Answer:
The next number in the sequence is: 13
Step-by-step explanation:
So we start with 3, then the pattern is:
subtract 9, add 18, subtract 8, add 16, subtract 7, add 14...
If we apply this to the pattern we can see that it works. You should be able to continue the pattern if needed.
Consider the function represented by 9x + 3y = 12 with x as the independent variable. How can this function be
written using function notation?
Of) = -
O F(x) = - 3x + 4
Of(x) = -x +
O fb) = - 3y+ 4
Answer:
f(x) = -3x + 4
Step-by-step explanation:
Step 1: Move the 9x over
3y = 12 - 9x
Step 2: Divide everything by 3
y = 4 - 3x
Step 3: Rearrange
y = -3x + 4
Step 4: Change y to f(x)
f(x) = -3x + 4
The base of a triangle is three times
the height. If the area is 150msquare,find the height.
Answer:
10m
Step-by-step explanation:
area = 1/2 base times height
x=height
3x=base
so
150=1/2(3x^2)
300=3x^2
100=x^2
10=x
so the height is 10 and the base is 30
Answer:
h = 10
Step-by-step explanation:
Hiiiiiii
[!] Urgent [!] Find the domain of the graphed function.
A cardboard box without a lid is to have a volume of 8,788 cm3. Find the dimensions that minimize the amount of cardboard used.
Answer:
x = y = 26 cm; z = 13 cm
Step-by-step explanation:
We can calculate the dimensions of the square base as
∛(2·8788) = 26 cm
the height of the box will be half of 26/2 which is 13 cm.
x = y = 26 cm; z = 13 cm
then the minimum area for the given volume can be calculated using what we call Lagrange multipliers, this makes it easier
area = xy +2(xz +yz)
But we were given the volume as 8788
Now we will make the partial derivatives of L to be in respect to the cordinates x, y, z, as well as λ to be equal to zero, then
L = xy +2(xz +yz) +λ(xyz -8788)
For x: we have
y+2z +λyz=0
For y we have
y: x +2z +λxz=0
For z we have 2x+2y +λxy=0............eqn(*)
For we have xyz -8788=0
If we simplify the partial derivative equation of y and x above then we have
λ = (y +2z)/(yz).
= 1/z +2/y............eqn(1)
λ = (x +2z)/(xz)
= 1/z +2/x.............eqn(2)
Set eqn(1 and 2) to equate we have
1/z +2/y = 1/z +2/x
x = y
From eqn(*) we can get z
λ = (2x +2y)/(xy) = 2/y +2/x
If we simplify we have
1/z +2y = 2/x +2/y
Then z = x/2
26/2 =13
Therefore,
x = y = 2z = ∛(2·8788)
X= 26
y = 26 cm
z = 13 cm
a) Al usar un microscopio el microscopio se amplía una célula 400 veces. Escribe el factor de ampliación como cociente o como escala.
b) La imagen de una célula usando dicho microscopio mide 1,5 mm ¿ Cuánto mide la célula en la realidad?
Answer:
x = 0,00375 mm
Step-by-step explanation:
a) El factor de ampliación es 400/1 es decir el tamaño real se verá ampliado 400 veces mediante el uso del microscopio
b) De acuerdo a lo establecido en la respuesta a la pregunta referida en a (anterior) podemos establecer una regla de tres, según:
Si al microscopio el tamaño de la célula es 1,5 mm, cual será el tamaño verdadero ( que es reducido 400 en relación al que veo en el microscopio)
Es decir 1,5 mm ⇒ 400
x (mm) ⇒ 1 (tamaño real de la célula)
Entonces
x = 1,5 /400
x = 0,00375 mm
I NEED HELP PLEASE, THANKS! :)
A rock is tossed from a height of 2 meters at an initial velocity of 30 m/s at an angle of 20° with the ground. Write parametric equations to represent the path of the rock. (Show work)
Answer:
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
Step-by-step explanation:
If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -
x = ( 30 cos 20° )( time ),
y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2
To determine " ( 30 cos 20° )( time ) " you would do the following calculations -
( x = 30 * 0.93... = ( About ) 28.01t
This represents our horizontal distance, respectively the vertical distance should be the following -
y = 30 * 0.34 - 4.9t^2,
( y = ( About ) 10.26t - 4.9t^2 + 2
In other words, our solution should be,
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
These are are parametric equations
confused on question in screenshot
Answer:
right triangle
Step-by-step explanation:
We can use the Pythagorean theorem to determine if this is a right triangle
a^2 + b^2 = c^2
13^2 + ( 8 sqrt(13)) ^2 = (sqrt(1001))^2
169 + 8^2 * 13 = 1001
169+64*13 = 1001
169+832=1001
1001 = 1001
Since this is true, this is a right triangle
Mia, Maya, and Maria are sisters. Mia's age is twice Maya's age and Maria is seven years younger than Mia. If Maria is 3 years old, how old are Mia and Maya?
Answer:
Mia:10 Maya:5 Maria:3
Step-by-step explanation:
3+7= 10= Mia's age
10÷2=5= Maya's age
Answer:
Mia - 10
Maya - 5
Maria - 3
An instructor asks students to rate their anxiety level on a scale of 1 to 100 (1 being low anxiety and 100 being high anxiety) just before the students take their final exam. The responses are shown below. Construct a relative frequency table for the instructor using five classes. Use the minimum value from the data set as the lower class limit for the first row, and use the lowest possible whole-number class width that will allow the table to account for all of the responses. Use integers or decimals for all answers.
48,50,71,58,56,55,53,70,63,74,64,33,34,39,49,60,65,84,54,58
Provide your answer below:
Lower Class Limit Upper Class Limit Relative Frequency
Answer:
The frequency table is shown below.
Step-by-step explanation:
The data set arranged ascending order is:
S = {33 , 34 , 39 , 48 , 49 , 50 , 53 , 54 , 55 , 56 , 58 , 58, 60 , 63 , 64 , 65 , 70 , 71 , 74 , 84}
It is asked to use the minimum value from the data set as the lower class limit for the first row.
So, the lower class limit for the first class interval is 33.
To determine the class width compute the range as follows:
[tex]\text{Range}=\text{Maximum}-\text{Minimum}[/tex]
[tex]=84-33\\=51[/tex]
The number of classes requires is 5.
The class width is:
[tex]\text{Class width}=\frac{Range}{5}=\frac{51}{2}=10.2\approx 10[/tex]
So, the class width is 10.
The classes are:
33 - 42
43 - 52
53 - 62
63 - 72
73 - 82
83 - 92
Compute the frequencies of each class as follows:
Class Interval Values Frequency
33 - 42 33 , 34 , 39 3
43 - 52 48 , 49 , 50 3
53 - 62 53 , 54 , 55 , 56 , 58 , 58, 60 7
63 - 72 63 , 64 , 65 , 70 , 71 5
73 - 82 74 1
83 - 92 84 1
TOTAL 20
Compute the relative frequencies as follows:
Class Interval Frequency Relative Frequency
33 - 42 3 [tex]\frac{3}{20}\times 100\%=15\%[/tex]
43 - 52 3 [tex]\frac{3}{20}\times 100\%=15\%[/tex]
53 - 62 7 [tex]\frac{7}{20}\times 100\%=35\%[/tex]
63 - 72 5 [tex]\frac{5}{20}\times 100\%=25\%[/tex]
73 - 82 1 [tex]\frac{1}{20}\times 100\%=5\%[/tex]
83 - 92 1 [tex]\frac{1}{20}\times 100\%=5\%[/tex]
TOTAL 20 100%
Find the missing side. Round your answer to the nearest tenth.
6a - 3c + a + 2b = what the answer
Answer:
7a+2b-3c
Step-by-step explanation:
6a+a = 7a
2b stays the same
-3c stays the same
Answer:
Hey mate, here is your answer. Hope it helps you.
7a-3c+2b
Step-by-step explanation:
6a+a-3c+2b
=7a-3c+2b
3c and 2b will be the same because the variables are different. They are not like terms.
Which of the following statements about feasible solutions to a linear programming problem is true?A. Min 4x + 3y + (2/3)z
B. Max 5x2 + 6y2
C. Max 5xy
D. Min (x1+x2)/3
Answer:
The answer is "Option A"
Step-by-step explanation:
The valid linear programming language equation can be defined as follows:
Equation:
[tex]\Rightarrow \ Min\ 4x + 3y + (\frac{2}{3})z[/tex]
The description of a linear equation can be defined as follows:
It is an algebraic expression whereby each term contains a single exponent, and a single direction consists in the linear interpolation of the equation.
Formula:
[tex]\to \boxed{y= mx+c}[/tex]
You want to test whether or not the following sample of 30 observations follows a normal distribution. The mean of the sample equals 11.83 and the standard deviation equals 4.53. 2 3 5 5 7 8 8 9 9 10 11 11 12 12 12 12 13 13 13 14 15 15 15 16 16 17 17 18 18 19 At the 5% level of significance, the conclusion of the test is that the a. data does not follow a normal distribution. b. null hypothesis cannot be rejected. c. sample data has no probability distribution. d. sample data is incorrect.
Answer:
b. null hypothesis cannot be rejected.
Step-by-step explanation:
At the 5% level of significance, the conclusion of the test is that the
The test statistic is 2 and the critical value is 7.815. Since the test statistic is less than the critical value, we can not reject the null hypothesis.
The graphs below are the same shape what is the equation of the blue graph
Answer:
B. g(x) = (x-2)^2 +1
Step-by-step explanation:
When you see this type of equation your get the variables H and K in a quadratic equation. In this case the (x-2)^2 +1 is your H. The (x-2)^2 +1 is your K.
For the H you always do the opposite so in this case instead of going to the left 2 times you go to the right 2 times (affects your x)
For the K you go up or down which in this case you go up one (affects your y)
And that's how you got your (2,1) as the center of the parabola
-Hope this helps :)
Please answer this correctly without making mistakes
Answer:
Question 2
Step-by-step explanation:
2) The time when she woke up was - 3° C
During nature walk, temperature got 3° C warmer than when she woke up.
So, temperature during nature walk = - 3 + 3 = 0° C
The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of six per hour.
(a) What is the probability that exactly three arrivals occur during a particular hour? (Round your answer to three decimal places.)
(b) What Is the probability that at least three people arrive during a particular hour? (Round your answer to three decimal places.)
(c) How many people do you expect to arrive during a 15-min period?
Answer:
a) P(x=3)=0.089
b) P(x≥3)=0.938
c) 1.5 arrivals
Step-by-step explanation:
Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.
The variable X is modeled by a Poisson process with a rate parameter of λ=6.
The probability of exactly k arrivals in a particular hour can be written as:
[tex]P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k![/tex]
a) The probability that exactly 3 arrivals occur during a particular hour is:
[tex]P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\[/tex]
b) The probability that at least 3 people arrive during a particular hour is:
[tex]P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938[/tex]
c) In this case, t=0.25, so we recalculate the parameter as:
[tex]\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5[/tex]
The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.
[tex]E(x)=\lambda=1.5[/tex]
I need help pls pls pls pls
Answer:
D. 4
Step-by-step explanation:
If he leaves the science assignments for the next day, he will spend zero hours on science assignments. This means that y is equal to 0. Plug this into the given equation and solve for x.
2x + y = 8
2x + 0 = 8
2x = 8
x = 4
Gerald can complete 4 math assignments.
A small regional carrier accepted 16 reservations for a particular flight with 12 seats. 8 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 48% chance, independently of each other.
A) Find the probability that overbooking occurs.
B) Find the probability that the flight has empty seats.
Answer:
a) 32.04% probability that overbooking occurs.
b) 40.79% probability that the flight has empty seats.
Step-by-step explanation:
For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. The probability of a passenger arriving is independent of other passengers. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [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]
And p is the probability of X happening.
Our variable of interest are the 8 reservations that went for the passengers with a 48% probability of arriving.
This means that [tex]n = 8, p = 0.48[/tex]
A) Find the probability that overbooking occurs.
12 seats, 8 of which are already occupied. So overbooking occurs if more than 4 of the reservated arrive.
[tex]P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{8,5}.(0.48)^{5}.(0.52)^{3} = 0.2006[/tex]
[tex]P(X = 6) = C_{8,6}.(0.48)^{6}.(0.52)^{2} = 0.0926[/tex]
[tex]P(X = 7) = C_{8,7}.(0.48)^{7}.(0.52)^{7} = 0.0244[/tex]
[tex]P(X = 8) = C_{8,5}.(0.48)^{8}.(0.52)^{0} = 0.0028[/tex]
[tex]P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2006 + 0.0926 + 0.0244 + 0.0028 = 0.3204[/tex]
32.04% probability that overbooking occurs.
B) Find the probability that the flight has empty seats.
Less than 4 of the booked passengers arrive.
To make it easier, i will use
[tex]P(X < 4) = 1 - (P(X = 4) + P(X > 4))[/tex]
From a), P(X > 4) = 0.3204
[tex]P(X = 4) = C_{8,4}.(0.48)^{4}.(0.52)^{4} = 0.2717[/tex]
[tex]P(X < 4) = 1 - (P(X = 4) + P(X > 4)) = 1 - (0.2717 + 0.3204) = 1 - 0.5921 = 0.4079[/tex]
40.79% probability that the flight has empty seats.
Joe hypothesizes that the students of an elite school will score higher than the general population. He records a sample mean equal to 568 and states the hypothesis as μ = 568 vs μ > 568. What type of test should Joe do?
Answer:
The test to be used is the right tailed test.
Step-by-step explanation:
The type of test joe should do would be a right tailed test. This is because;
A right tailed test which we sometimes call an upper test is where the hypothesis statement contains the greater than (>) symbol. This means that, the inequality points to the right. For example, we want to compare the the life of batteries before and after a manufacturing change.
If we want to know if the battery life of maybe 90 hours would be greater than the original, then our hypothesis statements might be:
Null hypothesis: (H0 = 90).
Alternative hypothesis: (H1) > 90.
In the null hypothesis, there are no changes, but in the alternative hypothesis, the battery life in hours has increased.
So, the most important factor here is that the alternative hypothesis (H1) is what determines if we have a right tailed test, not the null hypothesis.
Thus, the test to be used is the right tailed test.
Answer:
right tailed test.
Step-by-step explanation:
M/J Grade 8 Pre-Algebra-PT-FL-1205070-003
Answer:
Following are the description of the given course code:
Step-by-step explanation:
The given course code is Pre-Algebra, which is just an introduction arithmetic course programs to train high school in the Algebra 1. This course aims to strengthen required problem solving skills, datatypes, equations, as well as graphing.
In this course students start to see the "big picture" of maths but also understand that mathematical, algorithmic, and angular principles are intertwined to form a basis for higher mathematics education.The duration of this code is in year and it is divided into two levels. In this, code it includes PreK to 12 Education Courses , with the general mathematics .Answer:
A
Step-by-step explanation:
Conde Nast Traveler publishes a Gold List of the top hotels all over the world. The Broadmoor Hotel in Colorado Springs contains 700 rooms and is on the 2004 Gold List (Conde Nast Traveler, January 2004). Suppose Broadmoor's marketing group forecasts a demand of 670 rooms for the coming weekend. Assume that demand for the upcoming weekend is normally distributed with a standard deviation of 30.
a.What is the probability all the hotel's rooms will be rented (to 4 decimals)?
b. What is the probability 50 or more rooms will not be rented (to 4 decimals)?
Answer:
(a) The probability that all the hotel's rooms will be rented is 0.1587.
(b) The probability that 50 or more rooms will not be rented is 0.2514.
Step-by-step explanation:
We are given that the Broadmoor Hotel in Colorado Springs contains 700 rooms and is on the 2004 Gold List.
Suppose Broadmoor's marketing group forecasts a mean demand of 670 rooms for the coming weekend. Assume that demand for the upcoming weekend is normally distributed with a standard deviation of 30.
Let X = demand for rooms in the hotel
So, X ~ Normal([tex]\mu=670,\sigma^{2} =30^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean demand for the rooms = 670
[tex]\sigma[/tex] = standard deviation = 30
(a) The probability that all the hotel's rooms will be rented means that the demand is at least 700 = P(X [tex]\geq[/tex] 700)
P(X [tex]\geq[/tex] 700) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{700-670}{30}[/tex] ) = P(Z [tex]\geq[/tex] 1) = 1 - P(Z < 1)
= 1 - 0.8413 = 0.1587
The above probability is calculated by looking at the value of x = 1 in the z table which has an area of 0.8413.
(b) The probability that 50 or more rooms will not be rented is given by = P(X [tex]\leq[/tex] 650)
P(X [tex]\leq[/tex] 650) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{650-670}{30}[/tex] ) = P(Z [tex]\leq[/tex] -0.67) = 1 - P(Z < 0.67)
= 1 - 0.7486 = 0.2514
The above probability is calculated by looking at the value of x = 0.67 in the z table which has an area of 0.7486.
find the circumference of a circle with a diameter of 6 cm
Circumference = πd
~substitute → (π)(6 cm)
~simplify → 6π cm.
So the circumference of the circle shown here is 6π cm.
Answer:
18.85 cm
Step-by-step explanation:
The circumference of a circle has a formula.
Circumference = π × diameter
The diameter is 6 centimeters.
Circumference = π × 6
Circumference ≈ 18.85
The circumference of the circle is 18.85 centimeters.
what is the equation of the line that is parallel to the given line and passes through the point (2, 3) ? a. x + 2y = 4 b. x + 2y = 8 c. 2x + y =4 d. 2x + y = 8
Answer:
see explanations
Step-by-step explanation:
The given blue line has a slope of m = -1/2.
The line parallel to the given line passing through point (x0,y0)=(2,3) is given by the point-slope form:
(y-y0)=m(x-x0)
substitute values
(y-3) = (-1/2)(x-2)
Expand and transpose
y = (-1/2)x + 1 + 3
y = (-1/2)x + 4 ....................(1)
We choose the second equation b. x+2y=8 and convert to slope-intercept form:
2y=-x+8
y = (-1/2)x + 4, which is exactly equation (1)
So
b. x+2y=8 is the correct answer.
Answer:
b. x + 2y = 8
Step-by-step explanation:
The average life a manufacturer's blender is 5 years, with a standard deviation of 1 year. Assuming that the lives of these blenders follow approximately a normal distribution, find the probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.
Answer:
55.11% probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question:
[tex]\mu = 5, \sigma = 1, n = 9, s = \frac{1}{\sqrt{9}} = 0.3333[/tex]
Find the probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.
This is the pvalue of Z when X = 5.1 subtracted by the pvalue of Z when X = 4.5. So
X = 5.1
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{5.1 - 5}{0.3333}[/tex]
[tex]Z = 0.3[/tex]
[tex]Z = 0.3[/tex] has a pvalue of 0.6179
X = 4.5
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{4.5 - 5}{0.3333}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
0.6179 - 0.0668 = 0.5511
55.11% probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.
i am stuck on this please help!
Answer:
[tex]20 {x}^{3} - 36 {x}^{2} + 7x + 3[/tex]Solution,
[tex](5x + 1)(2x - 1)(2x - 3)[/tex]
[tex] = 5x(2x - 1) + 1(2x - 1) \times (2x - 3) \\ = (10 {x}^{2} - 5x + 2x - 1)(2x - 3) \\ = (10 {x}^{2} - 3x - 1)(2 x - 3) \\ = 10 {x}^{2} (2x - 3) - 3x(2 x - 3) - 1(2x - 3) \\ = 20 {x}^{3} - 30 {x}^{2} - 6 {x }^{2} + 9x - 2x + 3 \\ = 20 {x}^{3} - 36 {x}^{2} + 7x + 3[/tex]
Hope this helps..
Good luck on your assignment...
HELP ASAP WILL MARK BRAINIEST IF YOU ARE RIGHT !Which of the following represents a function?
Answer:
Option C.
Step-by-step explanation:
This is a function because all of the numbers have a partner, and none of them have more than one.
Example of Not a Function
Function Not a Function
-4 to 5 -4 to 5 <
9 to 7 -4 to 3 <
13 to 3 13 to 3 ^
-7 to 5 9 to 7 ^
-7 to 5 ^
Not a Function because of this
The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch. The width of a door is normally distributed with a mean of 23 7/8 inches and a standard deviation of 1/16 inch. Assume independence. a. Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. b. What is the probability that the width of the casing minus the width of the door exceeds 1/4 inch? c. What is the probability that the door does not fit in the casing?
Answer:
a) Mean = 0.125 inch
Standard deviation = 0.13975 inch
b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25) = 0.18673
c) Probability that the door does not fit in the casing = P(X < 0) = 0.18673
Step-by-step explanation:
Let the distribution of the width of the casing be X₁ (μ₁, σ₁²)
Let the distribution of the width of the door be X₂ (μ₂, σ₂²)
The distribution of the difference between the width of the casing and the width of the door = X = X₁ - X₂
when two independent normal distributions are combined in any manner, the resulting distribution is also a normal distribution with
Mean = Σλᵢμᵢ
λᵢ = coefficient of each disteibution in the manner that they are combined
μᵢ = Mean of each distribution
Combined variance = σ² = Σλᵢ²σᵢ²
λ₁ = 1, λ₂ = -1
μ₁ = 24 inches
μ₂ = 23 7/8 inches = 23.875 inches
σ₁² = (1/8)² = (1/64) = 0.015625
σ₂ ² = (1/16)² = (1/256) = 0.00390625
Combined mean = μ = 24 - 23.875 = 0.125 inch
Combined variance = σ² = (1² × 0.015625) + [(-1)² × 0.00390625] = 0.01953125
Standard deviation = √(Variance) = √(0.01953125) = 0.1397542486 = 0.13975 inch
b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25)
This is a normal distribution problem
Mean = μ = 0.125 inch
Standard deviation = σ = 0.13975 inch
We first normalize/standardize 0.25 inch
The standardized score of any value is that value minus the mean divided by the standard deviation.
z = (x - μ)/σ = (0.25 - 0.125)/0.13975 = 0.89
P(X > 0.25) = P(z > 0.89)
Checking the tables
P(x > 0.25) = P(z > 0.89) = 1 - P(z ≤ 0.89) = 1 - 0.81327 = 0.18673
c) Probability that the door does not fit in the casing
If X₂ > X₁, X < 0
P(X < 0)
We first normalize/standardize 0 inch
z = (x - μ)/σ = (0 - 0.125)/0.13975 = -0.89
P(X < 0) = P(z < -0.89)
Checking the tables
P(X < 0) = P(z < -0.89) = 0.18673
Hope this Helps!!!
how many solution does this equation have LOOK AT SCREENSHOT ATTACHED
Answer:
One solution
Step-by-step explanation:
99% of the time, linear equations (equations that have the first degree) have only one solution. However, it's always good to check.
6 - 3x = 12 - 6x
6 = 12 - 3x
-3x = -6
x = 2
As you can see, only one solution. Hope this helps!
The chi-square value for a one-tailed (lower tail) test when the level of significance is .1 and the sample size is 15 is a. 23.685. b. 6.571. c. 7.790. d. 21.064.
Answer:
The degrees of freedom are given by:
[tex] df =n-1= 15-1=14[/tex]
And if we look in the chi square distribution with 14 degrees of freedom and if we find a quantile who accumulates 0.1 of the area in the left we got:
[tex] \chi^2 = 7.790[/tex]
And then the best answer would be:
c. 7.790
Step-by-step explanation:
For this case we know that we are using a one tailed (lower tail) critical value using a significance level of [tex]\alpha=0.1[/tex] and for this case we know that the ample size is n=15. The degrees of freedom are given by:
[tex] df =n-1= 15-1=14[/tex]
And if we look in the chi square distribution with 14 degrees of freedom and if we find a quantile who accumulates 0.1 of the area in the left we got:
[tex] \chi^2 = 7.790[/tex]
And then the best answer would be:
c. 7.790
Five thousand tickets are sold at $1 each for a charity raffle. Tickets are to be drawn at random and monetary prizes awarded as follows: 1 prize of $800, 3 prizes of $200, 5 prizes of $50, and 20 prizes of $5. What is the expected value of this raffle if you buy 1 ticket?
Answer:
The expected value of this raffle if you buy 1 ticket is $0.41.
Step-by-step explanation:
The expected value of the raffle if we buy one ticket is the sum of the prizes multiplied by each of its probabilities.
This can be written as:
[tex]E(X)=\sum p_iX_i[/tex]
For example, the first prize is $800 and we have only 1 prize, that divided by the number of tickets gives us a probability of 1/5000.
If we do this with all the prizes, we can calculate the expected value of a ticket.
[tex]E(X)=\sum p_iX_i\\\\\\E(X)=\dfrac{1\cdot800+3\cdot200+5\cdot50+20\cdot20}{5000}\\\\\\E(X)=\dfrac{800+600+250+400}{5000}=\dfrac{2050}{5000}=0.41[/tex]