=ae^{m+n}\\\\\therefore f(m+n)=ae^{m}\times ae^{n}(\because x^{a+b}=x^{a}\times x^{b})\\\\\Rightarrow f(m+n)=f(m)\times f(n)[/tex]

Similarly

We can generalise the result for

[tex]f(x)=am^{x }[/tex] where a,m are real numbers

2)

[tex]f(m\cdot n)=f(m)+f(n)\\\\let\\f(x)=log(x)\\\therefore f(mn)=log(mn)=log(m)+log(n)\\\\\therefore f(mn)=f(m)+f(n)[/tex]

Answer: First option.

Step-by-step explanation:

It is important to remember that the line intersects the x-axis when [tex]y=0[/tex].

So, given the line [tex]3x+4y=12[/tex], you need to substitute [tex]y=0[/tex] into the equation and solve for "x", in order to find the x-intercept of this line.

Therefore, this is:

[tex]3x+4y=12\\\\3x+4(0)=12\\\\3x+0=12\\\\3x=12\\\\x=\frac{12}{3}\\\\x=4[/tex]

Therefore, this line intersects the x-axis at this point:

[tex](4,0)[/tex]

Answer: 0.5507

Step-by-step explanation:

The cumulative distribution function for exponential distribution:-

[tex]P(X\leq x)=1-e^{-\lambda x}[/tex]

Given : The life of a light bulb is exponentially distributed with a mean of 1,000 hours.

Then , [tex]\lambda=\dfrac{1}{1000}[/tex]

Then , the probability that the bulb will last less than 800 hours is given by :-

[tex]P(X\leq 800)=1-e^{\frac{-800}{1000}}\\\\=1-e^{-0.8}=0.5506710358\approx0.5507[/tex]

Hence, the probability that the bulb will last less than 800 hours = 0.5507

The probability that the bulb will last less than 800 hours is 0.550

An incandescent light bulb is an electric light with a wire filament heated to such a high temperature that it glows with visible light. The filament of  an incandescent light bulb is protected from oxidation with a glass or fused quartz bulb that is filled with inert gas or a vacuum. The exponential distribution is the probability distribution of the time between events in a Poisson point process. It is a particular case of the gamma distribution. While the arithmetic mean is the central value of a discrete set of numbers: specifically, the sum of the values divided by the number of values

The life of a light bulb is exponentially distributed with a mean of 1,000 hours. What is the probability that the bulb will last less than 800 hours?

What is the probability that the light bulb will have to replaced within 800 hours?

[tex]\lambda = \frac{1}{1000}[/tex]

[tex]P(x<=800) = 1 - e^{(-\lambda*x)} = 1-e^{[(-\frac{1}{1000} )*800]} = 1-e^{-0.8} = 1-0.449 = 0.550[/tex]

Grade:  5

Subject:  math

Chapter:  probability

Keywords:  mean, bulb, hours, exponential, light

Final answer:

To find the probability that the mean weight of a sample of 107 babies differs from the population mean by less than 40 grams, we can use the Central Limit Theorem. The probability is approximately 0.0441.

Explanation:

To find the probability that the mean weight of a sample of 107 babies differs from the population mean by less than 40 grams, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will be approximately normal as long as the sample size is large enough. In this case, our sample size is 107, which is considered large enough.

The formula to calculate the standard deviation of the sampling distribution of the sample mean is sigma/sqrt(n), where sigma is the population standard deviation and n is the sample size. Plugging in the values, we have 446/sqrt(107) = 42.96 grams.

To find the probability, we need to calculate the z-score for a difference of 40 grams from the population mean. The z-score formula is (x - mu) / (sigma / sqrt(n)), where x is the desired difference, mu is the population mean, sigma is the population standard deviation, and n is the sample size.

Plugging in the values, we have (40 - 3242) / (446/sqrt(107)) = -1.7. We want to find the probability that the z-score is less than -1.7. Using a standard normal distribution table or calculator, we find that the probability is approximately 0.0441.

To solve this question, we can use the Central Limit Theorem (CLT), which implies that the sampling distribution of the sample means will be normally distributed if the sample size is large enough (usually n > 30 is considered sufficient). Since we are dealing with a sample of 107 babies, which is quite large, we can safely use the CLT.
According to the CLT, the mean of the sample means will be equal to the population mean (μ), and the standard deviation of the sample means (often called the standard error, SE) will be equal to the population standard deviation (σ) divided by the square root of the sample size (n).
Given:
- The population mean (μ) = 3242 grams
- The population standard deviation (σ) = 446 grams
- The sample size (n) = 107 babies
First, we must calculate the standard error (SE):
SE = σ / √n
SE = 446 grams / √107
SE ≈ 446 grams / 10.344
SE ≈ 43.13 grams
Now, we want to find the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams. This means we are looking at weights from (μ - 40) grams to (μ + 40) grams.
So we will find the z-scores corresponding to (μ - 40) grams and (μ + 40) grams:
Z = (X - μ) / SE
For the lower limit (μ - 40 grams = 3202 grams):
Z Lower = (3202 - 3242) / 43.13
Z Lower ≈ -40 / 43.13
Z Lower ≈ -0.927
For the upper limit (μ + 40 grams = 3282 grams):
Z Upper = (3282 - 3242) / 43.13
Z Upper ≈ 40 / 43.13
Z Upper ≈ 0.927
Using the standard normal distribution (Z-distribution), we can find the probability that a Z-score falls between -0.927 and 0.927. These values correspond to the area under the standard normal curve between these two Z-scores.
If we look these values up in a Z-table, or use a calculator:
- The probability of Z being less than 0.927 is approximately 0.8238.
- The probability of Z being less than -0.927 is approximately 0.1762.
So, the probability of the mean weight being between μ - 40 grams and μ + 40 grams is the area between these two Z-scores, which can be found by subtracting the lower probability from the upper probability:
P(-0.927 < Z < 0.927) = P(Z < 0.927) - P(Z < -0.927)
P(-0.927 < Z < 0.927) = 0.8238 - 0.1762
P(-0.927 < Z < 0.927) = 0.6476
Rounding to four decimal places:
P(-0.927 < Z < 0.927) ≈ 0.6476
Therefore, the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams is approximately 0.6476, or 64.76%.