Answer:
a. Sample proportion ^p= 0.12
b. It is appropiate.
c. [0.0447;0.1953]
d. [^p ± d]
Step-by-step explanation:
Hello!
Given the information I'll assume that the variable of study has a binomial distribution:
X~Bi(n;ρ)
The sample data:
n= 50
"Success" x= 6
Sample proportion ^p= x/n = 6/50 = 0.12
Now, your study variable has a binomial distribution, but remember that the Central Limit Theorem states that given a big enough sample size (usually n≥ 30) you can approximate the sample proportion distribution to normal.
Since the sample is 50 you can apply the approximation, your sample proportion will have the following distribution:
^p≈ N( p; [p(1 - p)]/n)
With E(^p)= p and V(^p)= [p(1 - p)]/n.
This allows you to estimate the population proportion per Confidence Interval using the Z-distribution:
[^p±[tex]Z_{1-\alpha /2}[/tex]*√(^p(1 - ^p)/n)]
Since you are estimating the value of p, you'll use the estimated standard deviation (i.e. with the sample proportion instead of the population proportion)
to calculate the interval.
At level 90% the interval is:
[0.12±1.64*√([0.12(1 - 0.12)]/50)]
[0.0447;0.1953]
The margin of error (d) of an interval is half its amplitude (a)
if a= Upper bond - Low bond
then d= (Upper bond - Low bond)/2
d= (0.1953-0.0447)/2
d= 0.0753
And since the interval structure is "estimator" -/+ "margin of error" you can write it as:
[^p ± d]
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Wire electrical-discharge machining (WEDM) is a process used to manufacture conductive hard metal components. It uses a continuously moving wire that serves as an electrode. Coating on the wire electrode allows for cooling of the wire electrode core and provides an improved cutting performance. An article gave the following sample observations on total coating layer thickness (in µm) of eight wire electrodes used for WEDM. 21 16 29 36 42 25 24 25
Calculate a 99% CI for the standard deviation of the coating layer thickness distribution. (Round your answers to two decimal places.) , Is this interval valid whatever the nature of the distribution? Explain.
Yes, there are enough data points for this interval to be valid.
No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
To calculate a 99% confidence interval for the standard deviation of the coating layer thickness distribution, use the sample variance, sample size, and chi-square distribution. The coating layer thickness should be approximately normally distributed for the confidence interval to be valid.
Explanation:To calculate a 99% confidence interval for the standard deviation of the coating layer thickness distribution, we can use the chi-square distribution. First, we need to calculate the sample variance and sample size. Then, we can use the chi-square distribution table to find the critical values for a 99% confidence interval with (n-1) degrees of freedom. Finally, we can calculate the confidence interval using the formula:
CI = sqrt((n - 1) * s^2 / X^2)
where CI is the confidence interval, n is the sample size, s^2 is the sample variance, and X^2 is the critical value from the chi-square distribution.
In this case, the coating layer thickness should be approximately normally distributed for the confidence interval to be valid.
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In the given case, the answer No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
To calculate the 99% confidence interval (CI) for the standard deviation of the coating layer thickness distribution, we can use the chi-square distribution.
Given that the sample size is small (n = 8), we'll use the chi-square distribution with n - 1 = 7 degrees of freedom.
The formula for the confidence interval is:
[tex]\[ \left( \sqrt{\frac{(n-1)S^2}{\chi^2_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)S^2}{\chi^2_{1-\alpha/2, n-1}}} \right) \][/tex]
Given the sample observations: 21, 16, 29, 36, 42, 25, 24, 25, we first need to calculate the sample standard deviation:
[tex]\[ S = \sqrt{\frac{\sum_{i=1}^{n}(X_i - \bar{X})^2}{n-1}} \][/tex]
Let's perform the calculations:
[tex]\[ S = \sqrt{\frac{(21-27.625)^2 + (16-27.625)^2 + \ldots + (25-27.625)^2}{7}} \][/tex]
[tex]\[ S \approx 9.38 \][/tex]
Now, plug the values into the formula:
[tex]\[ \left( \sqrt{\frac{7 \times 9.38^2}{18.48}}, \sqrt{\frac{7 \times 9.38^2}{2.17}} \right) \][/tex]
[tex]\[ \left( \sqrt{26.97}, \sqrt{229.78} \right) \][/tex]
[tex]\[ \left( 5.19, 15.15 \right) \][/tex]
Therefore, the 99% confidence interval for the standard deviation of the coating layer thickness distribution is approximately [tex]\( (5.19, 15.15) \)[/tex] micrometers.
Therefore, the answer is: No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
Let X equal the thickness of spearmint gum manufactured for vending machines. Assume that the distribution of X is N(mu, sigma^2). The target thickness is 7.5 hundreds of an inch. We shall test the null hypothesis H_0: mu=7.5 against a two-sided alternative hypothesis, using 10 observations. Let m denote the sample mean and S^2 denote the sample variance.
Define the test statistic in terms of m and s.
Answer:
[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=\sqrt{10} (\frac{m-7.5}{s})[/tex]
Step-by-step explanation:
1) Notation
n=10 represent the sample size
[tex]\bar X=m[/tex] represent the sample mean
[tex]s[/tex] represent the sample standard deviation
m represent the margin of error
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 null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"
The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean for the population is 7.5 or no, the system of hypothesis would be:
Null hypothesis:[tex]\mu =7.5[/tex]
Alternative hypothesis:[tex]\mu \neq 7.5[/tex]
We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=(\sqrt{10})\frac{m-7.5}{s}[/tex]
and we have our statistic in terms of m (mean) and the sample standard deviation s.
A sample of 20 cigarettes is tested to determine nicotine content and the average value observed was 1.2 mg. Compute a 99 percent two-sided confidence interval for the mean nicotine content of a cigarette if it is known that the standard deviation of a cigarette’s nicotine content is σ = .2 mg.
Answer: [tex]1.0848<\mu<1.3152[/tex]
Step-by-step explanation:
Confidence interval for population mean is given by :-
[tex]\overline{x}-z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}< mu< \overline{x}+ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]z_{\alpha/2}[/tex] = two -tailed z-value for [tex]{\alpha[/tex] (significance level)
n= sample size .
[tex]\sigma[/tex] = Population standard deviation.
[tex]\overline{x}[/tex] = Sample mean
By considering the given information , we have
[tex]\sigma=0.2\text{ mg}[/tex]
[tex]\overline{x}=1.2\text{ mg}[/tex]
n= 20
[tex]\alpha=1-0.99=0.01[/tex]
Using z-value table ,
Two-tailed Critical z-value : [tex]z_{\alpha/2}=z_{0.005}=2.576[/tex]
The 99 percent two-sided confidence interval for the mean nicotine content of a cigarette will be :-
[tex]1.2- (2.576)\dfrac{0.2}{\sqrt{20}}<\mu<1.2+ (2.576)\dfrac{0.2}{\sqrt{20}}\\\\=1.2- 0.1152<\mu<1.2+ 0.1152\\\\=1.0848<\mu<1.3152 [/tex]
Hence, the 99 percent two-sided confidence interval for the mean nicotine content of a cigarette: [tex]1.0848<\mu<1.3152[/tex]
I am graphing something with 29 points. I know that you have to label the points with a capital letter. However, there are only 26 letters in the English alphabet, so what do I name the remaining 3 points?
Answer:
a₁, a₂, a₃, . . .
Step-by-step explanation:
When we name something in Mathematics it is always advisable to number them as [tex]$ a_1, a_2, a_3, ...$[/tex].
Because the alphabets are only 26 in number we might run out of notations.
The simple logic behind the notion of [tex]$ a_1, a_2, a_3, ... $[/tex] is that the numbers have no end and we can number as many variables we want using this logic.
In this problem, you can continue your notation from [tex]$ a_{27}, a_{28}, a_{29} $[/tex] so that you don't have to make a change in your figure.
A class of 10 students hang up their coats when they arrive at school. Just before recess, the teacher hands one coat selected at random to each child. What is the expected number of children who get his or her own coat?
The expected number of children who get his or her own coat among 10 students using expected deviation will be 6.
Probability is defined as the possibility of the occurrence of an event.
Probability lies between 0 and 1.
A low standard deviation suggests that data are grouped around the mean, whereas a large standard deviation shows that data are more dispersed.
Given that:
Number of students = 10
The probability for 10 students is 0.1
The pay-off table is as follows:
The expected deviation can be calculated as:
[tex]E(x) = 1\times0.1 + 2\times0.1+ 3\times0.1+ 4\times0.1 + 5\times0.1 + 6\times0.1 + 7\times0.1 + 8\times0.1 + 9\times0.1 + 10\times0.1[/tex]
E(x) = 5.5
The expected number of children who get his or her own coat is 6 when rounded off.
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The table showing the relation between the number of students to its probability is shown below.
Final answer:
The expected number of children who will get their own coat when the coats are handed out randomly to 10 students is 6, based on the concept of expected value in probability.
Explanation:
To calculate the expected number of children who get their own coat among 10 students using expected deviation:
First, assign a probability to each possible outcome (number of children getting their own coat). Since there are 10 students and each has an equal probability of getting their own coat, the probability for each outcome is 0.1.
Next, multiply each outcome by its respective probability and sum them up. This gives the expected value (E(x)).
E(x) = (1 * 0.1) + (2 * 0.1) + (3 * 0.1) + (4 * 0.1) + (5 * 0.1) + (6 * 0.1) + (7 * 0.1) + (8 * 0.1) + (9 * 0.1) + (10 * 0.1)
E(x) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 + 0.8 + 0.9 + 1.0
E(x) = 5.5
When rounded off, the expected number of children who get their own coat is 6.
An article in Fortune (September 21, 1992) claimed that nearly one-half of all engineers continue academic studies beyond the B.S. degree, ultimately receiving either an M.S. or a Ph.D. degree. Data from an article in Engineering Horizons (Spring 1990) indicated that 117 of 484 new engineering graduates were planning graduate study. Are the data from Engineering Horizons consistent with the claim reported by Fortune? Use a = 0.10 in reaching your conclusions. Find the P-value. Give your answer. The true proportion of engineering students planning graduate studies significantly different from 0.5 at a = 0.10. The P-value is less than (choose the least possible).
Answer:
Since the p–value is less than the significance level, the null hypothesis is rejected. The true proportion of engineering students planning graduate studies significantly different from 0.5 at α=0.10
Step-by-step explanation:
Please see attachment
In a poll conducted by the Gallup organization in April 2013, 48% of a random sample of 1022 adults in the U.S. responded that they felt that economic growth is more important than protecting the environment. Calculate and interpret a 95% confidence interval for the proportion of all U.S. adults in 5 April 2013 who felt that economic growth is more important than protecting the environment. Make sure to include all steps.
Answer: 95% confidence interval would be (0.449, 0.511)
Step-by-step explanation:
Since we have given that
n = 1022
p = 48% = 0.48
We need to find the 95% confidence interval first.
z = 1.96
Margin of error would be
[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.48\times 0.52}{1022}}\\\\=0.031[/tex]
95% confidence interval would be
[tex]p\pm 0.031\\\\=(0.48-0.031,0.48+0.031)\\\\=(0.449,0.511)[/tex]
It means true proportion who felt that economic growth is more important than protecting the environment is within 0.449 and 0.511 using 95% confidence.
If an experimenter conducts a t test for independent means and rejects the null hypothesis, the correct interpretation is that: a. the variance of one sample is so much larger than the variance of the other sample that the variances of the parent populations must not have been the same after all b. the mean of one sample is statistically the same as the mean of the other sample, so they probably come from populations with equal means c. the samples were from populations that were actually dependent rather than independent d. the mean of one sample is so far from the mean of the other sample that the samples must come from populations with different means
Answer: C
Step-by-step explanation:
Rejecting the null hypothesis means we've found a significant difference in the means. That means the probability that we'd see means so far apart by chance is less than our threshold of significance.
A dairy scientist is testing a new feed additive. She chooses 13 cows at random from a large population of cows. She randomly assigns nold = 8 to get the old diet, and nnew = 5 to get the new diet including the additive. The cows are housed in 13 separated pens and each gets separate feed, with or without additive as appropriate. After two weeks, she picks a day and milks each cow using standard procedures and records the milk produced in pounds. The data are below:Old Diet: 43, 51, 44, 47, 38, 46, 40, 35New Diet: 47, 75, 85, 100, 58Let µnew and µold be the population mean milk productions for the new and old diets, respectively. She wishes to test: H0 : µnew vs µold = 0 vs. HA : µnew vs µold 6= 0, using α = 0.05.(a) Are the two populations paired or independent?
Answer:
Step-by-step explanation:
Hello!
The objective of this experiment is to test if a new feed + additive generates a better production of milk in cows. For this, the owner selects 13 cows and randomly separates them into two groups.
Group 1 has 8 cows that receive the new feed + additive.
Group 2 has 5 cows that were fed with the old feed.
After two weeks of feeding the animals with the different feeds, the production of milk of each group was recorded so that they can be compared.
Since you have two separate groups to wich at random two different treatments were applied and later the variable was measured, these two samples/groups are independent and the proper test to compare the population means of the milk production in both groups is a pooled t.
I hope this helps!
Please help if so thank you And explain
Answer:
x = 34°
Step-by-step explanation:
Given AC and BD are perpendicular bisectors, we can say that at point E, there are 4 right angles [perpendicular bisectors intersect to create 4 90 degree angles].
Now, if we look at the triangle AED, we know that it is a right triangle, meaning that angle E is a right angle.
Also,
We know sum of 3 angles in a triangle is 180 degrees. Thus, we can write:
∠A + ∠E + ∠D = 180
Note: Angle A and Angle D are just the half part of the diagram. More exactly we can write:
∠EAD + ∠ADE + ∠DEA = 180
Given,
∠EAD = 56
∠DEA = 90
We now solve:
∠EAD + ∠ADE + ∠DEA = 180
56 + ∠ADE + 90 = 180
146 + ∠ADE = 180
146 + x = 180
x = 180 - 146
x = 34°
A particle moves in a straight line and has acceleration given by a(t) = 6t + 2. Its initial velocity is v(0) = −5 cm/s and its initial displacement is s(0) = 7 cm. Find its position function, s(t). SOLUTION Since v'(t) = a(t) = 6t + 2, antidifferentiation gives
Answer: The required position function is [tex]s(t)=t^3+t^2-5t+7.[/tex]
Step-by-step explanation: Given that a particle moves in a straight line and has acceleration given by
[tex]a(t)=6t+2.[/tex]
The initial velocity of the particle is v(0) = −5 cm/s and its initial displacement is s(0) = 7 cm.
We are to find the position function s(t).
We know that the acceleration function a(t) is the derivative of the velocity function v(t). So,
[tex]v^\prime(t)=a(t)\\\\\Rightarrow v^\prime(t)=6t+2\\\\\Rightarrow v(t)=\int (6t+2) dt\\\\ \Rightarrow v(t)=3t^2+2t+A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Also, the velocity function v(t) is the derivative of the position function s(t). So,
[tex]s^\prime(t)=v(t)\\\\\Rightarrow s^\prime(t)=3t^2+2t+A\\\\\Rightarrow s(t)=\int(3t^2+2t+A) dt \\\\\Rightarrow s(t)=t^3+t^2+At+B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
From equation (i), we get
[tex]v(0)=0+0+A\\\\\Rightarrow A=-5,~\textup{where A is a constant}[/tex]
and from equation (ii), we get
[tex]s(0)=0+0+0+B\\\\\Rightarrow B=7,~\textup{where B is a constant}.[/tex]
Substituting the values of A and B in equation (ii), we get
[tex]s(t)=t^3+t^2-5t+7.[/tex]
Thus, the required position function is [tex]s(t)=t^3+t^2-5t+7.[/tex]
Velocity: [tex]\( v(t) = 3t^2 + 2t - 5 \)[/tex] cm/s. Position: [tex]\( s(t) = t^3 + t^2 - 5t + 7 \)[/tex] cm, starting at 7 cm with initial velocity of -5 cm/s.
let's solve this step by step.
Given that ( a(t) = 6t + 2 ), we need to find the velocity function ( v(t) ) by integrating the acceleration function with respect to time.
1. **Find velocity function ( v(t) )**:
[tex]\[ a(t) = \frac{dv}{dt} \][/tex]
So, integrating ( a(t) ) with respect to ( t ) will give us ( v(t) ):
[tex]\[ \int a(t) \, dt = \int (6t + 2) \, dt \][/tex]
[tex]\[ v(t) = \int (6t + 2) \, dt = 3t^2 + 2t + C_1 \][/tex]
Here, ( C_1 ) is the constant of integration.
Given that ( v(0) = -5 ) cm/s, we can find the value of ( C_1 ):
[tex]\[ v(0) = 3(0)^2 + 2(0) + C_1 = C_1 = -5 \][/tex]
So, [tex]\( v(t) = 3t^2 + 2t - 5 \).[/tex]
2. **Find position function ( s(t) )**:
We know that velocity is the rate of change of displacement. So, we need to integrate the velocity function with respect to time to find the position function.
[tex]\[ v(t) = \frac{ds}{dt} \][/tex]
Integrating ( v(t) ) with respect to ( t ) will give us ( s(t) ):
[tex]\[ \int v(t) \, dt = \int (3t^2 + 2t - 5) \, dt \][/tex]
[tex]\[ s(t) = \int (3t^2 + 2t - 5) \, dt = t^3 + t^2 - 5t + C_2 \][/tex]
Here, [tex]\( C_2 \)[/tex] is the constant of integration.
Given that [tex]\( s(0) = 7 \)[/tex] cm, we can find the value of [tex]\( C_2 \):[/tex]
[tex]\[ s(0) = (0)^3 + (0)^2 - 5(0) + C_2 = C_2 = 7 \][/tex]
So, [tex]\( s(t) = t^3 + t^2 - 5t + 7 \).[/tex]
Therefore, the position function of the particle is [tex]\( s(t) = t^3 + t^2 - 5t + 7 \)[/tex]cm.
A NHANES report gives data for 654 women aged 20–29 years. The mean BMI of these 654 women was x¯=26.8 . We treated these data as an SRS from a normally distributed population with standard deviation ????=7.5 . (a) Suppose that we had an SRS of just 100 young women. What would be the margin of error for 95% confidence?
Answer: Margin of error would be 1.47 for 95% confidence.
Step-by-step explanation:
Since we have given that
Mean = 26.8
Standard deviation = 7.5
n = 100
We need to find the margin of error for 95% confidence.
So, z = 1.96
So, the margin of error would be
[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{7.5}{\sqrt{100}}\\\\=\dfrac{14.7}{10}\\\\=1.47[/tex]
Hence, margin of error would be 1.47 for 95% confidence.
The margin of error for a 95% confidence interval for the mean BMI of 100 young women sampled from the NHANES data with a standard deviation of 7.5 would be approximately ±1.47.
Explanation:The margin of error for a 95% confidence interval is determined using the standard deviation and the sample size. Given that you have a standard deviation (σ) of 7.5 and a sample size (n) of 100, you can calculate the standard error (SE) using the formula SE = σ/√n. To compute your 95% margin of error, multiply the standard error by the z-score associated with a 95% confidence level, which is 1.96.
Apply these to the formulas:
First calculate the Standard Error (SE) = σ/√n = 7.5/√100 = 7.5/10 = 0.75 Then, Margin of Error = z-score * SE = 1.96 * 0.75 ≈ 1.47.
So, the margin of error for a 95% confidence interval for the mean BMI of 100 young women sampled would be approximately ±1.47.
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An electrical power company is looking to expand into a new market. Before they commit to supplying the new area with electricity, they would like know the mean daily power usage for homes there. However, measuring the daily power usage of every home is not practical. Thus, an experiment must be designed where a sample of homes will have their daily power usage measured. Determine the required sample size to ensure that the 95% confidence interval for the population mean daily power usage is not larger than ±5 kWh. That is, determine the minimum sample size such that the error between the sample mean �" and population mean µ does not exceed 5 kWh, with 95% confidence. Based on historical trends, the population standard deviation can safely be assumed to be 50 kWh.
Answer:
atleast 385
Step-by-step explanation:
Given that an electrical power company is looking to expand into a new market. Before they commit to supplying the new area with electricity, they would like know the mean daily power usage for homes there.
Population std deviation = [tex]\sigma = 50[/tex]
Sample size =[tex]n[/tex]
STd error of sample mean = [tex]\frac{50}{\sqrt{n} }[/tex]
Margin of error for 95% would be Critical value ( std error)
Here since population std dev is known we can use Z critical value= 1.96
[tex]1.96*\frac{50}{\sqrt{n} }<5\\n>19.6^2\\n>384.16[/tex]
Sample size should be atleast 385
An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 45 hours. If a sample of 35 bulbs has an average life of 710 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.
Answer:
[693,38:725,62]hs
Step-by-step explanation:
Hello!
Your study variable is X: the lifespan of a light bulb. This variable is said to have an approximately normal distribution.
X≈N(μ;σ²)
Were
μ= populatiom mean
σ= 45 hs standard deviation
A sample of n= 35
To estimate the population mean with a confidence interval you have to use the Z statistic:
X{bar} ± [tex]Z_{1-\alpha /2}[/tex]*(σ/√n)
[tex]Z_{1-\alpha /2} = Z_{0.98} = 2.054[/tex]
710 ± 2.054*(45/√35
[693,38:725,62]hs
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You are dealt a hand of three cards, one at a time. Find the probability of each of the following. a) The first heart you get is the third card dealt. b) Your cards are all diamonds. c) You get no aces. d) You have at least one heart.
Final answer:
Calculating the probability of different card events when dealt a hand of three cards involves understanding basic principles of probability and operating with fractions. Each event requires calculating the probability of successive draws, considering that cards are dealt without replacement, impacting the probability of each subsequent draw.
Explanation:
Let's address each part of the question dealing with probabilities when being dealt a hand of three cards from a standard 52-card deck:
The first heart being the third card dealt: To have the first heart on the third card, the first two cards must be of any suit other than hearts. With 52 cards in the deck and 13 cards per suit, probabilities for the first two non-heart cards would be: P(non-heart first card) = 39/52 and P(non-heart second card given non-heart first) = 38/51. For the third card to be a heart, given the first two cards are not, the probability is P(heart third card given first two non-heart) = 13/50. Multiply these probabilities together to find the overall probability of this event.All cards are diamonds: For each card to be a diamond, the probability for each card dealt is P(diamond first card) = 13/52, P(diamond second card) = 12/51, and P(diamond third card) = 11/50. Multiply these three probabilities together for the overall probability.No aces: Since there are four aces in a deck, the probability for each card not being an ace is: P(no ace first card) = 48/52, P(no ace second card) = 47/51, and P(no ace third card) = 46/50. Multiply these three probabilities together for the overall probability.At least one heart: It is easier to calculate the probability of the opposite event that no hearts are dealt and subtract this from 1. P(no heart first card) = 39/52, P(no heart second card) = 38/51, and P(no heart third card) = 37/50. Multiply these three probabilities for the probability of no hearts, and subtract from 1 for the probability of getting at least one heart.Peter makes six cakes of 5 lb each three chocolate cakes and three vanilla cakes to serve at a party for 120 people. How many ounces of cake will each guest have? If the guest list increases to 150 guests, how many more cakes does Peter need to make?
Answer:
4 oz2 cakesStep-by-step explanation:
Since 6 cakes serve 120 people, each cake serves 120/6 = 20 people. Each cake weighs (5 lb)(16 oz/lb) = 80 oz. Then each person gets ...
(80 oz)/(20 persons) = 4 oz/person
__
150 servings will require 150/20 = 7.5 cakes. Peter already has 6 cakes, so needs to make 2 more.
5lb per cake *6 cakes= 30lb
30lb / 120 people = 0.25lb per person
150 people - 120 people = 30 people
0.25lb per person * 30 people = 7.5lb
5lb___1 cake
7.5lb___X=1.5 cakes
(7.5lb * 1 cake)/5lb = 1.5 cakes
An ant is moving on a numbered, horizontal line every second. The number ranges from −[infinity] to [infinity] . It moves to the left integer with a probability of 1/4 and to the right integer with a probability of 3/4. Suppose initially it starts at 0, so what is the probability that after 3 seconds it will be at 1?
The probability that the ant will be at 1 after 3 seconds is 49/64.
Explanation:To find the probability that the ant will be at 1 after 3 seconds, we need to consider all possible paths it can take. After each second, the ant can either move left with a probability of 1/4 or move right with a probability of 3/4.
Let's analyze all the possible paths:
The ant moves right in all three seconds - Probability = (3/4) * (3/4) * (3/4) = 27/64The ant moves right in the first two seconds and then moves left in the third second - Probability = (3/4) * (3/4) * (1/4) = 9/64The ant moves right in the first second, then moves left in the second second and right in the third second - Probability = (3/4) * (1/4) * (3/4) = 9/64The ant moves left in the first two seconds and then moves right in the third second - Probability = (1/4) * (1/4) * (3/4) = 3/64The ant moves left in all three seconds - Probability = (1/4) * (1/4) * (1/4) = 1/64Adding up the probabilities from each path, the total probability that the ant will be at 1 after 3 seconds is (27/64) + (9/64) + (9/64) + (3/64) + (1/64) = 49/64.
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Sergio and Lizeth have a very tight vacation budget. They plan to rent a car from a company that charges $75 a week plus $0.25 a mile. How many miles can they travel and still keep within their $200 budget?
Answer: 500 miles
Step-by-step explanation:
Given : Sergio and Lizeth have planned to rent a car from a company that charges $75 a week plus $0.25 a mile.
i.e. Fixed charge= $75
Rate per mile = $0.25
Let x denotes the number of miles.
Then, Total charges = Fixed charge+ Rate per mile x No. of miles traveled
= $75+ $0.25x
To keep budget within $200, we have following equation.
[tex]75+0.25x=200\\\\\Rightarrow\ 0.25=200-75\\\\\Rightarrow\ 0.25=125\\\\\Rightarrow\ x=\dfrac{125}{0.25}=\dfrac{12500}{25}=500[/tex]
Hence, they can travel 500 miles and still keep within their $200 budget.
The 20 colleges of interest to a high school senior include 8 that are expensive ( tuition more than 20,000 per year), 8 that are far from home( more than 200 miles away), and 7 that are both expensive and far from home. If the student decides to select a college that is not expensive and within 200 miles from home, how many selections are possible?
Answer: 6
Step-by-step explanation:
Let S= Total colleges
A = colleges are expensive.
B= colleges are far from home( more than 200 miles away).
Given : n(S)= 20
n(A)=8
n(B)=8
n(A∩B) =2
Then, the number of college that are not expensive and within 200 miles from home :-
[tex]n(A'\cap B')=n(S)-n(A\cup B)\\\\=20-(n(A)+n(B)-N(A\cap B))\ \ [\because\ n(A\cup B)=n(A)+n(B)-N(A\cap B)]\\\\=20-(8+8-2)\\\\=20-14=6[/tex]
i.e. the number of college that are not expensive and within 200 miles from home=6
Hence, the number of possible selections are 6 .
supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the dryers. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval.
The confidence interval for population mean (when population standard deviation is unknown) is given by :-
[tex]\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{s}{\sqrt{n}}[/tex]
, where n= sample size
[tex]\overline{x}[/tex] = Sample mean
s= sample size
t* = Critical value.
Given : n= 25
Degree of freedom : [tex]df=n-1=24[/tex]
[tex]\overline{x}= \$93.36[/tex]
[tex]s=\ $19.95[/tex]
Significance level for 98% confidence interval : [tex]\alpha=1-0.98=0.02[/tex]
Using t-distribution table ,
Two-tailed critical value for 98% confidence interval :
[tex]t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922[/tex]
⇒ The critical value that should be used in constructing the confidence interval = 2.4922
Then, the 95% confidence interval would be :-
[tex]93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu<93.36+(2.4922)\dfrac{19.95}{\sqrt{25}}[/tex]
[tex]=93.36-9.943878< \mu<93.36+9.943878[/tex]
[tex]=93.36-9.943878< \mu<93.36+9.943878[/tex]
[tex]=83.416122< \mu<103.303878\approx83.4161<\mu<103.3039[/tex]
Hence, the 98% confidence interval for the mean repair cost for the dryers. = [tex]83.4161<\mu<103.3039[/tex]
Evaluate: (2.4 x 104)(4.2 x 103)
Answer:
249.6×432.6=107,976.96
Answer:
Step-by-step explanation:
2.4*104=249.6
4.2*103=432.6
(249.6)(432.6)=107976.96
All a matter of simple multiplication. ;)
You roll a fair die three times. What is the probability of each of the following?
a) You roll all 4's.
b) You roll all even numbers.
c) None of your rolls gets a number divisible by 2.
d) You roll at least one 2.
e) The numbers you roll are not all 2's.
This solution calculates the probability of different outcomes when rolling a fair die three times. The results are obtained by defining the successful outcomes versus the total possible outcomes for each specific event.
Explanation:This question is about probability. A fair die has 6 equally likely outcomes. Let's address each part:
You roll all 4's: There's 1 chance in 6 to roll a 4. Since you're rolling the die three times, the probability is (1/6) * (1/6) * (1/6) = 1/216. You roll all even numbers: There are 3 even numbers on a die (2, 4, 6), so the probability is (3/6) * (3/6) * (3/6) = 1/8. None of your rolls gets a number divisible by 2: This is the same as rolling all odd numbers. There are 3 odd numbers on a die (1, 3, 5), so the probability is (3/6) * (3/6) * (3/6) = 1/8. You roll at least one 2: The opposite of this are outcomes without any 2, i.e., combinations of 1, 3, 4, 5, and 6. Thus, subtract combinations without any 2 from total possible combinations: 1 - [(5/6) * (5/6) * (5/6)] = 91/216. The numbers you roll are not all 2's: The only case when this does not happen is when you roll a 2 three times. So, if we subtract the probability of rolling three 2's from 1, we get 1 - (1/6) * (1/6) * (1/6) = 215/216.Learn more about probability here:
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The probability of rolling all 4's is 1/64. The probability of rolling all even numbers is 7/64. The probability of not rolling all 2's is 63/64.
Explanation:a) The probability of rolling all 4's is ⅛3, or 1/64. There is only one way to roll a 4 on a fair die and a total of 6 possible outcomes on each roll, so the probability is 1/6. Since the rolls are independent, the probability of getting a 4 on all three rolls is (1/6)(1/6)(1/6) = 1/64.
b) The probability of rolling all even numbers is also 1/64. There are 3 even numbers on a die (2, 4, and 6), so the probability of rolling an even number on any single roll is 3/6 or 1/2. Since the rolls are independent, the probability of getting an even number on all three rolls is (1/2)(1/2)(1/2) = 1/8. However, we need to subtract the probability of rolling all 4's from this, which is also 1/64. So the final probability is 1/8 - 1/64 = 7/64.
c) The probability of none of the rolls getting a number divisible by 2 is 1 - (1/2)(1/2)(1/2) = 1 - 1/8 = 7/8. This is the complement of rolling all even numbers.
d) The probability of rolling at least one 2 is 1 - the probability of rolling no 2's. The probability of not rolling a 2 on any single roll is 5/6, so the probability of not rolling a 2 on all three rolls is (5/6)(5/6)(5/6) = 125/216. Therefore, the probability of rolling at least one 2 is 1 - 125/216 = 91/216.
e) The probability of rolling all 2's is 1/64. Therefore, the probability of not rolling all 2's is 1 - 1/64 = 63/64.
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The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. Which is a reasonable range for this function?
F- {…1.71, 3, 4.29, 5.58, 6.87, …}
G- {4.29, 5.58, 6.87, …}
H- {1, 2, 3, …}
J- all positive real numbers
Final answer:
The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. The reasonable range for this function would be all positive real numbers.
Explanation:
The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. The reasonable range for this function depends on the context. Since the function represents the price of shipping, the range should be positive, as shipping cannot have a negative price.
Therefore, the reasonable range for this function would be J- all positive real numbers.
Final answer:
The reasonable range for the shipping cost function starts from 3 and includes values that increase by 1.29 for each additional item, corresponding to whole numbers of items shipped. Therefore, the answer is G - {4.29, 5.58, 6.87, ...}.
Explanation:
The function given is y = 1.29x + 3, which represents the price y that a website store charges for shipping x items. Since shipping cannot have a negative cost and the minimum number of items shipped is either zero or a positive integer, the reasonable range for this function would begin at the point where x is zero. Therefore, we start our range by calculating the shipping cost for zero items: 1.29(0) + 3 = 3. As x increases, the cost will also increase linearly according to the function. Hence, all subsequent shipping prices will be greater than 3.
Now, let's consider what the reasonable range for shipping items would be. It would be abnormal to have a fractional number of items shipped because items are discrete entities. Thus, the list of shipping prices should only include charges for whole numbers of items. So, the range should only include prices that correspond to whole numbers of items shipped.
As a result, the reasonable range would include values starting from y = 3 onwards at intervals of 1.29 times an integer value. Option G, which starts from 4.29 and increases at a constant rate of 1.29, represents these intervals since 4.29 is the price for shipping one item (1.29*1 + 3). Any positive number of items shipped will result in a corresponding shipping price that is greater than 3, and since it is discreetly incremented, the prices will form a sequence of specific numbers, not all positive real numbers. Therefore, option G - {4.29, 5.58, 6.87, ...} - is the most appropriate answer.
Can -2y = -6 be written in slope-intercept form? If so, what is it?
Answer:
y=3 or y-3=0
Step-by-step explanation:
-2y=-6
-2y÷(-2)=-6÷(-2)
y=-6÷(-2)
y=-6÷2
y=3
or
-2y=-6
-2y+6=0
-y+3=0
y-3=0
See picture for answer and solution steps.
A statistical program is recommended.
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data.
1.12 5.09 0.97 1.59 4.60 0.32 0.55 1.45
0.18 4.47 1.20 3.50 5.02 4.67 5.22 2.69
3.96 3.17 3.03 2.21 0.69 4.47 3.31 1.17
0.79 1.17 1.57 2.62 1.66 2.05
The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean are 2.484, 1.611, and 0.294, respectively. Does this data suggest that the true average percentage of organic matter in such soil is something other than 3%? Carry out a test of the appropriate hypotheses at significance level 0.10. [Note: A normal probability plot of the data shows an acceptable pattern in light of the reasonably large sample size.]
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)
t=
P-value=
Answer:
Which is the output of the formula =AND(12>6;6>3;3>9)?
A.
TRUE
B.
FALSE
C.
12
D.
9
Step-by-step explanation:
The police rounded up 35 people suspected of robbing a bank. Each person was assigned a number from 1 to 35. After a short investigation, three of them were arrested, numbers #14, #17 and #26. They made the following statements under intensive questioning: #14: I’m innocent. #17: I’m innocent. #26: #14 is guilty. If only one of these statements is true, who robbed the bank?
Answer:
If only one of these statements is true, #17 robbed the bank.
Step-by-step explanation:
1st Scenario: #26 tells the truth
If #26 is telling the truth, that means #14 is guilty and, consequently, lying. However, that would also mean that #17 is telling the truth and then more than one statement would be true.
2nd Scenario: #17 tells the truth
If #17 is telling the truth, #17 is innocent, but then again either #26 or #14 are lying and two statements would be true.
3rd Scenario: #14 tells the truth
If #14 is telling the truth, #14 is innocent and, consequently, #26 is lying. That leaves us with #17 claiming innocence, but since only #14 can be telling the truth, #17 is lying and robbed the bank.
In a random sample of 200 students, 55% indicated they have full-time jobs, while the other 45% have part-time jobs. Fifty of the 90 male students surveyed have a full-time job, and 60 of the females surveyed have a full-time job. What is the probability that a randomly selected student is female given they have a part-time job?
Answer:
The probability is [tex]\frac{5}{9}[/tex]
Step-by-step explanation:
The total number of students are 200.
number of full timers is 110 and number of part timers is 90.
number of male students is 90 and number of female students is 110.
Let the probability of part timers be P(B).
P(B) = [tex]\frac{90}{200}[/tex] = [tex]\frac{9}{20}[/tex]
Let the probability of female part timers be P(A)
P(A) = [tex]\frac{50}{200} = \frac{5}{20}[/tex]
now, the final probability is
= [tex]\frac{P(A)}{P(B)}[/tex]
=[tex]\frac{5/20}{9/20} = \frac{5}{9}[/tex]
A magazine currently has 8700 subscribers for its online web version. It is adding members at the rate of R(t) = 190e0.03t subscribers per month. If the proportion of members who remain subscribed t months from now is S(t) = e−0.06t, how many online subscribers will the magazine have three years from now? (Round your answer to the nearest integer.)
Number of subscriber the magazine will have after 3 years from now approximately be 8767
Solution:
Given that magazine currently has 8700 subscribers for its online web version
[tex]\begin{array}{l}{\mathrm{R}(\mathrm{t})=190 \mathrm{e}^{0.03 \mathrm{t}} \text { subscribers/month }} \\\\ {\mathrm{S}(\mathrm{t})=\mathrm{e}^{-0.06 \mathrm{t}}}\end{array}[/tex]
After 3 years, time(t) = 36 month
Total number of subscribers after 3 years from now :
Substitute "t" = 36
[tex]\begin{array}{l}{\mathrm{R}(36)=190 \mathrm{e}^{0.03 \times(36)}=190 \times(2.944)} \\\\ {\mathrm{R}(36) \approx 560} \\\\ {\mathrm{S}(36)=\mathrm{e}^{-0.06 \times(36)}=0.12}\end{array}[/tex]
Subscribers remaining = 0.12 x 560 = 67.2
The magazine currently has 8700 subscribers
Added Subscriber = 8700 + 560 = 9260
Remaining Subscriber = 8700 + 67.2 = 8767.2
Therefore number of subscriber the magazine will have after 3 years from now approximately be 8767
Please help me with question 24 and 26.
Answer:
Step-by-step explanation:
A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 48 specimens and counts the number of seeds in each. Use her sample results (mean = 36.9, standard deviation = 16.5) to find the 98% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
Answer: The open interval would be (31.4,42.5).
Step-by-step explanation:
Since we have given that
mean = 36.9
Standard deviation = 16.5
n = 48
At 98% confidence interval, z = 2.33
So, Interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=36.9\pm 2.33\dfrac{16.5}{\sqrt{48}}\\\\=36.9\pm 5.549\\\\=(36.9-5.5,36.9+5.6\\\\=(31.4,42.5)[/tex]
Hence, the open interval would be (31.4,42.5).
To find the 98% confidence interval for the number of seeds, use the formula for confidence intervals and the values given. The 98% confidence interval for the number of seeds is (31.1, 42.7).
Explanation:To find the 98% confidence interval for the number of seeds, we can use the formula:
Confidence Interval = mean ± (critical value) * (standard deviation / sqrt(sample size))
Since we want a 98% confidence interval, the critical value is found using the z-table. It is approximately 2.33.
Plugging in the values:
Confidence Interval = 36.9 ± (2.33) * (16.5 / sqrt(48))
Simplifying the expression gives us the 98% confidence interval for the number of seeds as (31.1, 42.7).
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