Divide total bags by total hours:
30 bags / 3 hours = 10 bags per hour
The graph to the right is the uniform density function for a friend whoThe graph to the right is the uniform density function for a friend who is x minutes late. Find the probability that the friend is at least 21 minutes late. is x minutes late. Find the probability that the friend is at least 21 minutes late.
Answer:
0.30
Step-by-step explanation:
Data provided in the question
Uniform density function for a friend = x minutes late
The Friend is at least 21 minutes late
Based on the above information, the probability that the friend is at least 21 minutes late is
[tex]= \frac{Total\ minutes - minimum\ minutes}{Total\ minutes}[/tex]
[tex]= \frac{30 - 21}{30}[/tex]
= 0.30
Based on the above formula we can easily find out the probability for the friend who is at least 21 minutes late
The probability of the friend to be at least 21 minutes late for the uniform density function shown in the graph is 0.30.
What is probability?Probability of an event is the ratio of number of favorable outcome to the total number of outcome of that event.
The graph is attached below shows the uniform density function for a friend who is x minutes late.
The probability that the friend is at least 21 minutes late-The friend is at least 21 minutes late. This means that the friend is 21 minutes late or more than it. 21 or more minutes goes from 21 to 30. Thus, the difference is,
[tex]d=30-21\\d=9[/tex]
The density of the graph is 1/30. The probability will be equal to the area under the curve.
In this, the length of the rectangle will be 9 and width will be 1/30 for the probability of at least 21 minutes late. The probability is,
[tex]P=9\times\dfrac{1}{30}\\P=0.30[/tex]
Thus, the probability of the friend to be at least 21 minutes late for the uniform density function shown in the graph is 0.30.
Learn more about the probability here;
https://brainly.com/question/24756209
In a study, 37% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to nuclear plant. Among 12 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health.
Answer:
The probability that when 12 adults are randomly selected, 3 or fewer are in excellent health is 0.2946
Step-by-step explanation:
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 12 * 0.37
= 4.44
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 12 * 0.37 * 0.63
= 2.7972
III.
standard deviation = sqrt( variance ) = sqrt(2.7972)
=1.6725
the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health
P( X < = 3) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 12 3 ) * 0.37^3 * ( 1- 0.37 ) ^9 + ( 12 2 ) * 0.37^2 * ( 1- 0.37 ) ^10 + ( 12 1 ) * 0.37^1 * ( 1- 0.37 ) ^11 + ( 12 0 ) * 0.37^0 * ( 1- 0.37 ) ^12
= 0.2947
1. Divide 6/13 by 6/12
A. 12/13
B. 1/12
C. 13/12
D.916
Answer:
C
Step-by-step explanation:
[tex]\frac{6}{13}[/tex]÷[tex]\frac{6}{12}[/tex] can also be [tex]\frac{6}{13}[/tex]×[tex]\frac{12}{3}[/tex] and [tex]\frac{6}{13}[/tex]×[tex]\frac{12}{3}[/tex]=13/12
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
\frac{6}{13}136 ÷\frac{6}{12}126 can also be \frac{6}{13}136 ×\frac{12}{3}312 and \frac{6}{13}136 ×\frac{12}{3}312 =13/12
A tank is filled with 1000 liters of pure water. Brine containing 0.04 kg of salt per liter enters the tank at 9 liters per minute. Another brine solution containing 0.05 kg of salt per liter enters the tank at 7 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 16 liters per minute. A. Determine the differential equation which describes this system. Let S(t)S(t) denote the number of kg of salt in the tank after tt minutes. Then
Answer:
The differential equation which describes the mixing process is [tex]\frac{dc_{salt,out}}{dt} + \frac{2}{125}\cdot c_{salt,out} = \frac{71}{100000}[/tex].
Step-by-step explanation:
The mixing process within the tank is modelled after the Principle of Mass Conservation, which states that:
[tex]\dot m_{salt,in} - \dot m_{salt,out} = \frac{dm_{tank}}{dt}[/tex]
Physically speaking, mass flow of salt is equal to the product of volume flow of water and salt concentration. Then:
[tex]\dot V_{water, in, 1}\cdot c_{salt, in,1} + \dot V_{water, in, 2} \cdot c_{salt,in, 2} - \dot V_{water, out}\cdot c_{salt, out} = V_{tank}\cdot \frac{dc_{salt,out}}{dt}[/tex]
Given that [tex]\dot V_{water, in, 1} = 9\,\frac{L}{min}[/tex], [tex]\dot V_{water, in, 2} = 7\,\frac{L}{min}[/tex], [tex]c_{salt,in,1} = 0.04\,\frac{kg}{L}[/tex], [tex]c_{salt, in, 2} = 0.05\,\frac{kg}{L}[/tex], [tex]\dot V_{water, out} = 16\,\frac{L}{min}[/tex] and [tex]V_{tank} = 1000\,L[/tex], the differential equation that describes the system is:
[tex]0.71 - 16\cdot c_{salt,out} = 1000\cdot \frac{dc_{salt,out}}{dt}[/tex]
[tex]1000\cdot \frac{dc_{salt, out}}{dt} + 16\cdot c_{salt, out} = 0.71[/tex]
[tex]\frac{dc_{salt,out}}{dt} + \frac{2}{125}\cdot c_{salt,out} = \frac{71}{100000}[/tex]
The estimator Yis a random variable that varies with different random samples; it has a probability distribution function that represents its sampling distribution, and mean and variance. Using the properties on expected values and variances of linear functions of random variables and sum operators, show that:
A. E(Y) = μ
B. Var(Y) σ2/N.
Answer:
Check Explanation
Step-by-step explanation:
According to the Central limit theorem, the population mean (μ) is approximately equal to the mean of sampling distribution (μₓ).
And the standard deviation of the sampling distribution (σₓ) is related to the population standard deviation (σ) through
Standard deviation of the sampling distribution = (Population standard deviation)/(√N)
where N = Sample size
σₓ = (σ/√N)
So, population mean (μ) = Mean of sampling distribution (μₓ)
Population Standard deviation = (Standard deviation of the sampling distribution) × √N
= σ × √N
A) The expected value of a given distribution is simply equal to the mean of that distribution.
Hence, the expected value of random variable Y thay varies with different samples is given as
E(Y) = Mean of sampling distribution = μₓ
But μₓ = μ
Hence, E(Y) = μ (Proved)
B) Var (Y) is given as the square of the random distribution's standard deviation.
Var (Y) = (standard deviation of the sampling distribution)²
= (σ/√N)²
= (σ²/N) (Proved)
Hope this Helps!!!
looking for the 2 correct answers please
Answer:
ROQQOPStep-by-step explanation:
The sine of an angle is equal to the cosine of its complement, and vice versa.
sin ∠QOP = cos ∠ROQ
cos ∠ROQ = sin ∠QOP
Harper's Index reported that the number of (Orange County, California) convicted drunk drivers whose sentence included a tour of the morgue was 506, of which only 1 became a repeat offender.
a. Suppose that of 1056 newly convicted drunk drivers, all were required to take a tour of the morgue. Let us assume that the probability of a repeat offender is still p= 1/596. Explain why the Poisson approximation to the binomial would be a good choice for r = number of repeat offenders out of 963 convicted drunk drivers who toured the morgue.
The Poisson approximation is good because n is large, p is small, and np < 10.The Poisson approximation is good because n is large, p is small, and np > 10. The Poisson approximation is good because n is large, p is large, and np < 10.The Poisson approximation is good because n is small, p is small, and np < 10. What is λ to the nearest tenth?
b. What is the probability that r = 0? (Use 4 decimal places.)
c. What is the probability that r > 1? (Use 4 decimal places.)
d. What is the probability that r > 2? (Use 4 decimal places.)
e. What is the probability that r > 3? (Use 4 decimal places.)
Answer:
a. The Poisson approximation is good because n is large, p is small, and np < 10.
The parameter of thr Poisson distribution is:
[tex]\lambda =np\approx1.6[/tex]
b. P(r=0)=0.2019
c. P(r>1)=0.4751
d. P(r>2)=0.2167
e. P(r>3)=0.0789
Step-by-step explanation:
a. The Poisson distribution is appropiate to represent binomial events with low probability and many repetitions (small p and large n).
The approximation that the Poisson distribution does to the real model is adequate if the product np is equal or lower than 10.
In this case, n=963 and p=1/596, so we have:
[tex]np=963*(1/596)\approx1.6[/tex]
The Poisson approximation is good because n is large, p is small, and np < 10.
The parameter of thr Poisson distribution is:
[tex]\lambda =np\approx1.6[/tex]
We can calculate the probability for k events as:
[tex]P(r=k)=\dfrac{\lambda^ke^{-\lambda}}{k!}[/tex]
b. P(r=0). We use the formula above with λ=1.6 and r=0.
[tex]P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\[/tex]
c. P(r>1). In this case, is simpler to calculate the complementary probability to P(r<=1), that is the sum of P(r=0) and P(r=1).
[tex]P(r>1)=1-P(r\leq1)=1-[P(r=0)+P(r=1)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\\\P(r>1)=1-(0.2019+0.3230)=1-0.5249=0.4751[/tex]
d. P(r>2)
[tex]P(r>2)=1-P(r\leq2)=1-[P(r=0)+P(r=1)+P(r=2)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\P(2)=1.6^{2} \cdot e^{-1.6}/2!=2.56*0.2019/2=0.2584\\\\\\P(r>2)=1-(0.2019+0.3230+0.2584)=1-0.7833=0.2167[/tex]
e. P(r>3)
[tex]P(r>3)=1-P(r\leq2)=1-[P(r=0)+P(r=1)+P(r=2)+P(r=3)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\P(2)=1.6^{2} \cdot e^{-1.6}/2!=2.56*0.2019/2=0.2584\\\\P(3)=1.6^{3} \cdot e^{-1.6}/3!=4.096*0.2019/6=0.1378\\\\\\P(r>3)=1-(0.2019+0.3230+0.2584+0.1378)=1-0.9211=0.0789[/tex]
(1,6) and (2,3) are on line?
Answer:
y = - 3x + 9 or (y - 6) = - 3(x - 1)
Step-by-step explanation:
I'm assuming you are looking for the equation of the line.
First, let's find the slope of the line, which we use equation (y1-y2) / (x1-x2)
x1: 1 x2: 2
y1: 6 y2: 3
(6-3) / (1-2) = 3 / -1 = -3
The slope of the line is -3.
The equation for slope-intercept form is y = mx + b, where m is slope and b is the y - intercept.
y = -3x+b
Now substitute a set of points in (any coordinates work). I'll use (1, 6)
6 = -3(1)+b
6 = - 3 + b
b = 9
So our equation in slope intercept form is y = - 3x + 9
You can also write this in point slope form, (y-y1) = m (x-x1) m is still slope
(y - 6) = - 3(x - 1)
Robert makes $951 gross income per week and keeps $762 of it after tax withholding. How many allowances has Robert claimed? For weekly income between 950 and 960, the number of withholding allowances claimed are: 0, 259 dollars; 1, 242 dollars; 2, 224 dollars; 3, 207 dollars; 4, 189 dollars; 5, 173 dollars; 6, 162 dollars; 7, 151 dollars; 8, 140 dollars. a. One b. Two c. Three d. Four
Answer:
Option D,four is correct
Step-by-step explanation:
The tax withholding from the gross income of $951 is the gross income itself minus the income after tax withholding i.e $189 ($951-$762)
The percentage of the withholding =189/951=20% approximately
Going by the multiple choices provided,option with 4,189 dollars seems to the correct option as that is the exact of the tax withholding on Robert's gross income and his earnings fall in between $950 and $960
The number of allowances that Robert has claimed is four. Option d is correct.
What is Tax withholding allowance?
A withholding tax is a tax that an employer deducts from an employee's paycheck and delivers it straight to the government (federal income tax).
From the given information:
The gross income per week for Robert = $951 The amount saved after removing tax allowance = $762The tax allowance = Gross income - savings amount
The tax allowance = $951 - $762
The tax allowance = $189
From the weekly income between 950 and 960 data, we can see that the number of allowances claimed related to the withholding allowances are:
0 → $259
1 → $242
2 → $224
3 → $207
4 → $189
5 → $173
6 → $162
7 → $151
8 → $140
Therefore, we can conclude that the number of allowances that Robert has claimed is four.
Learn more about tax withholding here:
https://brainly.com/question/25927192
Benson asked a group of 9th, 10th, 11th, and 12th graders the
number of hours they work per week in the summer. The means of
each group are provided below:
9th: 14.2 hours
10th: 18.8 hours
11th: 21.2 hours
12th: 24.9 hours
Which conclusion can be made from this data?
Answer:
YlThe higher the grader the more hours they can work.
Step-by-step explanation:
We can see that the higher the grader the more hours they can work ; which could mean less academic work if the work defined is that with which to earn money but if the work defined is academic it means more hours of academic work
what is the product a-3/7 ÷ 3-a/21
Answer:
the sum shown is (20/21)a -(1/7)
Step-by-step explanation:
As written, the sum is ...
[tex]a-\dfrac{\frac{3}{7}}{3}-\dfrac{a}{21}=\boxed{\dfrac{20}{21}a-\dfrac{1}{7}}[/tex]
__
We wonder if you mean the quotient ...
((a-3)/7)/((3-a)/21)
[tex]\dfrac{\left(\dfrac{a-3}{7}\right)}{\left(\dfrac{3-a}{21}\right)}=\dfrac{a-3}{7}\cdot\dfrac{21}{3-a}=\dfrac{-21(3-a)}{7(3-a)}=\boxed{-3}[/tex]
_____
Comment on the problem presentation
Parentheses are required when plain text is used to represent fractions. The symbols ÷, /, and "over" all mean the same thing: "divided by." The denominator is the next item in the expression. If arithmetic of any kind is involved in the denominator, parentheses are needed. This is the interpretation required by the Order of Operations.
When the expression is typeset, fraction bars and text formatting (superscript) serve to group items that require parentheses in plain text.
__
Please note that some authors make a distinction between the various forms of division symbol. Some use ÷ to mean ...
(everything to the left)/(everything to the right)
and they reserve / solely for use in fractions. Using this interpretation, your expression would be ...
(a -(3/7))/(3 -(a/21)) = (21a -9)/(63 -a)
That distinction is not supported by the Order of Operations.
An Archer shoots an arrow horizontally at 250 feet per second. The bullseye on the target and the arrow are initially at the same height. If the target is 60 feet from the archer, how far below the bullseye (in feet) will the arrow hit the target
Answer:
1.84feetStep-by-step explanation:
Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;
Range [tex]R = U\sqrt{\frac{H}{g} }[/tex]
U is the velocity of the arrow
H is the maximum height reached = distance below the bullseye reached by the arrow.
R is the horizontal distance covered i.e the distance of the target from the archer.
g is the acceleration due to gravity.
Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?
On substitution,
[tex]60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}[/tex]
Squaring both sides we have;
[tex](\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} = \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet[/tex]
The arrow will hit the target 1.84feet below the bullseye.
Answer:
8.7
Step-by-step explanation:
on edge . You're welcome
what is 1+1, this is for you to get points.
Answer:
2
Step-by-step explanation:
1+1 = 2
Hope this helps;)
Answer:
1 + 1 =2
Step-by-step explanation:
Hi, how are you?
Do u want to be my friend?
Also, I hope this answered your question?
In November Hillary drove 580 miles in her car the car travelled 33.5 miles for each gallon of petrol used
Petrol cost £1.09 per litre
1 gallon = 4.55 litres
Work out the cost of the petrol the car used in November
Answer:
T = £85.87
the cost of the petrol the car used in November is £85.87
Step-by-step explanation:
Given;
In November Hillary drove 580 miles in her car;
Distance travelled d = 580 miles
the car travelled 33.5 miles for each gallon of petrol used;
Fuel consumption rate r = 33.5 miles per gallon
Number of gallons N consumed by the car is;
N = distance travelled/fuel consumption rate
N = d/r = 580/33.5 = 17.3134 gallons
Given that;
Petrol cost £1.09 per litre
Cost per litre c = £1.09
1 gallon = 4.55 litres
Converting the amount of fuel used to litres;
N = 17.3134 gallons × 4.55 litres per gallon
N = 78.77612 litres
The total cost T = amount of fuel consumption N × fuel cost per litre c
T = N × c
T = 78.77612 litres × £1.09 per litre
T = £85.87
the cost of the petrol the car used in November is £85.87
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.(a) What is the velocity of the top of the ladder when the base is given below?7 feet away from the wall ft/sec20 feet away from the wall ft/sec24 feet away from the wall ft/sec(b) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. ft2/sec(c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall. rad/sec
Answer:
a) The height decreases at a rate of [tex]\frac{7}{12}[/tex] ft/sec.
b) The area increases at a rate of [tex]\frac{527}{24}[/tex] ft^2/sec
c) The angle is increasing at a rate of [tex]\frac{1}{12}[/tex] rad/sec
Step-by-step explanation:
Attached you will find a sketch of the situation. The ladder forms a triangle of base b and height h with the house. The key to any type of problem is to identify the formula we want to differentiate, by having in mind the rules of differentiation.
a) Using pythagorean theorem, we have that [tex] 25^2 = h^2+b^2[/tex]. From here, we have that
[tex]h^2 = 25^2-b^2[/tex]
if we differentiate with respecto to t (t is time), by implicit differentiation we get
[tex]2h \frac{dh}{dt} = -2b\frac{db}{dt}[/tex]
Then,
[tex]\frac{dh}{dt} = -\frac{b}{h}\frac{db}{dt}[/tex].
We are told that the base is increasing at a rate of 2 ft/s (that is the value of db/dt). Using the pythagorean theorem, when b = 7, then h = 24. So,
[tex]\frac{dh}{dt} = -\frac{2\cdot 7}{24}= \frac{-7}{12}[/tex]
b) The area of the triangle is given by
[tex]A = \frac{1}{2}bh[/tex]
By differentiating with respect to t, using the product formula we get
[tex] \frac{dA}{dt} = \frac{1}{2} (\frac{db}{dt}h+b\frac{dh}{dt})[/tex]
when b=7, we know that h=24 and dh/dt = -1/12. Then
[tex]\frac{dA}{dt} = \frac{1}{2}(2\cdot 24- 7\frac{7}{12}) = \frac{527}{24}[/tex]
c) Based on the drawing, we have that
[tex]\sin(\theta)= \frac{b}{25}[/tex]
If we differentiate with respect of t, and recalling that the derivative of sine is cosine, we get
[tex] \cos(\theta)\frac{d\theta}{dt}=\frac{1}{25}\frac{db}{dt}[/tex] or, by replacing the value of db/dt
[tex]\frac{d\theta}{dt}=\frac{2}{25\cos(\theta)}[/tex]
when b = 7, we have that h = 24, then [tex]\cos(\theta) = \frac{24}{25}[/tex], then
[tex]\frac{d\theta}{dt} = \frac{2}{25\frac{24}{25}} = \frac{2}{24} = \frac{1}{12}[/tex]
A consumer agency is investigating the blowout pressures of Soap Stone tires. A Soap Stone tire is said to blow out when it separates from the wheel rim due to impact forces usually caused by hitting a rock or a pothole in the road. A random sample of 29 Soap Stone tires were inflated to the recommended pressure, and then forces measured in foot-pounds were applied to each tire (1 foot-pound is the force of 1 pound dropped from a height of 1 foot). The customer complaint is that some Soap Stone tires blow out under small-impact forces, while other tires seem to be well made and don't have this fault. For the 29 test tires, the sample standard deviation of blowout forces was 1358 foot-pounds.
(a) Soap Stone claims its tires will blow out at an average pressure of 26,000 foot-pounds, with a standard deviation of 1020 foot-pounds. The average blowout force is not in question, but the variability of blowout forces is in question. Using a 0.1 level of significance, test the claim that the variance of blowout pressures is more than Soap Stone claims it is.
Classify the problem as being a Chi-square test of independence or homogeneity, Chi-square goodness-of-fit, Chi-square for testing or estimating σ2 or σ, F test for two variances, One-way ANOVA, or Two-way ANOVA, then perform the following.
(ii) Find the sample test statistic. (Use two decimal places.)
(iii) Find the P-value of the sample test statistic. (Use four decimal places.)
(b) Find a 99% confidence interval for the variance of blowout pressures, using the information from the random sample. (Use one decimal place.)
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: Impact force needed for Sap Stone tires to blow out. (foot-pounds)
n= 29, S= 1358 foot-pound
a)
Soap Stone claims that "The tires will blow out at an average pressure of μ= 26000 foot-pounds with a standard deviation of σ= 1020 foot-pounds.
According to the consumer's complaint, the variability of the blown out forces is greater than the value determined by the company.
I)
Then the parameter of interest is the population variance (or population standard deviation) and to test the consumer's complaint you have to conduct a Chi-Square test for σ².
σ²= (1020)²= 1040400 foot-pounds²
H₀: σ² ≤ 1040400
H₁: σ² > 1040400
α: 0.01
II)
[tex]X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}[/tex]
[tex]X^2_{H_0}= \frac{(n-1)S^2}{Sigma^2}= \frac{(29-1)*(1358)^2}{1040400} = 49.63[/tex]
III)
This test is one-tailed to the right and so is the p-value. This distribution has n-1= 29-1= 28 degrees of freedom, so you can calculate the p-value as:
P(X²₂₈≥49.63)= 1 - P(X²₂₈<49.63)= 1 - 0.99289= 0.00711
⇒ The p-value is less than the significance level so the test is significant at 1%. You can conclude that the population variance of the blowout forces is less than 1040400 foot-pounds², at the same level the population standard deviation of the blow out forces is less than 1020 foot-pounds.
b)
99% CI for the variance. Using the X² statistic you can calculate it as:
[tex][\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2}} ;\frac{(n-1)S^2}{X^2_{n-1;\alpha /2}} ][/tex]
[tex]X^2_{n-1;\alpha /2}= X^2_{28; 0.005}= 13.121[/tex]
[tex]X^2_{n-1;1-\alpha /2}= X^2_{28; 0.995}= 49.588[/tex]
[tex][\frac{28*(1358)^2}{49.588} ;\frac{28*(1358)^2}{13.121} ][/tex]
[1041312.253; 3935415.898] foot-pounds²
I hope this helps!
If h = 12 units and r = 4 units, what is the volume of the cone shown above? Use 3.14 for .
Answer:
200.96 units
Step-by-step explanation:
Use the formula for the volume of a cone [tex]V=\pi r^{2} \frac{h}{3}[/tex]
Plug in the values ([tex]\pi[/tex]=3.14) and multiply them all out
Answer:
≈ 201
Step-by-step explanation:
V= πr²h/3
V= 3.14*4²*12/3= 200.96 ≈ 201
what measurement do you use for surface area
Answer:
Step-by-step explanation:
Surface Area is the combined area of all two-dimensional surfaces of a shape. Just like ordinary area, the units are the squares of the units of length (that is, if a shape's sides are measured in meters, then the shape's area is measured in square meters
According to the image whats the answer? 80 points brainliest
Answer:
Hey!
Your answer is 150 square meters!
Step-by-step explanation:
13*3=39 (the tilted face)
12*3=36 (the flat face)
5*3=15 (the base)
TO FIND THE TRINGLULAR AREA:
1/2 base x height...
1/2 x 5 x 12 = 30
(WE DOUBLE THIS BECAUSE THERE ARE TWO TRIANGLES)
SO 60...
ADD THESE TOGETHER...
39 + 36 + 15 + 30 + 30...
GIVES US 150 square metres
HOPE THIS HELPS!!Answer:
150 square meters
Step-by-step explanation:
Suppose the average driving distance for last year's Player's Champion Golf Tournament in Ponte Vedra, FL, was 292.5 yards with a standard deviation of 14.2 yards. A random sample of 60 drives was selected from a total of 4,244 drives that were hit during this tournament. What is the probability that the sample average was 289 yards or less?
Answer:
The Probability that the sample average was 289 yards or less
P(x⁻≤ 289) = P( Z≤ -1.909) = 0.0287
Step-by-step explanation:
step(i):-
Mean of the Population = 292.5 yards
Standard deviation of the Population = 14.2 yards
sample size 'n' =60 drives
N = 4244 drives
Step(ii):-
Let X⁻ be random sample average
[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]
Let X⁻ = 289
[tex]Z = \frac{289 -292.5}{\frac{14,2}{\sqrt{60} } }[/tex]
Z = - 1.909
The Probability that the sample average was 289 yards or less
P(x⁻≤ 289) = P( Z≤ -1.909)
= 0.5 -A(1.909)
= 0.5 -0.4713
= 0.0287
Conclusion:-
The Probability that the sample average was 289 yards or less = 0.0287
Solve x^2 + 5x+6 = 0
Answer:
X=-2or,-3
Step-by-step explanation:
X^2+5x+6=0
or,x^2+(2+3)x+6=0
or,x^2+2x+3x+6=0
or,x(x+2)+3(x+2)=0
or,(x+3) (x+2)=0
Either,
x+3=0 x+2=0
or,x=-3 or,x=-2
Therefore,the value if x is -2 or -3 .
I HOPE IT WILL HELP YOU
Determine whether b can be written as a linear combination of Bold a Subscript Bold 1a1, Bold a Subscript Bold 2a2, and Bold a Subscript Bold 3a3. In other words, determine whether weights x 1x1, x 2x2, and x 3x3 exist, such that x 1x1Bold a Subscript Bold 1a1plus+x 2x2Bold a Subscript Bold 2a2plus+x 3x3Bold a Subscript Bold 3a3equals=b. Determine the weights x 1x1, x 2x2, and x 3x3 if possible.
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
b can be written as a linear combination of [tex]a_1 \ and \ a_2[/tex]
b
The values of [tex]x_1 = 4 \ and \ x_2 = 2[/tex]
Step-by-step explanation:
From the question we are told that
[tex]x_1 a_1 +x_2 a_2 = b[/tex]
Where [tex]a_ 1 = (4, 5,-4)[/tex], [tex]a_2 = (-4 , 3, 3)[/tex] and [tex]b = (8,26 , -10)[/tex]
So
[tex]x_1 ( 4, 5,-4) + x_2 (-4 , 3, 3) = (8,26 , -10)[/tex]
[tex]4x_1, 5x_1,-4x_1 + -4x_2 , 3x_2, 3x_2 = (8,26 , -10)[/tex]
=> [tex]4x_1 -4x_2 =8[/tex]
[tex]x_1 -x_2 =2 ---(1)[/tex]
=> [tex]5x_1 + 3x_2 = 26 --- (2)[/tex]
=> [tex]-4x_1 + 3x_2 = -10 ---(3)[/tex]
Now multiplying equation 1 by 3 and adding the product to equation 2
[tex].\ \ \ 3x_1 -3x_2 = 6\\+ \ \ 5x_1 + 3x_2 = 26 \\=> \ \ \ 8x_1 = 32[/tex]
=> [tex]x_1 = 4[/tex]
substituting [tex]x_1[/tex] into equation 1
[tex]4 - x_2 =2[/tex]
[tex]x_2 =2[/tex]
Now to test substitute [tex]x_1 \ and \ x_2[/tex] into equation 3
[tex]-4(4) + 3(2) = -10[/tex]
[tex]-10 = -10[/tex]
Since LHS = RHS then there exist values [tex]x_1 = 4 \ and \ x_2 = 2[/tex] such that
[tex]x_1 a_1 +x_2 a_2 = b[/tex]
Hence b can be written as a linear combination of [tex]a_1 \ and \ a_2[/tex]
Between what two integers is square root 54?
Answer:
7 and 8
Step-by-step explanation:
54 is between 7^2 = 49 and 8^2 = 64.
The square roots have the same relationship.
49 < 54 < 64
√49 < √54 < √64
7 < √54 < 8
Find the missing length, c, in the right triangle below. Round to the nearest tenth, if necessary. a. 76.3 in. b. 5.5 in. c. 3.5 in. d. 8.7 in.
Answer:
i think it's c.)3.5 i may be wrong
Step-by-step explanation:
what is an equation of the line that passes through the point (-3,-1) and is perpendicular to the line y=1/2x - 3
Answer:
Step-by-step explanation: gradient of perpendicular line is
-2
y+1/x+3=-2
y+1=-2x-6
y=-2x-7
What is the volume of a cube with a side length of 14 cm.?
Answer:
V =2744 cm^3
Step-by-step explanation:
The volume of a cube is given by
V = s^3 where s is the side length
V = 14^3
V =2744 cm^3
Answer: 2,744 cm³
Step-by-step explanation: Since the length, width, and height of a cube are all equal, we can find the volume of a cube by multiplying side × side × side.
So we can find the volume of a cube using the formula s³.
Notice that we have a side length of 14cm.
So plugging into the formula, we have (14 cm)³ or
(14 cm)(14 cm)(14 cm) which is 2,744 cm³.
So the volume of the cube is 2,744 cm³.
The time it takes for a planet to complete its orbit around a particular star is called the? planet's sidereal year. The sidereal year of a planet is related to the distance the planet is from the star. The accompanying data show the distances of the planets from a particular star and their sidereal years. Complete parts? (a) through? (e).
I figured out what
(a) is already.
(b) Determine the correlation between distance and sidereal year.
(c) Compute the? least-squares regression line.
(d) Plot the residuals against the distance from the star.
(e) Do you think the? least-squares regression line is a good? model?
Planet
Distance from the? Star, x?(millions of? miles)
Sidereal? Year, y
Planet 1
36
0.22
Planet 2
67
0.62
Planet 3
93
1.00
Planet 4
142
1.86
Planet 5
483
11.8
Planet 6
887
29.5
Planet 7
? 1,785
84.0
Planet 8
? 2,797
165.0
Planet 9
?3,675
248.0
Answer:
(a) See below
(b) r = 0.9879
(c) y = -12.629 + 0.0654x
(d) See below
(e) No.
Step-by-step explanation:
(a) Plot the data
I used Excel to plot your data and got the graph in Fig 1 below.
(b) Correlation coefficient
One formula for the correlation coefficient is
[tex]r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}[/tex]
The calculation is not difficult, but it is tedious.
(i) Calculate the intermediate numbers
We can display them in a table.
x y xy x² y²
36 0.22 7.92 1296 0.05
67 0.62 42.21 4489 0.40
93 1.00 93.00 20164 3.46
433 11.8 5699.4 233289 139.24
887 29.3 25989.1 786769 858.49
1785 82.0 146370 3186225 6724
2797 163.0 455911 7823209 26569
3675 248.0 911400 13505625 61504
9965 537.81 1545776.75 25569715 95799.63
(ii) Calculate the correlation coefficient
[tex]r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}[/tex]
(c) Regression line
The equation for the regression line is
y = a + bx where
[tex]a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} - \dfrac{9\times 1545776.75 - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}[/tex]
(d) Residuals
Insert the values of x into the regression equation to get the estimated values of y.
Then take the difference between the actual and estimated values to get the residuals.
x y Estimated Residual
36 0.22 -10 10
67 0.62 -8 9
93 1.00 -7 8
142 1.86 -3 5
433 11.8 19 - 7
887 29.3 45 -16
1785 82.0 104 -22
2797 163.0 170 - 7
3675 248.0 228 20
(e) Suitability of regression line
A linear model would have the residuals scattered randomly above and below a horizontal line.
Instead, they appear to lie along a parabola (Fig. 2).
This suggests that linear regression is not a good model for the data.
What is the range of g?
Answer:
{-7, -4, -1, 3, 7}
Step-by-step explanation:
The range is the list of y-coordinates of the points:
range = {-7, -4, -1, 3, 7}
A single card is drawn at random from a standard 52 card check. Work out in its simplest form
Answer:
1/52
Step-by-step explanation:
Find all solutions of the equation in the interval [0, 2π).
Answer:
x = pi/6
x = 11pi/6
x = 5pi/6
x =7pi/6
Step-by-step explanation:
2 sec^2 (x) + tan ^2 (x) -3 =0
We know tan^2(x) = sec^2 (x) -1
2 sec^2 (x) +sec^2(x) -1 -3 =0
Combine like terms
3 sec^2(x) -4 = 0
Add 4 to each side
3 sec^2 (x) = 4
Divide by 3
sec^2 (x) = 4/3
Take the square root of each side
sqrt(sec^2 (x)) = ±sqrt(4/3)
sec(x) = ±sqrt(4)/sqrt(3)
sec(x) = ±2 /sqrt(3)
Take the inverse sec on each side
sec^-1 sec(x) = sec^-1(±2 /sqrt(3))
x = pi/6 + 2 pi n where n is an integer
x = 11pi/6 + 2 pi n
x = 5pi/6 + 2 pi n
x =7pi/6 + 2 pi n
We only want the solutions between 0 and 2pi