Answer:
The answer is a population parameter.
Step-by-step explanation:
Population can include people, but other examples include objects, event, businesses, and so on. Population is the entire pool from which statistical sample is drawn.
A parameter is a value that describes a characteristics of an entire population, such as population mean, because you can almost never measure an entire population, you usually don't know the real value of a parameter.
Consider all possible sample of size N that can be drawn from a given population (either with or without replacement). For example, we can compute a statistics (such as the mean and the standard deviation ) that will vary from sample to sample. In this manner we obtain a distribution of statistics that is called Sampling distribution.
In statistics, a sampling distribution is the theoretical distribution of a sample statistic that arises from drawing all possible samples of a specific size from a population. It helps to quantify the variability and predictability of sample statistics when used as estimates for population parameters.
Explanation:A sampling distribution refers to the "distribution of a sample statistic". This is option C from your list. This term describes the probability distribution of a statistic based on a random sample. For example, if we study random samples of a certain size from any population, the mean score will form a distribution. This is the sampling distribution of the mean. Similarly, variance, standard deviations and other statistics also have sampling distributions. The purpose of a sampling distribution is to quantify the variation and uncertainty that arises when we use sample statistics (like the mean) to estimate population parameters (like the population mean).
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For each of the finite geometric series given below, indicate the number of terms in the sum and find the sum. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation.
A. 5+5(0.15)+5(0.15)2+⋯+5(0.15)14
number of terms =
value of sum =
B. 5(0.15)4+5(0.15)5+5(0.15)6+⋯+5(0.15)9
number of terms =
value of sum =
Answer:
a)Number of terms =15
Value of sum =
(b)Number of terms =6
Value of sum
Step-by-step explanation:
A.
nth term,
a=5, r=
Simce the bases are the same, the powers are equal.
n-1=14
n=14+1=15
Number of terms =
15
Value of sum =
B.
nth term,
a=,
Simce the bases are the same, the powers are equal.
4+n-1=9
n=9+1-4=6
Number of terms =
6
Value of sum
NOTE: For the sum, we desire an expression that gives the exact value, rather than entering an approximation
The proportion of baby boys born in the United States has historically been 0.508. You choose an SRS of 50 newborn babies and find that 45% are boys. Do ALL calculations to 5 decimal places before rounding.
Answer:
[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]
[tex]p_v =P(z<-0.82)=0.206[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508
Step-by-step explanation:
Data given and notation
n=50 represent the random sample taken
[tex]\hat p=0.45[/tex] estimated proportion of newborn boys babies
[tex]p_o=0.508[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95 (asumed)
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We assume that we want to check if the true proportion is less than 0.508.
Null hypothesis:[tex]p\geq 0.508[/tex]
Alternative hypothesis:[tex]p < 0.508[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-0.82)=0.206[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508
It is concluded that the null hypothesis H₀ is not rejected. There is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.
What is the z test statistic for one sample proportion?Suppose that we have:
n = sample size[tex]\hat{p}[/tex] = sample proportionp₀ = population proportion (hypothesised)Then, the z test statistic for one sample proportion is:
[tex]Z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}[/tex]
Making the hypothesis and performing the test, assuming that we want to test if the population mean proportion is < 0.508 at the level of significance 0.05.
(1) Null and Alternative HypothesesThe following null and alternative hypotheses for the population proportion needs to be tested:
[tex]H_0: p \geq 0.508 \\\\ H_a: p & < & 0.508 \end{array}[/tex]
This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.
(2) Rejection RegionBased on the information provided, the significance level is [tex]\alpha = 0.05[/tex], and the critical value for a left-tailed test is [tex]z_c = -1.64[/tex]
The rejection region for this left-tailed test is [tex]R = \{z: z < -1.645\}[/tex]
(3) Test StatisticsThe z-statistic is computed as follows:
[tex]z & = & \displaystyle \frac{\hat p - p_0}{\sqrt{ \displaystyle\frac{p_0(1-p_0)}{n}}} \\\\& = & \displaystyle \frac{0.45 - 0.508}{\sqrt{ \displaystyle\frac{ 0.508(1 - 0.508)}{50}}} \\\\ & = & -0.82[/tex]
(4) Decision about the null hypothesisSince it is observed that [tex]z = -0.82 \ge z_c = -1.645[/tex], it is then concluded that the null hypothesis is not rejected.
Thus, it is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.
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Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 41 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6 comma 683 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than StartFraction 6 comma 683 l b Over 41 EndFraction equals 163 lb. What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.
Answer:
The probability that the plane is oveloaded is P=0.9983.
The pilot should take out the baggage and send it in another plain or have less passengers in the plain to not overload.
Step-by-step explanation:
The aircraft will be overloaded if the mean weight of the passengers is greater than 163 lb.
If the plane is full, we have 41 men in the plane. This is our sample size.
The weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.
So the mean of the sample is 180.5 lb (equal to the population mean).
The standard deviation is:
[tex]\sigma=\frac{\sigma}{\sqrt{N}} =\frac{38.2}{\sqrt{41}}=\frac{38.2}{6.4} =5.97[/tex]
Then, we can calculate the z value for x=163 lb.
[tex]z=\frac{x-\mu}{\sigma}=\frac{163-180.5}{5.97}=\frac{-17.5}{5.97}= -2.93[/tex]
The probability that the mean weight of the men in the airplane is below 163 lb is P=0.0017
[tex]P(\bar X<163)=P(z<-2.93)=0.00169[/tex]
Then the probability that the plane is oveloaded is P=0.9983:
[tex]P(overloaded)=1-P(X<163)=1-0.0017=0.9983[/tex]
The pilot should take out the baggage or have less passengers in the plain to not overload.
Minute Maid states that a bottle of juice contains 473 mL. Consumer groups are interested in determining if the bottles contain less than the amount stated on the label. To test their claim, they sample 30 bottles. The sample mean was 472mL and the standard deviation is 0.2. What does mu represent here? Group of answer choices The average contents of all bottles of juice in the population, which is 472mL. The average contents of all bottles of juice in the sample, which is unknown. The average contents of all bottles of juice in the sample, which is 472mL. The average contents of all bottles of juice in the population, which is unknown.
Answer:
The average contents of all bottles of juice in the population
[tex]\mu = 473\text{ mL}[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 473 mL
Sample mean, [tex]\bar{x}[/tex] = 472 mL
Sample size, n = 30
Sample standard deviation, s = 0.2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 473\text{ mL}\\H_A: \mu < 473\text{ mL}[/tex]
Representation of [tex]\mu[/tex]
It is the population parameter for mean.Thus, it represents the average contents of all bottles of juice in the population, which is 473 mL.The average juice a bottle contain is the mean value of the juice.
[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population, which is 472mL.
The given parameters are:
[tex]\mathbf{n = 30}[/tex] --- the sample size
[tex]\mathbf{\sigma = 0.2}[/tex] --- the standard deviation
[tex]\mathbf{\bar x = 472}[/tex] --- the sample mean
[tex]\mathbf{\mu = 473}[/tex] --- the population mean
The above highlights means that:
The parameter [tex]\mathbf{\mu }[/tex] represents the population mean
This means that:
[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population.
From the question, the value is given as: 473
Hence, the true statement is:
[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population, which is 473mL.
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At a large bank, account balances are normally distributed with a mean of $1,637.52 and a standard deviation of $623.16. What is the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650?
Answer:
[tex]P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)[/tex]
And we can use the complement rule and we got:
[tex]P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the bank account balances of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(1637.52,623.16)[/tex]
Where [tex]\mu=1637.52[/tex] and [tex]\sigma=623.16[/tex]
Since the distribution of X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using this formula we got:
[tex]P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)[/tex]
And we can use the complement rule and we got:
[tex]P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344[/tex]
Answer: the probability is 0.49
Step-by-step explanation:
Since the account balances at the large bank are normally distributed.
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = account balances.
µ = mean account balance.
σ = standard deviation
From the information given,
µ = $1,637.52
σ = $623.16
We want to find the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650. It is expressed as
P(x > 1650) = 1 - P(x ≤ 1650)
For x = 1650,
z = (1650 - 1637.52)/623.16 = 0.02
Looking at the normal distribution table, the probability corresponding to the z score is 0.51
P(x > 1650) = 1 - 0.51 = 0.49
A teen chewing bubble gum blows a huge bubble, the volume of which satisfies the differential equation:dV/dt=2(V)^(0.5)Where t is the time in seconds after the teen start to blow the bubble. If the bubble pops as soon as it reaches 529 cubic centimeters in volume, how many seconds does it take for the bubble to pop? Assume that the bubble had no volume when the teen first started blowing.
The ODE is separable:
[tex]\dfrac{\mathrm dV}{\mathrm dt}=2\sqrt V\implies\dfrac{\mathrm dV}{2\sqrt V}=\mathrm dt[/tex]
Integrate both sides to get
[tex]\sqrt V=t+C[/tex]
Assuming the bubble starts with zero volume, so that [tex]V(0)=0[/tex], we find
[tex]\sqrt0=0+C\implies C=0[/tex]
Then the volume of the bubble at time [tex]t[/tex] is
[tex]V(t)=t^2[/tex]
and so the time it take for the bubble to reach a volume of 629 cubic cm is
[tex]t^2=529\implies t=23[/tex]
or 23 seconds after the teen first starts blowing the bubble.
A relief fund is set up to collect donations for the families affected by recent storms. A random sample of 400 people shows that 28% of those 200 who were contacted by telephone actually made contributions compared to only 18% of the 200 who received first class mail requests. Which is the correct 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail
Answer:
[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181[/tex]
[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182[/tex]
And the 95% confidence interval would be given (0.0181;0.181).
We are confident at 95% that the difference between the two proportions is between [tex]0.0181 \leq p_B -p_A \leq 0.182[/tex]
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
[tex]p_A[/tex] represent the real population proportion for telephone
[tex]\hat p_A =0.28[/tex] represent the estimated proportion for telephone
[tex]n_A=200[/tex] is the sample size required for telephone
[tex]p_B[/tex] represent the real population proportion for mail
[tex]\hat p_B =0.18[/tex] represent the estimated proportion for mail
[tex]n_B=200[/tex] is the sample size required for mail
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181[/tex]
[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182[/tex]
And the 95% confidence interval would be given (0.0181;0.181).
We are confident at 95% that the difference between the two proportions is between [tex]0.0181 \leq p_B -p_A \leq 0.182[/tex]
The 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail is between 1.14% and 18.86%.
Explanation:To calculate a 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail, we use the formula for comparing two proportions.
We are given that 28% of the 200 contacted by telephone donated, which is a proportion of p1 = 0.28, and 18% of the 200 contacted by first class mail donated, which is a proportion of p2 = 0.18. Each group size is n1 = n2 = 200.
The formula for the standard error of the difference between two independent proportions is:
SE = sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)
So, the standard error (SE) for our example is:
SE = sqrt(0.28(1-0.28)/200 + 0.18(1-0.18)/200) = sqrt(0.001248 + 0.000792) = sqrt(0.00204)
The Z-score for a 95% confidence interval is approximately 1.96. Therefore, the margin of error (MOE) is calculated as:
MOE = 1.96 * SE
We then compute the confidence interval as:
(p1 - p2) ± MOE = (0.28 - 0.18) ± 1.96 * sqrt(0.00204)
Calculate the MOE:
MOE = 1.96 * sqrt(0.00204) ≈ 1.96 * 0.0452 ≈ 0.0886
Then the 95% confidence interval is:
(0.10 ± 0.0886) which is (0.0114, 0.1886)
This means that we are 95% confident that the true difference in the proportion of people who would donate when contacted by telephone versus first class mail is between 1.14% and 18.86%.
HELP PLEASE 100 POINTS! Review the diagram below. Apply the properties of angles to solve for the missing angles. Angle x is ____ degrees. And please show how you got it.
Step-by-step explanation:
<y+60+58=180 (angle sum property)
<y+118=180
<y=180-118=62
<ACB+<ACD=180(degree of a line )
58+x=180
x=122
(btw u cant get a 100 points by answering one question)
In a city, the distance between the library and the police station is 3 miles less than twice the distance between the police
station and the fire station. The distance between the library and the police station is 5 miles. How far apart are the police
station and the fire station?
0
O
O
O
1 mile
3 miles
4 miles
6 miles
Answer: the distance between the the police station and the fire station is 4 miles.
Step-by-step explanation:
Let x represent the distance between the library and the police station.
Let y represent the distance between the the police station and the fires station.
In the city, the distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station. This is expressed as
x = 2y - 3
The distance between the library and the police station is 5 miles. This means that
5 = 2y - 3
2y = 5 + 3 = 8
y = 8/2 = 4
2. The Hereford Cattle Society says that the mean weight of a one-year-old Hereford bull is 1135 pounds, with a standard deviation of 97 pounds. Suppose 40 bulls are randomly selected and loaded on a train car. Find the probability their combined weight exceeds 46000 pounds. (Hint: The combined weight exceeds 46000 pounds if the average weight exceeds 46000 40 = 1150 pounds.)
Answer:
16.35% probability their combined weight exceeds 46000 pounds.
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are 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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 1135, \sigma = 97, n = 40, s = \frac{97}{\sqrt{40}} = 15.34[/tex]
Find the probability their combined weight exceeds 46000 pounds.
This is 1 subtracted by the pvalue of Z when [tex]X = \frac{46400}{40} = 1150[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1150 - 1135}{15.34}[/tex]
[tex]Z = 0.98[/tex]
[tex]Z = 0.98[/tex] has a pvalue of 0.8365
1 - 0.8365 = 0.1635
16.35% probability their combined weight exceeds 46000 pounds.
Final answer:
The probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds is approximately 16.35%. This calculation uses the Z-score related to the sample mean weight exceeding 1,150 pounds.
Explanation:
To find the probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds, we first need to determine if the average weight per bull exceeds 1,150 pounds (since 46,000 ÷ 40 = 1,150). The mean weight of a Hereford bull is given as 1,135 pounds with a standard deviation of 97 pounds. Using the Central Limit Theorem, we can find the mean and standard deviation for the sample mean weight of 40 bulls.
The standard deviation of the sample mean (σ_x-bar) is σ ÷ √{n} = 97 pounds ÷ √{40}, which equals approximately 15.34 pounds. To find the Z-score, we use the formula Z = (X - μ) ÷ σ_x-bar, where X is 1,150 pounds, μ (the mean) is 1,135 pounds, and σ_x-bar is 15.34 pounds, yielding a Z-score of about 0.98.
Using the standard normal distribution table, a Z-score of 0.98 corresponds to a probability of approximately 0.8365. However, since we want the probability that the sample mean is greater than 1,150 pounds, we need to look at the upper tail, which is 1 - 0.8365 = 0.1635. Therefore, there's a 16.35% chance the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds.
Find the function r that satisfies the given condition. r'(t) = (e^t, sin t, sec^2 t): r(0) = (2, 2, 2) r(t) = ()
Answer:
r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2)
Step-by-step explanation:
A primitive of e^t is e^t+c, since r(0) has 2 in its first cooridnate, then
e^0+c = 2
1+c = 2
c = 1
Thus, the first coordinate of r(t) is e^t + 1.
A primitive of sin(t) is -cos(t) + c (remember that the derivate of cos(t) is -sin(t)). SInce r(0) in its second coordinate is 2, then
-cos(0)+c = 2
-1+c = 2
c = 3
Therefore, in the second coordinate r(t) is equal to -cos(t)+3.
Now, lets see the last coordinate.
A primitive of sec²(t) is tan(t)+c (you can check this by derivating tan(t) = sin(t)/cos(t) using the divition rule and the property that cos²(t)+sin²(t) = 1 for all t). Since in its third coordinate r(0) is also 2, then we have that
2 = tan(0)+c = sin(0)/cos(0) + c = 0/1 + c = 0
Thus, c = 2
As a consecuence, the third coordinate of r(t) is tan(t) + 2.
As a result, r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2).
To find the function r(t) that satisfies the given condition, integrate each component of r'(t) and solve for the constants using the initial condition.
Explanation:To find the function r(t) that satisfies the condition r'(t) = (e^t, sin t, sec^2 t) and r(0) = (2, 2, 2), we need to integrate each component of r'(t) with respect to t.
The integral of e^t with respect to t is e^t + C_1, where C_1 is a constant. The integral of sin t with respect to t is -cos t + C_2, where C_2 is a constant.The integral of sec^2 t with respect to t is tan t + C_3, where C_3 is a constant.Combining these integrals, we get r(t) = (e^t + C_1, -cos t + C_2, tan t + C_3). Substituting the initial condition r(0) = (2, 2, 2), we can solve for the constants and obtain the function r(t) = (e^t + 2, -cos t + 2, tan t + 2).
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A city park employee collected 1,850 cents in nickels, dimes, and quarters at the bottom of a wishing well. There were 30 nickels, and a combined total of 110 dimes and quarters. How many dimes and quarters were at the bottom of the wishing well?
dimes ??
quarters ???
can someone tell me how to solve this problem, ive been getting it wrong. I think I've missed a step maybe.
Answer: there are 70 dimes and 40 quarters.
Step-by-step explanation:
A dime is 10 cents
A nickel is 5 cents
A quarter is 25 cents
Let x represent the number of dimes at the bottom.
Let y represent the number of quarters at the bottom.
The city park employee collected 1,850 cents in nickels, dimes, and quarters at the bottom of a wishing well. If there were 30 nickels, it means that
10x + 25y + 30 × 5 = 1850
10x + 25y = 1850 - 150
10x + 25y = 1700- - - - - - - - -1
There is a combined total of 110 dimes and quarters. It means that
x + y = 110
Substituting x = 110 - y into equation 1, it becomes
10(110 - y) + 25y = 1700
1100 - 10y + 25y = 1700
1100 + 15y = 1700
15y = 1700 - 1100
15y = 600
y = 600/15
y = 40
x = 110 - y = 110 - 40
x = 70
Use the properties of limits to help decide whether the limit exists. If the limit exists, find its value. ModifyingBelow lim With x right arrow infinity StartFraction 6 x cubed plus 5 x minus 7 Over 6 x Superscript 4 Baseline minus 4 x cubed minus 9 EndFraction
Answer:
The value of given limit problem is 0.
Step-by-step explanation:
The given limit problem is
[tex]lim_{x\rightarrow \infty}\dfrac{6x^3+5x-7}{6x^4-4x^3-9}[/tex]
We need to find the value of given limit problem.
Divide the numerator and denominator by the leading term of the denominator, i.e., [tex]x^4[/tex]
[tex]lim_{x\rightarrow \infty}\dfrac{\frac{6x^3+5x-7}{x^4}}{\frac{6x^4-4x^3-9}{x^4}}[/tex]
[tex]lim_{x\rightarrow \infty}\dfrac{\frac{6}{x}+\frac{5}{x^3}-\frac{7}{x^4}}{6-\frac{4}{x}-\frac{9}{x^4}}[/tex]
Apply limit.
[tex]\dfrac{\frac{6}{ \infty}+\frac{5}{ \infty}-\frac{7}{ \infty}}{6-\frac{4}{ \infty}-\frac{9}{ \infty}}[/tex]
We know that [tex]\frac{1}{\infty}=0[/tex].
[tex]\dfrac{0+0-0}{6-0-0}[/tex]
[tex]\dfrac{0}{6}[/tex]
[tex]0[/tex]
Hence, the value of given limit is 0.
The limit of the expression (6x^3 + 5x - 7)/(6x^4 - 4x^3 - 9) as x approaches infinity is 0.
Explanation:In mathematics, when we are asked to find the limit of an expression as x approaches infinity, we can use a technique known as the 'highest powers' method. This involves dividing all terms in the function by the highest power of x in the denominator. In this expression, the highest power of x in the denominator is x4. So, we'll divide all terms by x4.
The given expression is: (6x3+5x-7) / (6x4-4x3-9). Dividing all terms by x4, we get: (6/x+5/x3-7/x4) / (6-4/x-9/x4)
As x approaches infinity, all terms that have x in the denominator go to 0. So, our expression simplifies to (0/6) = 0. Hence, the limit of the given expression as x approaches infinity is 0.
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Dave’s Automatic Door, referred to in Exercise 29, installs automatic garage door openers. Based on a sample, following are the times, in minutes, required to install 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.
The question is not complete and the full question says;
Calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.
Answer:
A) Range = 30 minutes
B) Mean = 38
C) Mean Deviation = 7.2
D) This is well written in the explanation.
Step-by-step explanation:
A) In statistics, Range = Largest value - Smallest value. From the question, the highest time is 54 minutes while the smallest time is 24 minutes.
Thus; Range = 54 - 24 = 30 minutes
B) In statistics,
Mean = Σx/n
Where n is the number of times occurring and Σx is the sum of all the times occurring
Thus,
Σx = 28 + 32 + 24 + 46 + 44 + 40 + 54 + 38 + 32 + 42 = 380
n = 10
Thus, Mean(x') = 380/10 = 38
C) Mean deviation is given as;
M.D = [Σ(x-x')]/n
Thus, Σ(x-x') = (28-38) + (32-38) + (24-38) + (46-38) + (44-38) + (40-38) + (54-38) + (38-38) + (32-38) + (42-38) = 72
So, M.D = 72/10 = 7.2
D) The range of the times is 30 minutes.
The average time required to open one door is 38 minutes.
The number of minutes the time deviates on average from the mean of 38 minutes is 7.2 minutes
The average time to install an automatic garage door opener, based on the provided data, is approximately 38 minutes, though individual times can vary.
Explanation:The subject of this question is Mathematics, specifically statistics. The question provides a series of data points representing the time, in minutes, that it took to install 10 automatic garage door openers. To find the average installation time, you would add up all of the times and then divide by the number of data points, which in this case is 10.
The data points are: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42. So, average installation time = (28+32+24+46+44+40+54+38+32+42) / 10 = 38 minutes.
This suggests that on average, it takes about 38 minutes to install an automatic garage door opener. But remember, this is an average. Individual installation times can vary greatly, as seen in the range of times provided.
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The mean and the standard deviation of the sampled population are 225.8and 17.8 respectively. What is the mean and standard deviation of the sample mean when n = 25?
Answer:
Mean and standard deviation of the sample mean are 225.8 and 3.56 respectively.
Step-by-step explanation:
The mean (μₓ) and standard deviation of the sample mean (σₓ) are related to the mean (μ) and standard deviation of the population (σ) through the following relationship
μₓ = μ = 225.8
σₓ = σ/√n = 17.8/√25 = 17.8/5 = 3.56
A simple random sample of 26 precipitation amounts has a standard deviation of 0.18. Find the test statistic, using a 0.01 significance level to test a claim that the standard deviation of all precipitation amounts is equal to 0.25
Answer:
[tex] t=(26-1) [\frac{0.18}{0.25}]^2 =12.96[/tex]
What is the critical value for the test statistic at an α = 0.01 significance level?
Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.
We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be [tex]\Chi^2 =10.520[/tex]
And the right critical value would be : [tex]\Chi^2 =46.927[/tex]
And the rejection zone would be: [tex] \chi^2 < 10.52 \cup \chi^2 >46.927[/tex]
Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.
Step-by-step explanation:
Previous concepts and notation
The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".
n = 26 sample size
s= 0.18
[tex]\sigma_o =0.25[/tex] the value that we want to test
[tex]p_v [/tex] represent the p value for the test
t represent the statistic
[tex]\alpha=0.01[/tex] significance level
State the null and alternative hypothesis
On this case we want to check if the population standard deviation is equal to 0.25, so the system of hypothesis are:
H0: [tex]\sigma =0.25[/tex]
H1: [tex]\sigma \neq 0.25[/tex]
In order to check the hypothesis we need to calculate the statistic given by the following formula:
[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]
This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.
What is the value of your test statistic?
Now we have everything to replace into the formula for the statistic and we got:
[tex] t=(26-1) [\frac{0.18}{0.25}]^2 =12.96[/tex]
What is the critical value for the test statistic at an α = 0.01 significance level?
Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.
We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be [tex]\Chi^2 =10.520[/tex]
And the right critical value would be : [tex]\Chi^2 =46.927[/tex]
And the rejection zone would be: [tex] \chi^2 < 10.52 \cup \chi^2 >46.927[/tex]
Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.
At the 0.01 significance level, we fail to reject the null hypothesis, meaning there is not enough evidence to refute the claim that the standard deviation is 0.25.
To test the claim that the standard deviation of all precipitation amounts is equal to 0.25 using a 0.01 significance level, we use the chi-square test for standard deviation.
The null hypothesis (H0) states that the population standard deviation (σ) is 0.25. The alternative hypothesis (Ha) states that σ is not equal to 0.25.
The test statistic for a chi-square test is calculated using the formula:
Chi-square (χ²) = (n - 1)s² / σ₀²
Where:
n is the sample size (26)s is the sample standard deviation (0.18)σ₀ is the hypothesised population standard deviation (0.25)Substituting the values:
χ² = (26 - 1)(0.18)² / (0.25)²
Calculating this:
χ² = 25 * 0.0324 / 0.0625 = 12.96
This χ² value is compared to the critical values from the chi-square distribution table with (n-1) or 25 degrees of freedom at a 0.01 significance level. The two-tailed critical values for 25 degrees of freedom at 0.01 significance level are approximately 10.85 and 46.93.
Since 10.85 < 12.96 < 46.93, we fail to reject the null hypothesis.
Hence, at the 0.01 significance level, there is not enough evidence to reject the claim that the standard deviation of all precipitation amounts is 0.25.
A brewery's filling machine is adjusted to fill bottles with a mean of 32.7 oz. of ale and a variance of 0.003. Periodically, a bottle is checked and the amount of ale noted.
(a) Assuming the amount of fill is normally distributed, what is the probability that the next randomly checked bottle contains more than 32.73 oz? (Give your answer correct to four decimal places.)
(b) Let's say you buy 95 bottles of this ale for a party. How many bottles would you expect to find containing more than 32.73 oz. of ale? (Round your answer up to the nearest whole number.) bottles You may need to use the appropriate table in Appendix B to answer this question.
Answer:
(a) P(X>32.73) = 0.2912
(b) Out of 95 bottles, 28 would contain more than 32.73 oz of ale.
Step-by-step explanation:
(a) The amount of fill is normally distributed so we will calculate the z-score and then use it to find the probability using the normal distribution probability table.
Let the amount of fill be denoted by X. The z-score can be computed using the formula:
z = (X - μ)/σ
where μ = mean value of fill
σ = standard deviation of value of fill = √Variance
P(X>32.73) = 1 - P(X<32.73)
= 1 - P((X-μ)/σ < (32.73 - 32.7)/√0.003)
= 1 - P(z<0.55)
Using the normal distribution table in Appendix B, we can see the probability at z=0.55 is 0.7088. So,
P(X>32.73) = 1 - 0.7088
P(X>32.73) = 0.2912
(b) We are buying 95 bottles and we need to calculate how many of them contain more than 32.73 oz. For that, we will multiply the total number of bottles by the probability of finding more than 32.73 oz which we have calculated in (a).
95 * 0.2912 = 27.664
Rounding off to a whole number we get 28 bottles.
Out of 95 bottles, 28 would contain more than 32.73 oz of ale.
A company is considering producing some new Gameboy electronic games. Based on past records, management believes that there is a 70 percent chance that each of these will be successful and a 30 percent chance of failure. Market research may be used to revise these probabilities. In the past, the successful products were predicted to be successful based on market research 90 percent of the time. However, for products that failed, the market research predicted these would be successes 20 percent of the time. If market research is performed for a new product, what is the probability that the results indicate an unsuccessful market for the product and the product is actually unsuccessful
The probability that the market research predicts an unsuccessful market for the product and the product is actually unsuccessful is 24 percent. This is calculated using the concept of conditional probability and multiplying the probability of the product actually failing (30%) with the probability of market research correctly predicting a failure (80%).
Explanation:To calculate the probability that the market research indicates an unsuccessful market for the product and the product is actually unsuccessful, we first need to understand the concept of conditional probability, given certain conditions.
Here, there is a 30 percent chance that a product will fail. The market research incorrectly predicts success for failed products 20 percent of the time. Therefore, the market research predicts failure for failed products 80 percent of the time (100% - 20%).
By multiplying these probabilities together, we can find the joint probability that a product is unsuccessful and market research also predicts it to be unsuccessful.
The result is 0.30 x 0.80 = 0.24. Therefore, there is a 24 percent chance that the market research will predict a failure and the product will also actually be a failure.
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The probability that market research indicates an unsuccessful market and the product is unsuccessful is 24%.
To calculate the probability that the results indicate an unsuccessful market for the product and that the product is unsuccessful, we need to use Bayes' Theorem. Initially, we have the probability of a product being successful, P(Success) = 0.70, and the probability of a product being a failure, P(Failure) = 0.30. Furthermore, if the product is successful, market research predicts it successful 90% of the time. If the product is a failure, market research incorrectly predicts it as a success 20% of the time, meaning it correctly predicts failure 80% of the time (100% - 20%).
We are interested in P(Market Research Failure and Actual Failure). This is the probability that the market research indicates a failure and the product indeed fails. Using the probabilities provided:
P(Market Research Successful | Success) = 0.90P(Market Research Successful | Failure) = 0.20P(Market Research Failure | Success) = 0.10P(Market Research Failure | Failure) = 0.80Therefore, P(Market Research Failure and Actual Failure) can be calculated as:
P(Market Research Failure and Actual Failure) = P(Market Research Failure | Failure) x P(Failure) = 0.80 x 0.30 = 0.24 or 24%.
Suppose that you randomly select freshman from the study until you find one who replies "yes." You are interested in the number of freshmen you must ask. What is the probability that you will need to ask fewer than three freshmen
Answer:
Incomplete question
Complete question: in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies "yes." You are interested in the number of freshmen you must ask. What is the probability that you will need to ask fewer than three freshmen?
Answer: P = 0.08656
Step-by-step explanation:
Using the concept of geometric distribution
P(X=x) = p(1-p)^1-x
Where x is the number of freshmen asked and p is the probability of success
Given probability of success as 71.3%
Therefore, P = 0.713
Probability that you'll need to ask less than 3 freshmen is given as
P = p(1-p)^1-x
x = 3 and p = 0.713
P = 0.713(1-0.713)^1-3
P = 0.713(0.287)^-2
P =0.713×12.140
P = 8.656%
P = 0.08656
Executives at a large multinational company are being accused of spending too much time on email and not enough time with their employees whom they supervise. In order to see if this is true, the VP of Human Resources decides to do a comparison of the number of minutes per day that the top 100 executives spend on e-mail compared to the total minutes spent by everyone in the company. The average minutes spent per day by each employee was 96 minutes. The average for just the top 100 executives was 91 minutes with a standard deviation of 25 minutes. On average, are the Executives spending more time on e-mail than all the employees? If you wanted to use the the 1% level of significance to test your hypothesis, what would be the correct critical value?
Answer:
We conclude that there is not enough evidence to support the claim that company is spending too much time on email.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 96 minutes
Sample mean, [tex]\bar{x}[/tex] = 91 minutes
Sample size, n = 100
Alpha, α = 0.05
Sample standard deviation, s = 25 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \leq 96\text{ minutes}\\H_A: \mu > 96\text{ minutes}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{91 - 96}{\frac{25}{\sqrt{100}} } = -2[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 99 degree of freedom } = 1.6603[/tex]
Since,
The calculated test statistic is less than the critical value, we fail to reject the null hypothesis and accept it.
Thus, we conclude that there is not enough evidence to support the claim that company is spending too much time on email.
Suppose that the problem is modified to specify that the chairs produced should accompany the tables so that of one-third of the tables produced each must have 6 accompanying chairs, of another one-third of the tables each must have 4 accompanying chairs, and of the other one-third of the tables each must have 2 accompanying chairs. How would this constraint be written
Answer:
C = 4T
Step-by-step explanation:
The question is not completed, the complete question and the solution is attached in the file below
help, ill mark brainliest
Answer:
Marcus rents the car for 5 days
Step-by-step explanation:
For 5 days:
Plan A would equal $150 (30x5)
Plan B would equal $140 (125+15)
Plan b is cheaper only when Marcus rents the car for 5 days.
The daily sales at a convenience store produce a normal distribution with a mean of $1,250 and a standard deviation of $125. The probability that the sales on a given day at this store are less than $1,310 is:
Answer:
0.6844 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $1,250
Standard Deviation, σ = $125
We are given that the distribution of daily sales is a bell like shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find
P(sales less than $1,310)
[tex]P( x < 1310) = P( z < \displaystyle\frac{1310 - 1250}{125}) = P(z < 0.48)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 1310) =0.6844= 68.44\%[/tex]
0.6844 is the probability that sales on a given day at this store are less than $1,310.
The probability that sales on a given day at this store are less than $1,310 is [tex]68.44\%[/tex]
Probability:It is given that, mean [tex]\mu=1250[/tex] and deviation [tex]\sigma=125[/tex]
The z- score is given as,
[tex]z-score=\frac{x-\mu}{\sigma} \\\\z=\frac{1310-1250}{125}=0.48 \\\\[/tex]
We have to find probability that the sales on a given day at this store are less than $1,310
[tex]P(x < 1310)=P(z < 0.48)[/tex]
From z- value table.
[tex]P(x < 1310)=68.44\%[/tex]
The probability that sales on a given day at this store are less than $1,310 is [tex]68.44\%[/tex]
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Ian has decided to buy a new car for $22,000 and agreed to make monthly payments for three years at
6.5% compounded monthly.
a. How much is each payment?
b. How much total interest will he pay over the life of the loan?
Answer:
Step-by-step explanation:
a) We would apply the periodic interest rate formula which is expressed as
P = a/[{(1+r)^n]-1}/{r(1+r)^n}]
Where
P represents the monthly payments.
a represents the amount of the loan
r represents the annual rate.
n represents number of monthly payments. Therefore
a = $22000
r = 0.065/12 = 0.0054
n = 12 × 3 = 36
Therefore,
P = 22000/[{(1+0.0054)^36]-1}/{0.0054(1+0.0054)^36}]
22000/[{(1.0054)^36]-1}/{0.0054(1.0054)^36}]
P = 22000/{1.214 -1}/[0.0054(1.214)]
P = 22000/(0.214/0.0065556)
P = 22000/32.64
P = $674
b) The total amount that he would pay in 3 years is
674 × 36 = 24265
total interest paid over the life of the loan is
24265 - 22000 = $2265
You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from overfishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day
The question given is incomplete, I googled and got the complete question as below:
You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from over-fishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day. Use these data to address the following questions. Show your work.
a. What is the probability that your last pot will have the necessary 7 crabs?
b. What is the probability that your last pot will be empty?
Answer:
a. Probability = 0.0083
b. Probability = 0.0907
Step-by-step explanation:
This is Poisson distribution with parameter λ=2.4
a)
The probability that your last pot will have the necessary 7 crabs is calculated below:
P(X=7)= {e-2.4*2.47/7!} = 0.0083
b)
The probability that your last pot will be empty is calculated as:
P(X=0)= {e-2.4*2.40/0!} = 0.0907
The probability of getting 7 crabs in the last pot is approximately 0.0082, while the probability of the last pot being empty is roughly 0.0907.
To solve this problem, we need to use the Poisson distribution. The Poisson distribution is a probability distribution that can be used to predict the number of events occurring within a fixed interval of time or space. Given that the average (λ) number of crabs per pot is 2.4, we can proceed to solve for the probabilities.
a. Probability of having 7 crabs in the last pot:
Calculate the average number of crabs per pot: 2.4 crabs.Using a Poisson distribution with an average of 2.4, find the probability of getting 7 crabs: P(X=7) = [tex]e^{-2.4}*\frac{(2.4^7)}{7!}[/tex]Calculate the probability, which is approximately 0.0082 or 0.82%.b. Probability of the last pot being empty:
Using the same Poisson distribution, find the probability of getting 0 crabs: P(X=0) = [tex]e^{-2.4}*\frac{(2.4^0)}{0!}[/tex]Calculate the probability, which is approximately 0.0907 or 9.07%.The complete question is
You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from overfishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day
Use these data to address the following questions. Show your work.
a. What is the probability that your last pot will have the necessary 7 crabs?
b. What is the probability that your last pot will be empty?
Zenich Corp manufactures laptops. The waiting time is 72 minutes before the start of production and the manufacturing lead time is 160 minutes per laptop. What is its manufacturing cycle efficiency? (Round to the nearest whole percent.)
Answer:
The Manufacturing cycle efficiency = 69 %
Step-by-step explanation:
The waiting time is = 72 minute
The manufacturing lead time is ( [tex]T_{lead}[/tex] )= 160 minutes
Total cycle time ( [tex]T_{c}[/tex] ) = 160 + 72 = 232 minutes
Manufacturing cycle efficiency = [tex]\frac{T_{lead} }{T_{c} }[/tex]
⇒ Manufacturing cycle efficiency = [tex]\frac{160}{232}[/tex]
⇒ Manufacturing cycle efficiency = 69 %
Therefore, the Manufacturing cycle efficiency = 69 %
The distribution of water fleas (Daphnia) in a given water pond is fairly random and the population density is fairly constant. The average number of water fleas caught by sweeping the water a single time with a standard net is 3.7 individuals. If tomorrow a net will be used once in the pond what is the probability of catching: a) 5 individuals?b) at least 2 individuals?
Answer:
(a) The probability of catching 5 individuals in the pond is 0.1429.
(b) The probability of catching at least 2 individuals in the pond is 0.8838.
Step-by-step explanation:
Let X = number of water fleas caught by sweeping the water a single time.
The random variable X follows a Poisson distribution with parameter λ = 3.7.
The probability mass function of the Poisson distribution is:
[tex]P(X=x)=\frac{e^{-3.7}3.7^{x}}{x!}[/tex]
(a)
Compute the value of P (X = 5) as follows:
[tex]P(X=5)=\frac{e^{-3.7}3.7^{5}}{5!}=\frac{17.1443}{120} =0.142869\approx0.1429[/tex]
Thus, the probability of catching 5 individuals in the pond is 0.1429.
(b)
Compute the value of P (X ≥ 2) as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)
[tex]=1-\frac{e^{-3.7}3.7^{0}}{0!}-\frac{e^{-3.7}3.7^{1}}{1!}\\=1-0.0247-0.0915\\=0.8838[/tex]
Thus, the probability of catching at least 2 individuals in the pond is 0.8838.
What is the future value of $1,720 in 14 years assuming an interest rate of 7.25 percent compounded semiannually? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.) Future value
Answer:
The future value of $1,720 in 14 years assuming an interest rate of 7.25 percent compounded semiannually is $4,661.61
Step-by-step explanation:
The compound interest formula is given by:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.
In this problem, we have that
Semianually is twice a year, so [tex]n = 2[/tex].
Also, [tex]P = 1720, t = 14, r = 0.0725[/tex]
So
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
[tex]A = 1720*(1 + \frac{0.0725}{2})^{2*14}[/tex]
[tex]A = 4661.61[/tex]
The future value of $1,720 in 14 years assuming an interest rate of 7.25 percent compounded semiannually is $4,661.61
The future value of $1,720 in 14 years, given a 7.25% interest rate compounded semiannually, can be calculated using the formula for compound interest: FV = PV * (1 + rate/n)^(nt). Plugging in the values, the future value comes out to be $3,490.91.
Explanation:We can find the future value using the formula: Future Value = Present Value * (1 + rate/ n)^(n*t), where the Present Value is $1,720, the rate is 7.25% (or 0.0725 in decimal form), n is 2 (since it's compounded semiannually), and t, the time period, is 14 years.
So Future Value = $1,720 * (1 + 0.0725 / 2) ^ (2 * 14) .
Therefore, the future value of $1,720 in 14 years, with an interest rate of 7.25 percent compounded semiannually, would be $3,490.91
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3/5 of your class earned an A or a B last week. 1/4 of those teammates earned an A. what fraction of you class earned an A? Possible answers could be 4/9, 7/20, 2/1, 3/20 but which one is it?
Answer:
3/20
Step-by-step explanation:
3/5 earned an A or B = "3/5 of the class"1/4 of 3/5 earned an A = "3/5 of class divided by 4"[tex]\frac{3}{5}[/tex] ÷ 4 = [tex]\frac{3}{5}[/tex] [tex]* \frac{1}{4}[/tex] = [tex]\frac{3 * 1}{5 * 4}[/tex] = [tex]\frac{3}{20}[/tex]
To calculate the fraction of the class that earned an A, multiply 3/5 by 1/4, which equals 3/20. Thus, 3/20 of the class earned an A. This is the correct answer among the given options.
Explanation:Calculating the Fraction of a Class That Earned an A
To find out what fraction of the class earned an A, we start with the fact that 3/5 of the class earned an A or a B.
We also know that 1/4 of those students earned an A.
To determine the fraction of the entire class that earned an A, we multiply these two fractions together:
Fraction of class that earned an A = (3/5) × (1/4)
This multiplication gives us:
(3 × 1) / (5 × 4) = 3/20
Therefore, 3/20 of the class earned an A.
This fraction cannot be reduced further and it is also one of the options provided, confirming that it is the correct answer.
what is the area of a triangle expressed as a monomial
Answer:
[tex] 28x^2 [/tex]
Step-by-step explanation:
[tex]area \:o f \: \triangle = \frac{1}{2} \times 8x \times 7x \\ = 4x \times 7x \\ = 28 {x}^{2} [/tex]