Answer:
The base is a square of side 9.19 cm and the height is 7.66 cm
[tex]C_m=126.58\ cents[/tex]
Step-by-step explanation:
Optimization
We'll use simple techniques to find the optimum values that minimize the cost function given in the problem. Since the restriction is an equality, the derivative will come handy to find the critical points and then we'll prove they are a minimum.
First, we consider the shape of the rectangular bin has a square base and no top. Let x be the side of the base, thus the Area of the base is
[tex]A_b=x^2[/tex]
Let y be the height of the box, thus each one of the four lateral sides of the box is a rectangle with sides x and y and the total lateral area is
[tex]A_s=4xy[/tex]
The cost of the material used to manufacture the box is 0.5 cents per square centimeter of the base and 0.3 cents per square centimeter of the sides, thus the total cost to produce one box is
[tex]C(x,y)=0.5x^2+0.3\cdot 4xy[/tex]
[tex]C(x,y)=0.5x^2+1.2xy[/tex]
Note the cost is a two-variable function. We need to have it expressed as a single variable function. To achieve that, we use the volume provided as [tex]646 cm^3[/tex]. The volume of the box is the base times the height
[tex]V=x^2y[/tex]
Using the value of the volume we have
[tex]x^2y=646[/tex]
Solving for y
[tex]\displaystyle y=\frac{646}{x^2}[/tex]
Replacing into the cost function, it only depends on one variable
[tex]\displaystyle C(x)=0.5x^2+1.2x\cdot \frac{646}{x^2}[/tex]
Operating
[tex]\displaystyle C(x)=0.5x^2+ \frac{775.2}{x}[/tex]
Taking the first derivative
[tex]\displaystyle C'(x)=x-\frac{775.2}{x^2}[/tex]
Equating to 0
[tex]\displaystyle x-\frac{775.2}{x^2}=0[/tex]
Solving
[tex]\displaystyle x=\sqrt[3]{775.2}[/tex]
[tex]x=9.19\ cm[/tex]
Now find the height
[tex]\displaystyle y=\frac{646}{9.19^2}[/tex]
[tex]y=7.66\ cm[/tex]
Find the second derivative
[tex]\displaystyle C''(x)=1+\frac{1550.4}{x^3}[/tex]
Since this value is positive, for all x positive, the function has a minimum at the critical point.
Thus, the minimum cost is
[tex]\displaystyle C_m=0.5\cdot 9.19^2+ \frac{775.2}{9.19}[/tex]
[tex]\boxed{C_m=126.58\ cents}[/tex]
Answer:
126.58 cents or $1.27
Step-by-step explanation:
the math from above is correct they just want the answers in dollars
(Urgent! please help!) Find the constant of proportionality for the ratio of cost to number of bananas from the table.
______
Answer: 0.25
Step-by-step explanation:
I did the quiz
Answer:
0.25
Step-by-step explanation:
Hope this helps, have a nice day
4 pints, 2 quarts, 8 cups, 1 gallon which one dose not belong and why
Answer:
1 gallon does not belong
Step-by-step explanation:
4 pints = 2 quarts
2 quarts = 8 cups
8 cups does not equal 1 gallon, 16 cups does
A tire manufacturer has a 60,000 mile warranty for tread life. The manufacturer considers the overall tire quality to be acceptable if less than 5% are worn out at 60,000 miles. The manufacturer tests 250 tires that have been used for 60,000 miles. They find that 3.6% of them are worn out. With this data, we test the following hypotheses. H0: The proportion of tires that are worn out after 60,000 miles is equal to 0.05. Ha: The proportion of tires that are worn out after 60,000 miles is less than 0.05.
What assumption about the sample underlies the hypothesis test?
a.The sample comes from a population of tires with 60,000 miles where 5% of tires are worn out.
b.The sample comes from a population of tires with 60,000 miles where 3.6% of the tires are worn out.
c.The sample comes from a population of tires where 250 are used for 60,000 miles.
d.The sample comes from a population of 250 tires that have a 60,000 mile warranty.
Answer:
(A)The sample comes from a population of tires with 60,000 miles where 5% of tires are worn out.
Step-by-step explanation:
The tire manufacturer has a 60,000 mile warranty for tread life. To verify the authenticity of its claim, it picks a sample of 250 tires (out of the population of tires) that have been used for 60,000 miles.
Given the hypothesis:
[tex]H_0[/tex]: The proportion of tires that are worn out after 60,000 miles is equal to 0.05.
[tex]H_a[/tex]: The proportion of tires that are worn out after 60,000 miles is less than 0.05.
It is assumed that 5% of the tires in the sample are worn out and the hypothesis is formulated to prove or disprove this assumption.
Can someone please work this out. With step by step instructions. Thank you
Answer:
12m^3
Step-by-step explanation:
Basically, you are looking for the volume of the Oil. In order to do that you simply need to use the rectangular prism volume formula: lwh
lwh=V
(5)(2)(1.2)=V
10(1.2)=V
12=V
Half of the sum of 32 and 2
Answer:
17
Step-by-step explanation:
32+2=34
34/2= 17
Half of the sum of 32 and twice a number 'x' expressed as '2ans' would be calculated by adding 32 and 2x and then dividing by 2, resulting in the expression 16 + x.
The question asks to find half of the sum of 32 and an unspecified number (mentioned as '2ans'). Assuming '2ans' means twice the number 'ans', which can be represented as 2x, where 'x' is the particular value of 'ans'. First, calculate the sum of 32 and 2x, and then divide that sum by two to find half of it.
The steps to solve this are:
Calculate the sum: 32 + 2x.
To find half of the sum, divide by 2: (32 + 2x) / 2.
Simplify the expression: 16 + x.
Therefore, half of the sum of 32 and twice a number 'x' (2x) is 16 + x.
What is the name of an angle that is formed by one side of an triangle and the extension of an adjacent side?
Answer:
an exterior angle is the correct answer.....
Answer:
A
Step-by-step explanation:
75% of b is 12. What is b
Answer:
B=16
Step-by-step explanation:
If 75% is 12, 100% would be 16. B = 16!
Answer:16
Step-by-step explanation:
12/3=4 4=25%of b 4+12=16
A movie theater has a seating capacity of 317. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. There are half as many adults as there are children. If the total ticket sales was $ 2296, How many children, students, and adults attended?
Answer:
There are 154 children, 77 adults and 86 students in attendance
Step-by-step explanation:
Given
Seat capacity = 317
Total tickets = $2296
Charges is as follows;
Children = $5.00
Students = $7.00
Adults = $12.00
Required
Number of children, students and adults
Let A, C and S represent adults, children and students respectively.
So,
From sales of tickets, we have the following:
12A + 5C + 7S = 2296 --- Equation 1
From attendance, we have
A + C + S = 317 --- Equation 2
Given that, there are half as many adults as there are children.
So, A = ½C
Substitute ½C for A in equation 1 and 2
12A + 5C + 7S = 2296 becomes
12 * ½C + 5C + 7S = 2296
6C + 5C + 7S = 2296
11C + 7S = 2296 ---- Equation 3
A + C + S = 317
½C + C + S = 317
Multiply through by 2
2(½C + C + S) = 2 * 317
C +2C + 2S = 634
3C + 2S = 634 ----- Equation 4
We'll solve equations 3 and 4, simultaneously.
First, write out the two equations.
11C + 7S = 2296 ---- (3)
3C + 2S = 634 ------ (4)
Using elimination method to eliminate S.
Multiply (3) by 2 and multiply (4) by 7.
2 (11C + 7S = 2296 )
22C + 14S = 4592 ---- Equation 5
7 (3C + 2S = 634 )
21C + 14S = 4438 ------ Equation 6
Subtract (6) from (5)
22C + 14S = 4592
21C + 14S = 4438
---------------------------
22C - 21C + 14S - 14S = 4592 - 4438
C = 154
Recall that A = ½C
So, A = ½ * 154
A = 77
Recall equation 2
A + C + S = 317
Make S the subject of formula
S = 317 - A - C
Substituton 77 for A and 154 for C
So,
S = 317 - 77 - 154
S = 86
Hence, there are 154 children, 77 adults and 86 students in attendance
The Federal Pell Grant Program gives grants to low-income undergraduate students. According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, the average Pell grant award for 2007-2008 was $2,600. We wonder if the mean amount is different this year for Pell grant recipients at San Jose State University. Suppose that we randomly select 50 Pell grant recipients from San Jose State University. For these 50 students, the mean Pell grant award is $2,450 with a standard deviation of $600. Let μ = the mean amount of Pell grant awards received by San Jose State University Pell grant recipients this year. We test the following hypotheses. H 0: μ = 2600 H a: μ ≠ 2600 The sample size is greater than 30, so a t-model is a good fit for the sampling distribution. Use this information to answer the next two questions. Flag this Question Question 12.5 pts What is the t-test statistic? If necessary, round to two decimal places.
Answer: the t-test statistic is - 1.77
Step-by-step explanation:
For the null hypothesis,
H 0: μ = 2600
For the alternative hypothesis,
H a: μ ≠ 2600
This is a two tailed test
Since no population standard deviation is given, the distribution is a student's t.
Since n = 50,
Degrees of freedom, df = n - 1 = 50 - 1 = 49
t = (x - µ)/(s/√n)
Where
x = sample mean = $2450
µ = population mean = $2600
s = samples standard deviation = $600
Therefore,
t = (2450 - 2600)/(600/√50) = - 1.77
The t-test statistic calculated for the given data comparing the mean Pell grant award at San Jose State University to the national average for 2007-2008 is approximately -1.77.
Explanation:The t-test statistic is calculated using the formula: t = (sample mean - population mean) / (standard deviation / sqrt(sample size)). In this case, the sample mean is the mean Pell grant award for the 50 students from San Jose State University ($2,450), the population mean is the average Pell grant award according to the National Postsecondary Student Aid Study for 2007-2008 ($2,600), the standard deviation is $600, and the sample size is 50. Substitute these values into the formula to get: t = ($2,450 - $2,600) / ($600 / sqrt(50)) = -$150 / ($600 / 7.071) = -$150 / 84.85 = -1.77. So the t-test statistic is approximately -1.77, comparing the mean Pell grant award at San Jose State University to the national average for 2007-2008.
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If the standard deviation of a random variable X is 20 and a random sample of size nequals19 is obtained, what is the standard deviation of the sampling distribution of the sample mean?
The standard deviation of the sampling distribution of the sample mean is 4.587 and this can be determined by using the formula of the standard deviation of the sample mean.
Given :
The standard deviation of a random variable X is 20 and a random sample of size n equals 19.
The formula of the standard deviation of the sample mean can be used to determine the standard deviation of the sampling distribution of the sample mean.
The standard deviation of the sample mean is given by:
[tex]\sigma_m=\dfrac{\sigma}{\sqrt{n} }[/tex] --- (1)
Now put the value of [tex]\sigma[/tex] that is 20 and the value n that is 19 in the equation (1).
[tex]\sigma_m = \dfrac{20}{\sqrt{19} }[/tex]
[tex]\sigma_m = \dfrac{20}{4.36}[/tex]
[tex]\sigma_m = 4.587[/tex]
So, the standard deviation of the sampling distribution of the sample mean is 4.587.
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The standard deviation of the sampling distribution of the sample mean, also referred to as the standard error, for this scenario is approximately 4.58.
Explanation:The standard deviation of a sampling distribution of the sample mean, also known as the standard error, can be calculated using the formula Standard Error = σ/√n, where σ is the population standard deviation and n is the sample size. In your case, the standard deviation (σ) of the random variable X is 20, and your sample size (n) is 19.
Plugging these values into the formula, we get Standard Error = 20/√19, which can be approximated as 4.58.
Therefore, the standard deviation of the sampling distribution of the sample mean for this scenario is approximately 4.58.
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Theo solved n + 5 = 11 for the unknown. The steps he
took are shown.
He started with 5 tiles.
Then he added tiles as he counted up to 11.
He realized he added 6 more tiles to his original 5 to make
11. So he found the unknown to be 6.
Answer:
what is the question?
Step-by-step explanation:
Suppose x is a normally distributed random variable with muμequals=1616 and sigmaσequals=22. Find each of the following probabilities. a. P(xgreater than or equals≥17.517.5) b. P(xless than or equals≤1212) c. P(16.7816.78 less than or equalsxless than or equals≤20.4620.46) d. P(11.4811.48less than or equals≤xless than or equals≤19.0619.06)
Answer:
Step-by-step explanation:
Since x is a normally distributed random variable, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = the random variable
µ = mean
σ = standard deviation
From the information given,
µ = 16
σ = 2
a. P(x ≥ 17.5) = 1 - (x < 17.5)
For x < 17.5
z = (17.5 - 16)/2 = 0.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.77
P(x ≥ 17.5) = 1 - 0.77 = 0.23
b. P(x ≤ 12)
z = (12 - 16)/2 = - 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.023
P(x ≤ 12) = 0.023
c) P(16.78 ≤ x ≤ 20.46)
For x = 16.78,
z = (16.78 - 16)/2 = 0.39
Looking at the normal distribution table, the probability corresponding to the z score is 0.65
For x = 20.46,
z = (20.46 - 16)/2 = 2.23
Looking at the normal distribution table, the probability corresponding to the z score is 0.987
Therefore,
P(16.78 ≤ x ≤ 20.46) = 0.987 - 0.65 = 0.337
d) P(11.48 ≤ x ≤ 19.06)
For x = 11.48,
z = (11.48 - 16)/2 = - 2.26
Looking at the normal distribution table, the probability corresponding to the z score is 0.012
For x = 19.06,
z = (19.06 - 16)/2 = 1.53
Looking at the normal distribution table, the probability corresponding to the z score is 0.94
Therefore,
P(11.48 ≤ x ≤ 19.06) = 0.94 - 0.012 = 0.928
Ronique recorded the number of games that the school softball team won each year for the past 7 years: 31, 24, 32, 22, 34, 28, 38. Ronique's team won at least
games in 75% of the seasons that they played.
Answer:
156
Step-by-step explanation:
Total number of games played by Ronique's team is ≤ 279(round off).
What is at least?
" At least represents the minimum quantity of the number . it can be more than the given condition."
According to the question,
Number of games team won each year for the past 7 years are
31, 24, 32, 22, 34, 28, 38.
Total number of games won by team = 31 + 24 + 32+22 + 34 + 28 + 38
= 209
Team won at least 75% of the seasons they played
'x' represents the total number of seasons they played
Therefore,
75% of x ≤ 209
⇒[tex]\frac{75}{100} (x)\leq 209[/tex]
⇒ [tex]x \leq \frac{(209)(100)}{75}[/tex]
⇒ [tex]x\leq 278.666..[/tex]
⇒[tex]x\leq 279 (round off)[/tex]
Hence, total number of games played by Ronique's team is ≤ 279(round off).
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Exhibit 13-2 Source of Variation Sum of Squares Degrees of Freedom Mean Square F Between treatments 2,073.6 4 Between blocks 6,000.0 5 1,200 Error 20 288 Total 29 The test statistic to test the null hypothesis equals _____. a. 4.17 b. 28.8 c. 1.8 d. .432
Answer:
option C is correct,the null hypothesis equals 1.8
Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu. Assume that the population has a normal distribution. Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students.
Answer:
[tex]95-2.76\frac{6.6}{\sqrt{30}}=91.674[/tex]
[tex]95+2.76\frac{6.6}{\sqrt{30}}=98.326[/tex]
We can say at 99% confidence that the true mean is between (91.674;98.326)
Step-by-step explanation:
Data given
[tex]\bar X=95[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=6.6 represent the sample standard deviation
n=30 represent the sample size
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
We need to find the critical value [tex]t_{\alpha/2}[/tex] and we need to find first the degrees of freedom, given by:
[tex]df=n-1=30-1=29[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that [tex]t_{\alpha/2}=2.76[/tex]
And replacing we got:
[tex]95-2.76\frac{6.6}{\sqrt{30}}=91.674[/tex]
[tex]95+2.76\frac{6.6}{\sqrt{30}}=98.326[/tex]
We can say at 99% confidence that the true mean is between (91.674;98.326)
The daily sales at a convenience store have a mean of $1350 and a standard deviation of $150. The mean of the sampling distribution of the mean sales of a sample of 25 days for this convenience store is:
Answer:
The mean of the sampling distribution = $1350
Step-by-step explanation:
Given -
Population mean [tex]\boldsymbol{(\nu)}[/tex] = $1350
Standard deviation = $150
Sample size ( n ) = 25
The mean of the sampling distribution = population mean
[tex]\boldsymbol{(\nu)_{M}} = \boldsymbol{(\nu)}[/tex]
The mean of the sampling distribution = $1350
A delivery truck was purchased for $60,000 and is expected to be used for 5 years and 100,000 miles. The truck’s residual value is $10,000. By the end of the first year, the truck has been driven 16,000 miles. What is the depreciation expense in the first year using activity-based depreciation?
Answer:
Step-by-step explanation:
Cost of Truck = $60,000
Expected use = 100,000 miles
Residual value = $10,000
Depreciation per mile = ( Cost of Truck- Residual value)/ Estimated use
= ( 60,000-10,000)/100,000
= 50,000/100,000
= $0.5
Miles driven in first year = 16,000
Depreciation expense for first year = Miles driven in first year x Depreciation per mile
= 16,000 x 0.5
= $8,000
The depreciation expense in the first year using activity-based depreciation is $ 8,000.
What is Depreciation ?An asset loses value over time as a result of use, damage, or obsolescence. Depreciation is the measurement for this decline.
Depreciation, or a decline in asset value, can be brought on by a variety of other variables, such as bad market conditions, etc.
Calculation of Depreciation for first year
Purchase Cost of Truck = $60,000
Estimated lifetime use = 100,000 miles
Salvage value = $10,000
Depreciation per mile = ( Cost of Truck- Residual value)/ Estimated use
= ( 60,000-10,000)/100,000
= 50,000/100,000
= $0.5
Miles driven in first year = 16,000
Depreciation for 1st year = Miles drove in first year x Depreciation per mile
= 16,000 x 0.5
= $8,000
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A photo of a mosquito in a science book is magnified to 635% of the mosquito's actual size.If the mosquito is 16 millimetres long,what is the length of the mosquito in the picture.
Answer:
The length of the mosquito in the picture is 117.6 mm.
Step-by-step explanation:
This question can be solved using a rule of three.
The real size of the mosquito is 16 mm, which is 100% = 1.
In the picture, the size is magnified 635%, so it is 100+635 = 735% = 7.35. So
16mm - 1
x mm - 7.35
x = 7.35*16 = 117.6
The length of the mosquito in the picture is 117.6 mm.
Final answer:
To find the magnified length of a mosquito in a picture, multiply its actual size (16 mm) by the magnification factor (6.35), resulting in a magnified length of 101.6 millimeters.
Explanation:
The question involves calculating the magnified length of a mosquito in a picture, given that the magnification is 635% of its actual size and the actual size is 16 millimeters. First, understand that 635% magnification means the mosquito's image is 6.35 times its actual size. To find the magnified length, multiply the actual length by the magnification factor.
Magnified length = Actual length × Magnification factor
= 16 mm × 6.35
= 101.6 mm
Therefore, the length of the mosquito in the picture is 101.6 millimeters.
Lucia hit a golf ball 240 feet. How many yards did she hit the ball?
A) 80 yards
B) 60 yards
C) 120 yards
D) 300 yards
Answer:
80 yards
Step-by-step explanation:
There are 3 feet in a yard, therefore you take 240 feet divided by 3 to get how many yards there are, and in your case it is 80 yards.
The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.6 pounds and standard deviation 1.9 pounds. A turkey farmer wants to provide a money-back guarantee that her 6-week poults will weigh at least a certain amount. What weight should she guarantee so that she will have to give her customer's money back only 1% of the time?
Answer:
She should guarantee a weight of 4.18 pounds.
Step-by-step explanation:
Problems of normally distributed samples can be solved using 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.
In this problem, we have that:
[tex]\mu = 8.6, \sigma = 1.9[/tex]
What weight should she guarantee so that she will have to give her customer's money back only 1% of the time?
She should guarantee the 1st percentile of weights, which is X when Z has a pvalue of 0.01. So it is X when Z = -2.327.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.327 = \frac{X - 8.6}{1.9}[/tex]
[tex]X - 8.6 = -2.327*1.9[/tex]
[tex]X = 4.18[/tex]
She should guarantee a weight of 4.18 pounds.
The turkey farmer should guarantee that her 6-week poults will weigh at least 4.2 pounds to ensure that she will have to give money back only 1% of the time.
Given:
- Mean weight [tex](\( \mu \))[/tex] of poults: 8. 6 pounds
- Standard deviation [tex](\( \sigma \))[/tex] of poults: 1.9 pounds
1. Use the z-score corresponding to the 1st percentile of the normal distribution, which is approximately [tex]\( z_{0.01} \approx -2.3263 \)[/tex].
2. Calculate the guaranteed weight X:
[tex]\[ X = \mu + z_{0.01} \cdot \sigma \] \[ X = 8.6 + (-2.3263) \cdot 1.9 \] \[ X \approx 4.2 \text{ pounds} \][/tex]
Therefore, the turkey farmer should guarantee that her 6-week poults will weigh at least 4.2 pounds to ensure that she will have to give her customers' money back only 1% of the time. This ensures that 99% of the poults will weigh at least 4.2 pounds.
hernando is choosing a random number between 0 and 9 state the number of successful outcomes for choosing an even number
There are five successful outcomes for choosing an even number between 0 and 9, which are 0, 2, 4, 6, and 8.
Hernando is choosing a random number between 0 and 9. The even numbers in this range are 0, 2, 4, 6, and 8. To determine the number of successful outcomes for choosing an even number, we simply count the even numbers listed.
There are five even numbers between 0 and 9 which are: 0, 2, 4, 6, and 8. Therefore, there are five successful outcomes for choosing an even number.
Varios estudiantes de una escuela pintaron 15 pancartas en 3 horas. Cuantas horas les tomará a los estudiantes pintar 60 pancartas
Answer:
The time it will take to paint 60 banners is 12 hours.
Step-by-step explanation:
The question is:
Several students from one school painted 15 banners in 3 hours. How many hours will it take for students to paint 60 banners
Solution:
The Unitary method is a mathematical procedure to determine the value of 1 unit from n unit values.
It is provided that the time it takes to paint 15 banners is, t = 3 hours.
Then the time it should have taken to paint 1 banner is, [tex]\frac{3}{15}=\frac{1}{5}\ hours[/tex].
Compute the time it will take to paint 60 banners as follows:
Time = [tex]60\times \frac{1}{5}=12[/tex]
Thus, the time it will take to paint 60 banners is 12 hours.
Rearrange this to make a the subject
Answer:
[tex]a = \frac{w + 4 - 3b}{6} [/tex]
Step-by-step explanation:
[tex]w = 3(2a + b) - 4 \\ w + 4 = 6a + 3b \\ w + 4 - 3b = 6a \\ \frac{w + 4 - 3b}{6} = \frac{6a}{6} \\ a = \frac{w + 4 - 3b}{6} [/tex]
Thanks
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An equilateral triangular plate with sides 10 m is submerged vertically in water so that the base is even with the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1000 kg/m3.)
Answer:
1225000
Step-by-step explanation:
The hydrostatic force against one side of the submerged equilateral triangle can be obtained by integrating the product of the fluid pressure and the area of an infinitesimally small strip on the triangle from base to top, yielding a result of 49,000 Newton.
Explanation:To express the hydrostatic force against one side of the equilateral triangular plate as an integral, we need to start with the equation for the pressure at any point in a fluid: P = ρgh. This is the product of the water density (ρ), gravitational acceleration (g), and height inside the fluid (h).
Next, we multiply this by the area of an infinitesimally small horizontal strip on our triangular plate, and then integrate to account for the entire side of the triangle, from the base to the top. In this problem, the area of the strip depends on the height, and for an equilateral triangle with a side length of 10m, it can be expressed as 10-2y.
Therefore, the hydrostatic force F becomes the integral ∫(0 to 10) ρg(10 - y)dy, where ρ is the weight density of water 1000 kg/m3, g stands for the acceleration due to gravity 9.8 m/s2, and y is the height varying from 0 to 10 m. Evaluating this integral, we get F = 49,000 Newton.
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Ben has a square garden that has sides measuring X feet, Ben wants to increase every side of his garden by 11 feet in order to plant more vegetables after Ben increased his garden it will have a perimeter of 104 feet, write an equation that can use the find X
We have been given that Ben has a square garden that has sides measuring X feet, Ben wants to increase every side of his garden by 11 feet in order to plant more vegetables. So each side of garden would be [tex]X+11[/tex].
We are also told that after Ben increased his garden it will have a perimeter of 104 feet.
We know that perimeter of square is 4 times the side length. So perimeter of garden after enlargement would be [tex]4(X+11)[/tex].
Now we will equate perimeter of new garden with 104 as:
[tex]4(X+11)=104[/tex]
Therefore, the equation [tex]4(X+11)=104[/tex] can be used to solve for X.
[tex]\frac{4(X+11)}{4}=\frac{104}{4}[/tex]
[tex]X+11=26[/tex]
[tex]X+11-11=26-11[/tex]
[tex]X=15[/tex]
Therefore, the value of X is 15 feet.
Design a regular grammar to generate the set of all integers beginning with the digit 3 such that the digits are consecutive and odd. If a digit is 9, its following digit (if present) will be 1. The set of valid strings is {3, 35, 357, 35791, 357913, ...}
Answer:
The grammar has start symbol S, terminals are 1,3,5,7 an 9 ; and the variables are S, A, B, C and D
Step-by-step explanation:
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There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In the second game you win if you can throw more than 80 % heads. For each game would you rather flip the coin 30 times or 300 times?
a) 300 times for each game
b. 30 times for each game
c) ç30 times for the first game and 300 times for the second
"d) 300 times for the first game and 30 times for the second
Answer:
d) 300 times for the first game and 30 times for the second
Step-by-step explanation:
We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.
As the coin is flipped more than one time and calculated the proportion, we have to use the sampling distribution of the sampling proportions.
The mean and standard deviation of this sampling distribution is:
[tex]\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}[/tex]
We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.
The probability of getting a proportion within this interval can be calculated as:
[tex]P(0.45<x<0.55)=P(z_L<z<z_H)[/tex]
referring the z values to the z-score of the standard normal distirbution.
We can calculate this values of z as:
[tex]z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)<0[/tex]
If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.
With this information, our chances of winning increase with the number of trials. We prefer for this game the option of 300 games.
For the second game, we win if we get a proportion over 80%.
The probability of winning is:
[tex]P(p>0.8)=P(z>z^*)[/tex]
The z value is calculated as before:
[tex]z^*=\dfrac{p^*-\mu_p}{\sigma_p}=\dfrac{p^*-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p^*-p)>0[/tex]
As (p*-p)=0.8-0.5=0.3>0, the value z* increase with the number of trials (N).
If our chances of winnings depend on P(z>z*), they become lower as z* increases.
Then, we can conclude that our chances of winning decrease with the increase of the number of trials.
We prefer the option of 30 trials for this game.
Final answer:
For the first game where 45% to 55% heads are needed, flipping the coin 300 times is beneficial due to the law of large numbers. For the second game requiring over 80% heads, flipping only 30 times is better to have a higher chance of deviation from the theoretical probability. So the correct option is d.
Explanation:
In determining whether you would rather flip a coin 30 times or 300 times for each game, one must consider the law of large numbers, which implies that as the number of trials increases, the experimental probabilities tend to get closer to the theoretical probabilities. In this case, a fair coin has a 50% chance of landing on heads on any given flip.
For the first game, where you win if you land between 45% and 55% heads, it's advantageous to flip the coin more times. This is because, with a larger number of flips (like 300), the results are more likely to converge to the expected probability, making it more probable you'll land within that range. Hence, you should choose 300 flips for the first game.
For the second game, where you win if you land more than 80% heads, you would prefer fewer flips, such as 30. It's less likely for a fair coin to consistently land on heads as the number of flips increases, so reducing the number of flips increases the chance of a more significant deviation from the expected probability.
The correct answer, therefore, would be option d: 300 times for the first game and 30 times for the second game.
The amount of time a certain brand of light bulb lasts is normally distribued with a mean of 1400 hours and a standard deviation of 55 hours. Using the empirical rule, what percentage of light bulbs last between 1345 hours and 1455 hours?
Answer:
By the Empirical Rule, 68% of light bulbs last between 1345 hours and 1455 hours
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 1400 hours
Standard deviation = 55 hours
Using the empirical rule, what percentage of light bulbs last between 1345 hours and 1455 hours?
1345 = 1400 - 1*55
So 1345 is one standard deviation below the mean.
1455 = 1400 + 1*55
So 1455 is one standard deviation above the mean.
By the Empirical Rule, 68% of light bulbs last between 1345 hours and 1455 hours
Solve for x; X/3 = 2/9
Step-by-step explanation:
x/ 3 = 2 / 9
Cross multiply
9x = 6
x = 6 / 9
x = 2 / 3
The number 20 is no less the difference of 4 times the number C and 8
The mathematical statement translates to an inequality 20 ≤ 4C - 8, and solving this results in C ≥ 7.
Explanation:The question provides a mathematical statement which can be translated into an equation. The statement 'The number 20 is no less the difference of 4 times the number C and 8' translates into the equation 20 ≤ 4C - 8. In this equation, 4 times a number C minus 8 can result in a value that is at least 20. To find the value for C, you would need to isolate C. This is done by first adding 8 to both sides to get 28 ≤ 4C, then dividing by 4 to get C ≥ 7. C, therefore, could be any value that is 7 or greater.
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