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]
Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 5.00. What should be the length of B⃗ so that the magnitude of their vector sum is 9.00?
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.
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?
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
Further explanationAn 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]
Learn moreLearn more about bulb https://brainly.com/question/9584045Learn more about hours https://brainly.com/question/1032699Learn more about mean https://brainly.com/question/3777157Answer details
Grade: 5
Subject: math
Chapter: probability
Keywords: mean, bulb, hours, exponential, light
Certain functions obey the property f(m + n)=f(m)f(n) for all positive numbers and n.Can you think of a function that obeys this property? Hint: Functions that obey this property ill be familiar from ordinary pre-calculus algebra courses. Same question, but this time, the property is f(mn) f(m) +f(n) . (Note, don't expect f to be integer-valued. The hint from the first part applies here too.) m
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]
The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion
s = 4 sin(πt) + 3 cos(πt),
where t is measured in seconds. (Round your answers to two decimal places.)
A. Find the average velocity during each time period.
(i) [1, 2]
(ii) [1, 1.1]
(iii) [1, 1.01]
(iv) [1, 1.001]
B. Estimate the instantaneous velocity of the particle when t = 1.
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:
Suppose babies born in a large hospital have a mean weight of 3242 grams, and a standard deviation of 446 grams. If 107 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams? Round your answer to four decimal places.
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%.
Kelly price sells college textbooks on commission. She gets 8 % on the first $7500 of sales, 16% on the next $7500 of sales, and 20% on sales over $ 15,000. In July of 1997 $9500, ShelbyShelby's sales total was
What was ShelbyShelby's gross commission for July 1997?
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:
Calculate the commission on the first $7500 at 8%, which is $7500 * 0.08 = $600.Now, since Shelby's total sales are $9500, we subtract the first tier limit of $7500 to find the remaining sales amount, which is $9500 - $7500 = $2000.Next, calculate the commission on the remaining $2000 at the second tier rate of 16%, which is $2000 * 0.16 = $320.Add the two commission amounts together to find the total commission: $600 + $320 = $920.Therefore, Shelby’s gross commission for July 1997 is $920.
Guesstimate how much total time do Americans spend driving in one year? Express the answer in hours, years and lifetimes.
How would one go about guestimating a problem like this.
Answer:
1,460 hours per year.
113,880 hours in a lifetime.
Step-by-step explanation:
Great question, it is always good to ask away and get rid of any doubts that you may be having.
Let's assume that on an average day Americans drive 4 hours a day. This includes going and coming from work as well as any groceries they might need to do. Since there are 365 days in a year we can multiply this by the hours a day driven.
[tex]365*4 = 1,460[/tex]
We can "guesstimate" that on average an American drives 1,460 hours per year.
Based on my research the average lifetime of an American is 78 years old. We can now multiply this by the number of hours driven in a year to have an idea of the number of hours driven in a lifetime.
[tex]1,460*78 = 113,880[/tex]
On average we "guesstimate" that an American drives 113,880 hours in a lifetime.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Question 15 of 20 (1 point) View problem in a pop-up Assume that the mean systolie blood pressure of normal adults is 120 millimeters of mercury (mm Hg) and the standard deviation is 5.6. Assume the variable is normally distributed. Use a graphing calculator and round the answers to four decimal places. Part 1 If an individual is selected, find the probability that the individual's pressure will be between 120.4 and 121.9 mm Hg. P120A
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
A total of 363 were sold for the school play. They were either adult tickets or student tickets. The number of student tickets sold was two times the number of adult tickets sold. How many adult tickets were sold?
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:
Divide both sides of the equation by 3:x = 363 / 3x = 121Therefore, 121 adult tickets were sold.
In a certain country, the true probability of a baby being a boy is 0.522. Among the next five randomly selected births in the country, what is the probability that at least one of them is a girl?
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.
In terms of explaining the probability of assignment to trial arms in consent forms, ICH notes that is should be included...True or False
Answer: True, hope this helps! :)
[ Complex Analysis ]
Find the laurent series for 1/(1-Z2) about Z0=1.
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].
a b c or d pls helpp
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).
Call a household prosperous if its income exceeds $100,000. Call the household educated if the householder completed college. Select an American household at random, and let A be the event that the selected household is prosperous and B the event that it is educated. According to the Current Population Survey, P(A)=0.138, P(B)=0.261, and the probability that a household is both prosperous and educated is P(A and B)=0.082. What is the probability P(A or B) that the household selected is either prosperous or educated?
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.
Calculate the area of the surface S. S is the portion of the cone (x^2/4)+(y^2/4)=(z^2/9) that lies between z=4 and z=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]
The scores for today’s math quiz are 75, 95, 60, 75, 95, and 80. Explain the steps needed to create a histogram for the data.
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.
Question 3: What is the x-intercept of the line 3x+4y=12?
Question 3 options:
(4, 0)
(3, 0)
(0, 4)
(0, 3)
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]
120 hardness measurements are made on a large slab of steel with an average of Rockwell C value of 39 and a standard deviation of 4.0. What percent of measurements should fall between 35 and 45?
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.
List all the bijections from A into B.
a) A={q,r,s} and B={2,3,4}
b) A={1,2,3,4} and B={5,6,7,8}
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]
Assume that on a standardized test of 100 questions, a person has a probability of 80% of answering any particular question correctly. Find the probability of answering between 74 and 84 questions
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
A bag contains three types of coins i.e. one rupee coins, 50 paisa coins and 25 paisa coins totalling 175 coins. If the total value of the coins of each kind be the same, the total amount in the bag is
a) Rs 300 b) Rs 175 c) Rs 126 d) Rs 75
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.
Last year, a person wrote 119 checks. Let the random variable x represent the number of checks he wrote in one day, and assume that it has a Poisson distribution. What is the mean number of checks written per day? What is the standard deviation? What is the variance?
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
Engineers want to design seats in commercial aircraft so that they are wide enough to fit 90% of all males. (Accommodating 100% of males would require very wide seats that would be much too expensive.) Men have hip breadths that are normally distributed with a mean of 14.9 in. and a standard deviation of 0.9 in. Find Upper P 90. That is, find the hip breadth for men that separates the smallest 90% from the largest 10%.
Answer:
16.053 inches
Step-by-step explanation:
Many such probability questions are easily answered by a suitable calculator or spreadsheet.
4,000 students take an exam and get scores with a mean of 68 and a standard deviation of 26.
Approximately how many students scored between 16 and 120? Hint: Empirical Rule: 68%, 95%, 99.7%
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.
In a survey of adults who follow more than one sport, 30% listed football as their favorite sport. You survey 15 adults who follow more than one sport. What is the probability rounded to the nearest tenth that fewer than 4 of them will say that football is their favorite sport?
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
_________________ "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"
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.
Explanation: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.
Learn more about Test-Retest Reliability here:https://brainly.com/question/32360703
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Complete the square x2 – 6x + 8.
Question 3 options:
(x+3)2 – 1
(x-3)2 – 1
(x-3)2
(x-6)2 - 28
[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.
Devin signs a note for a discounted loan agreeing to pay $2,700 in 6 months at a discount rate of 7%. Determine the amount of the discount and the proceeds to Devin.
Discount = $
Proceeds = $
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
Do the ratios 4ft./6ft. and 12sec/18sec form a proportion? Why or why not?
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.
[4] Answer the questions about the two galaxies below.
(a) Find the velocity of a galaxy if it is 8.22 Mpc away.
(b) Find the distance to a galaxy if it is moving away at a speed of 50,000 km/s.
Note: be sure to always use the formula sheet.
Answer: 592 km/s, 694 Mpc
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]