Answer:
(a)f(x)=0.8869x
(b)g(x)=143.1126x
(c)g(f(x))=126.9266x
(d)g(f(1000))=126926.6 Yen
Step-by-step explanation:
Given on a particular day
One U.S. dollar could purchase 0.8869 EurosOne Euro could purchase 143.1126 yen(a)If x represents the number of Dollars
Since one can purchase 0.8869 Euro with 1 USD, the function f(x) is a direct relationship where x is dollars and f(x) is in Euros.
f(x)=0.8869x(b)If x represents the number of Euros
Since one can purchase 143.1126 yen with 1 Euros, the function g(x) is a direct relationship where x is Euros and g(x) is in Yen.
g(x)=143.1126x(c)Given:
g(x)=143.1126xf(x)=0.8869xg(f(x))=143.1126(0.8869x)
g(f(x))=126.9266x(d)g(f(1000))
g(f(x))=126.9266xg(f(1000))=126.9266 X 1000 =126926.6 YenLori gets an offer from another bank that is also paying 6% on CD’s, but is compounding interest daily. How much will the CD be worth in 10 years?
Answer:
you have to include how much the cd is worth first
Step-by-step explanation:
you would do
cd price x 1.6 (to the power of 10)
Establish which of the following statements are true. (a) A sequence is convergent if and only if all of its subsequences are convergent. (b) A sequence is bounded if and only if all of its subsequences are bounded. (c) A sequence is monotonic if and only if all of its subsequences are monotonic. (d) A sequence is divergent if and only if all of its subsequences are divergent.
Answer:
Statement A - True.
Statement B - False.
Statement C - True.
Statement D - False.
Step-by-step explanation:
(a) A sequence is convergent if and only if all of its subsequences are convergent - this statement is correct.
(b) A sequence is bounded if and only if all of its subsequences are bounded - this statement is incorrect.
(c) A sequence is monotonic if and only if all of its subsequences are monotonic - this statement is correct.
(d) A sequence is divergent if and only if all of its subsequences are divergent - this statement is incorrect.
Final answer:
In the context of mathematical sequences, statements (a), (b), and (c) regarding convergence, boundedness, and monotonicity are true as all subsequences follow the property of the original sequence. However, statement (d) is false because even divergent sequences can have convergent subsequences.
Explanation:
When analyzing the behavior of sequences and series in mathematics, it's important to understand various characteristics such as convergence, boundedness, monotonicity, and divergence. We can assess the truthfulness of the given statements based on these characteristics.
Assessment of the Statements:
(a) True: A sequence is convergent if and only if all of its subsequences are convergent. This is a fundamental property of convergent sequences, implying that if the original sequence approaches a specific value, every subsequence will also approach that same value.
(b) True: A sequence is bounded if and only if all of its subsequences are bounded. Every subsequence of a bounded sequence also has to be bounded, because it cannot exceed the bounds set by the original sequence.
(c) True: A sequence is monotonic if and only if all of its subsequences are monotonic. Regardless of which elements are chosen to form a subsequence, if the original sequence preserves its direction of progression (either increasing or decreasing), so will the subsequences.
(d) False: A sequence is divergent if and only if all of its subsequences are divergent. Here we have a counterexample: consider the sequence (-1)ⁿ, which diverges. However, it has convergent subsequences, such as the constant subsequences of all 1s or all -1s.
Therefore, statements (a), (b), and (c) are true, while statement (d) is false. Understanding the conditions for convergence, boundedness, and monotonicity is key to studying the behaviors of sequences and leveraging properties like the comparison test for evaluating whether a series converges or diverges.
Given the function g(x)=x^2+10x+23g(x)=x
2
+10x+23, determine the average rate of change of the function over the interval -8\le x \le -4−8≤x≤−4.
The average rate of change of the function g(x) over the interval from x = -8 to x = -4 is -2.
The average rate of change of a function over a certain interval is similar to finding the slope of the secant line that passes through the points on the graph of the function corresponding to the end points of the interval. In this case, for the function g(x) = x^2 + 10x + 23, we want to find the average rate of change over the interval from x = -8 to x = -4.
To do this, we calculate the change in g(x) divided by the change in x (\(\Delta x\)).
The value of the function at x = -8 is g(-8) = (-8)^2 + 10(-8) + 23 = 64 - 80 + 23 = 7.
The value of the function at x = -4 is g(-4) = (-4)^2 + 10(-4) + 23 = 16 - 40 + 23 = -1.
Now, the average rate of change is given by the formula:
Average rate of change = (g(-4) - g(-8)) / (-4 - (-8))
= (-1 - 7) / (-4 + 8) = -8 / 4 = -2
So, the average rate of change of the function g(x) over the interval from x = -8 to x = -4 is -2.
find the sum of 253,965 and 1,563,001 write the answer in words
Answer and Step-by-step explanation:
We can easily add up these two numbers:
1,563,001
+ 253,965
___________
1,816,966
Now, we need to write out the answer is words:
"One million, eight hundred sixteen thousand, nine hundred and sixty-six"
Hope this helps!
Answer:
Step-by-step explanation:
1,563,001
+ 253,965
___________
1,816,966
In words:
"One million Eight Hundred Sixteen Thousan Nine hundred and. Sixty Six"
-1/2(-3y+10) what would be the answer?
Answer:
3/2y * -5
Step-by-step explanation:
-1/2 * -3y= 3/2
-1/2* 10 = -5
6. Suppose Steve goes fishing near the nuclear power plant at Hawkins. He’s interested in catching King Salmons and Walleyes. Assume the following: • All species of fish in the lake have weights that are normally distributed. • The weight of the King Salmons are i.i.d. ∼ Normal with µK = 150 lbs and σK = 10 lbs. Let K be the weight of a randomly caught King Salmon. • The weight of the Walleyes are i.i.d. ∼ Normal with µW = 51 lbs and σW = 9 lbs. Let W be the weight of a randomly caught Walleye. (a) (3 points) Suppose Steve catches 4 King Salmons at random. What is the probability that the total weight of the King Salmons caught is greater than 575 lbs?
Answer:
89.44% probability that the total weight of the King Salmons caught is greater than 575 lbs
Step-by-step explanation:
To solve this question, we need 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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For sums of size n from a population, the mean is [tex]\mu*n[/tex] and the standard deviation is [tex]\sigma\sqrt{n}[/tex]
The weight of the King Salmons are i.i.d. ∼ Normal with µK = 150 lbs and σK = 10 lbs. 4 king salmons.
So [tex]\mu = 4*150 = 600, \sigma = 10\sqrt{4} = 20[/tex]
What is the probability that the total weight of the King Salmons caught is greater than 575 lbs?
This is 1 subtracted by the pvalue of Z when X = 575. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{575 - 600}{20}[/tex]
[tex]Z = -1.25[/tex]
[tex]Z = -1.25[/tex] has a pvalue of 0.1056
1 - 0.1056 = 0.8944
89.44% probability that the total weight of the King Salmons caught is greater than 575 lbs
The common fruit fly Drosophila melanogaster is the most studied organism in genetic research because it is small, easy to grow, and reproduces rapidly. The length of the thorax (where the wings and legs attach) in a population of male fruit flies is approximately Normal with mean 0.800 millimeters (mm) and standard deviation 0.078 mm. Draw a Normal curve on which this mean and standard deviation are correctly located. (Hint: Draw an unlabeled Normal curve, locate one, two and three standard deviations away from the mean, then add number labels on the horizontal axis.)
Answer:
The Normal curve with the mean and standard deviations is shown below.
Step-by-step explanation:
According to the Empirical Rule in a normal distribution with mean µ and standard-deviation σ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be broken into three parts:
68% data falls within 1 standard deviation of the mean. That is P (µ - σ ≤ X ≤ µ + σ) = 0.68. 95% data falls within 2 standard deviations of the mean. That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95. 99.7% data falls within 3 standard deviations of the mean. That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.The length of the thorax in a population of male fruit flies is approximately Normal.
The mean is, µ = 0.800 mm and the standard deviation is, σ = 0.078 mm.
Then:
68% data falls within 1 standard deviation of the mean. That is P (µ - σ ≤ X ≤ µ + σ) = P (0.722 ≤ X ≤ 0.878) = 0.68.95% data falls within 2 standard deviations of the mean. That is P (µ - 2σ ≤ X ≤ µ + 2σ) = P (0.644 ≤ X ≤ 0.956) = 0.95.99.7% data falls within 3 standard deviations of the mean. That is P (µ - 3σ ≤ X ≤ µ + 3σ) = P (0.566 ≤ X ≤ 1.034) = 0.997.The Normal curve with the mean and standard deviations is shown below.
The net of a triangular pyramid is shown.
The net of a triangular pyramid.
[Not drawn to scale]
Each triangle in the net has a base length that measures 6 inches and a height that measures 4 inches. What is the surface area of the pyramid that can be formed from this net?
12 inches squared
24 inches squared
36 inches squared
48 inches square
Answer:
24
Step-by-step explanation:
because you need to multiply 6x4x3 divided by 3
Answer:
Bb 24in2
Step-by-step explanation:
It is correct
correct on E D G E N U I T Y
A clothing business finds there is a linear relationship between the number of shirts, Q ,it can sell and the price, P , it can charge per shirt. In particular, historical data shows that 4000 shirts can be sold at a price of $ 133 , while 27000 shirts can be sold at a price of $ 41 . Give a linear equation in the form P = a Q + b that gives the price P they can charge for Q shirts. (This is called a demand function.)
Answer:
P = -0.004Q + 149
Step-by-step explanation:
The general form of the linear equation is:
[tex]P=aQ+b[/tex]
The slope of the equation (a) can be found by using the two given points (4,000; $133) and (27,000; $41)
[tex]a = \frac{\$41-\$133}{27,000-4,000}\\a=-0.004[/tex]
Applying the point (4,000; $133) to the equation below yields in the linear equation for Price as a function of the number of shirts:
[tex]P-P_0=a(Q-Q_0)\\P-133=-0.004(Q-4,000)\\P = -0.004Q+149[/tex]
The linear equation is:
P = -0.004Q + 149
An insurance company selected a random sample of 500 clients under 18 years of age and found that 180 of them had had an accident in the previous year. A random sample of 600 clients aged 18 and older was also selected and 150 of them had had an accident in the past year. We want to estimate how much the accident proportions differ between the two age groups.
Answer: The accident proportion differ between the two age groups is 0.11.
Step-by-step explanation:
Since we have given that
Number of clients under 18 years of age = 500
Number of clients had had an accident = 180
Proportion of client had an accident = [tex]\dfrac{180}{500}=\dfrac{18}{50}[/tex]
Number of clients aged 18 and older = 600
Number of clients had had an accident = 150
Proportion of client had an accident = [tex]\dfrac{150}{600}=\dfrac{15}{60}[/tex]
So, According to question, we get that
Difference between the two age groups proportions would be :
[tex]\dfrac{18}{50}-\dfrac{15}{60}\\\\=\dfrac{108-75}{300}\\\\=\dfrac{33}{300}\\\\=\dfrac{11}{100}[/tex]
Hence, the accident proportion differ between the two age groups is 0.11.
When sampling without replacement from a finite population of size N, the following formula is used to find the standard deviation of the population of sample means: σ = However, when the sample size n, is smaller than 5% of the population size, N, the finite population correction factor, , can be omitted. Explain in your own words why this is reasonable. For N = 200, find the values of the finite population correction factor when the sample size is 10%, 5%, 3%, 1% of the population, respectively. What do you notice?
Answer:
Check below for the required explanations
Step-by-step explanation:
The population correction factor is given by the formula :
PCF = [(N-n)/(N-1)]^1/2..........(1)
a) When the sample size is smaller than 5% of N. That is n < 0.05N
If n = 0.05N is substituted into the PCF formula, PCF will be approximately 1.
For a value of 1, PCF can be safely ignored.
b) N = 200
i) n = 10% N
n = 0.1 × 200 = 20
Substitute n = 20 into equation (1)
PCF = (200-20)/(200-1)]^1/2
PCF = 0.95
ii) n = 5% N
n = 0.05× 200 = 10
Substitute n = 20 into equation (1)
PCF = (200-10)/(200-1)]^1/2
PCF = 0.98
iii) i) n = 3% N
n = 0.03 × 200 = 6
Substitute n = 20 into equation (1)
PCF = (200-6)/(200-1)]^1/2
PCF = 0.99
iiii) n = 1% N
n = 0.01 × 200 = 2
Substitute n = 20 into equation (1)
PCF = (200-2)/(200-1)]^1/2
PCF = 0.998(approx. =1)
It is noticed that the smaller the sample size, the closer the population correction factor to unity. At 1% of the population, the population correction factor is negligible
a board is 77.47 centimeters long. How long is the board in inches. 1 inch is 2.54
Answer:
30.5 in
Step-by-step explanation:
Answer:
30.5 inches
Step-by-step explanation:
Since the board is 77.47 cm and 1 inch is 2.54 cm you have to divide 77.47 by 2.54.
what more ounces our pounds?help if you see this please i will give you 25 points
Answer:
Ounces is smaller than lbs
Step-by-step explanation:
16 ounces = 1 lbs
Ounces is smaller than lbs
Answer:
An ounce is the smallest unit for measuring weight, a pound is a larger unit, and a ton is the largest unit
Step-by-step explanation:
Please give me brainliest
A consumer protection agency is testing a sample of cans of tomato soup from a company. If they find evidence that the average level of the chemical bisphenol A (BPA) in tomato soup from this company is greater than 100 ppb (parts per billion), they will recall all the soup and sue the company.
(a) State the null and alternative hypotheses.
(b) Using the context of the problem, what would a Type I Error be in this situation?
(c) Using the context of the problem, what would a Type II Error be in this situation?
(a) Null Hypothesis (H0): μ ≤ 100 ppb (The average level of BPA in the tomato soup is less than or equal to 100 ppb)
Alternative Hypothesis (H1): μ > 100 ppb (The average level of BPA in the tomato soup is greater than 100 ppb)
(b) A Type I Error would occur if the agency finds evidence that the average level of BPA is greater than 100 ppb (rejects H0) when, in fact, the true average level is less than or equal to 100 ppb.
(c) A Type II Error would occur if the agency fails to find evidence that the average level of BPA is greater than 100 ppb (fails to reject H0) when, in fact, the true average level is greater
(a) Null Hypothesis (H₀): The average level of BPA in the company's tomato soup is less than or equal to 100 ppb.
Mathematically: μ ≤ 100 ppb
Alternative Hypothesis (H₁):
The average level of BPA in the company's tomato soup is greater than 100 ppb.
Mathematically: μ > 100 ppb
(b) A Type I error occurs when we reject the null hypothesis when it is actually true. In this context, a Type I error would mean that the consumer protection agency concludes that the average BPA level is greater than 100 ppb and recalls the soup and sues the company, when in reality, the average BPA level is less than or equal to 100 ppb. This would result in unnecessary product recall and legal action against the company.
(c) A Type II error occurs when we fail to reject the null hypothesis when it is actually false. In this context, a Type II error would mean that the consumer protection agency concludes that the average BPA level is less than or equal to 100 ppb and does not recall the soup or sue the company, when in reality, the average BPA level is greater than 100 ppb. This would allow unsafe products to remain on the market, potentially harming consumers.
True or False: The mean and the average are the same number/
Answer:
true
Step-by-step explanation:
Bo had a container of flour that weighed 32 ounces. On Monday, he added 47 ounces of flour. During the week, he made cookies and waffles with some of the flour. At the end of the week, the container weighed 61 ounces. Estimate how many ounces of flour Bo used during the week
Answer:
~20 oz
Step-by-step explanation:
Originally, Bo had 32 ounces of flour, which is around 30 ounces. Then, he added in 47 ounces, which is about 50. So, now the new total is about 30 + 50 = 80 ounces.
During the week, he used up some of those ~80 ounces of flour. He used up enough so that at the end of the week, there was only 61 ounces, or approximately 60 oz, left. The difference between these two numbers must be how much Bo used:
80 - 60 = 20 ounces
Thus, the answer is about 20 ounces.
Hope this helps!
Bo started with 79 ounces of flour (32 ounces original and 47 ounces added), and ended with 61 ounces remaining, so he used an estimated 18 ounces of flour during the week.
Explanation:To determine how much flour Bo used during the week, we need to find out how much the total weight of flour was at the beginning of the week, and then subtract the weight of the flour that was left at the end of the week.
At first, Bo's container had 32 ounces of flour. On Monday, he added 47 more ounces. Therefore, the weight of the flour at the beginning of the week was 32 + 47 = 79 ounces.
At the end of the week, the leftover flour weighed 61 ounces. So, to estimate how much flour Bo used, we subtract the leftover weight from the original total weight: 79 ounces - 61 ounces = 18 ounces.
So, Bo used an estimated 18 ounces of flour during the week.
Learn more about Estimation here:https://brainly.com/question/16131717
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Subtract the fractions and reduce to lowest terms. 1/ 3 − 1/ 12
3/ 4+2/ 5=1 3/ 20
3 forths + 2 fifths = 1 and 3 twentyiths
Hope i helped! :)
Answer:
.25 or 1/4
Step-by-step explanation:
1/3-1/12 is .25 or if you turn it into a fraction it is 1/4.Because one quarter out of a dollar equals .25 cents.
An online clothing company decides to investigate whether offering their customers a coupon upon completion of their first purchase will encourage them to make a second purchase. To do so, the company programs the website to randomly select 100 first time customers. Sixty of these customers are randomly selected to receive a coupon for $5 off their next purchase, to be made in the next 30 days. The other 40 customers are not offered a coupon. The table below shows the number of customers in each group that made a second purchase within 30 days of their first purchase.
Based upon the table, is “yes, made a second purchase” independent of “yes, being sent a coupon”?
A) Yes, exactly half of the customers made a second purchase and half did not.
(B) Yes, the largest count in the table comes from those who were sent a coupon and made a second purchase within 30 days.
(C) No, the probability of making a second purchase is not equal to the probability of making a second purchase given that a coupon was sent.
(D) No, the probability of making a second purchase is the same whether or not a coupon was sent.
(E) It is impossible to draw a conclusion about independence because a coupon was not sent to exactly half of the customers.
Answer:
(C) No, the probability of making a second purchase is not equal to the probability of making a second purchase given that a coupon was sent.
Step-by-step explanation:
Let A = the customer makes a second purchase within 30 days and let B = customer is sent a coupon. Events A and B are independent if P(A) = P(A | B).
P(A) = P(the customer makes a second purchase within 30 days) = \frac{50}{100} = 0.5
100
50
=0.5
P(A | B) = P(the customer makes a second purchase within 30 days | customer is sent a coupon) = \frac{34}{60} = 0.567
60
34
=0.567
Because P(A) ≠ P(A | B) making a second purchase is not independent of being sent a coupon.
Tammy picks apples at an orchard. She earns $3.10 for each hour she works and $2.30 for each bushel of apples she picks. Her goal is to earn at least $100 this week.
Write an inequality that will help Tammy determine the number of hours (h) and bushels (b) she needs to reach her goal.
Answer:
3.10h + 2.30b ≥ 100
Step-by-step explanation:
Tammy earns $3.10 for each hour,
Let h represents hour,
So total earning for h number of hours will be 3.10h
Tammy earns $2.30 for each bushel of apples
Let b represents bushel of apples,
So total earning for b number of bushels will be 2.30b
Tammy's goal is to earn at least $100 this week
Therefore, to reach her goal the inequality will be
3.10h + 2.30b ≥ 100
Where h is the number of hour and b is the number of bushels.
Isabella's car used 3 gallons of gas to drive 48 miles. At what rate does her car use gas in gallons per mile?
Answer:
0.0625 gallons per mile
Step-by-step explanation:
The average ticket price for a Spring Training baseball game is $29.89, with a standard deviation of $5.28. In a random sample of 40 Spring Training tickets, find the probability that the mean ticket price exceeds $33
Answer:
0.01% probability that the mean ticket price exceeds $33
Step-by-step explanation:
To solve this question, we need 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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 29.89, \sigma = 5.28, n = 40, s = \frac{5.28}{\sqrt{40}} = 0.8348[/tex]
Find the probability that the mean ticket price exceeds $33
This is 1 subtracted by the pvalue of Z when X = 33. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{33 - 29.89}{0.8348}[/tex]
[tex]Z = 3.73[/tex]
[tex]Z = 3.73[/tex] has a pvalue of 0.9999
1 - 0.9999 = 0.0001
0.01% probability that the mean ticket price exceeds $33
Answer:
[tex]P(\bar X >33)[/tex]
And we can use the z score formula given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 33 we got:
[tex] z = \frac{33-29.89}{\frac{5.28}{\sqrt{40}}}= 3.725[/tex]
And we can use the complement rule and we got:
[tex] P(Z>3.725)= 1-P(Z<3.725)= 1-0.9999= 0.0001[/tex]
Step-by-step explanation:
Previous conepts
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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
We know the following properties for the variable of interest:
[tex]\mu = 29.89 , \sigma=5.28 [/tex]
We select a sample size of n = 40>30. From the central limit theorem we know that 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 for this case we want to find this probability:
[tex]P(\bar X >33)[/tex]
And we can use the z score formula given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 33 we got:
[tex] z = \frac{33-29.89}{\frac{5.28}{\sqrt{40}}}= 3.725[/tex]
And we can use the complement rule and we got:
[tex] P(Z>3.725)= 1-P(Z<3.725)= 1-0.9999= 0.0001[/tex]
el área de un rectángulo mide 15 unidades cuadradas. si un lado de rectángulo mide 3.75 unidades de largo cuánta unidad de mide el perímetro de rectángulo?
Translation from Google
the area of a rectangle measures 15 square units. if one side of a rectangle is 3.75 units long how much unit is the perimeter of the rectangle?
Answer:
15.5 unidades
Step-by-step explanation:
Area of rectangle is given by the product of its length and width and expressed as
A=lw
Where l is length, w is width and A is area.
Similarly, perimeter is given by
P=2(l+w)
Given that A=15 and w=3.75
15=3.75l
l=15÷3.75=4 unidades
Now perimeter will be
P=2(4+3.75)=2(7.75)=15.5 units
What is [4]{7^3} in exponential form?
Answer:
1.372 × 10^3
Step-by-step explanation:
The line graph contains an error. Study the graph carefully and use complete sentences to describe the error.
Answer:
The scale doesn't start at 0. One could argue the maximum value should be 7, but this is a formatting choice.
Step-by-step explanation:
Answer:
The scale doesn't start at 0. One could argue the maximum value should be 7, but this is a formatting choice.
Step-by-step explanation:
translate the following into an equation. Nine is the sum of 1 and two times a number
Answer:
9 = 1 + 2n
Step-by-step explanation:
A sum is indicated with a plus sign (+). Two times a number is represented by the product of 2 and the variable used to represent the number. Here, we have chosen "n" (for "number"). Then the sum is ...
1 + 2n
"Is" in this context means "equals", so we have ...
9 = 1 + 2n
In a box there are 145 apples.
Some of the apples are red and the rest are green.
The ratio of red to green apples is 2:3
How many green apples are there?
Answer:
87 green apples
Step-by-step explanation:
Red apples : green apples = 2 : 3
Number of red apples = 2x
Number of green apples = 3x
Total apples = 145
3x + 2x = 145
5x = 145
x = 145/5
x = 29
Number of green apples = 3x = 3 * 29 = 87
If the mean GPA among students is 3.25 with a standard deviation of 1.75, and we select a random sample of 300 people, at what value for the sample mean would be greater than exactly 95% of possible sample means?
To find the sample mean greater than 95% of possible sample means for a population mean of 3.25 and standard deviation of 1.75 with a sample size of 300, apply the Central Limit Theorem to calculate the standard error and use the 95th percentile z-score to compute the desired sample mean.
Explanation:To find the sample mean that is greater than exactly 95% of possible sample means given a population mean (μ) of 3.25 and a standard deviation (σ) of 1.75 for a sample size (n) of 300, we use the concept of a z-score in conjunction with the Central Limit Theorem.
The Central Limit Theorem states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. Since we have a large sample size of 300, we can assume normality. The formula for the standard error of the mean (SEM) is given by σ divided by the square root of n, SEM = σ / √n. In this case, SEM would be 1.75 / √300.
To find the specific sample mean that would exceed 95% of the sample means, we need to find the z-score that corresponds to the 95th percentile, which is typically around 1.645 for a one-tailed test (since we're looking for means greater than a value). We then multiply this z-score by the SEM and add it to the population mean. The calculation is as follows:
Calculate SEM: SEM = 1.75 / √300 Find the z-score corresponding to the 95th percentile: z ≈ 1.645 Calculate the sample mean: sample mean = μ + (z * SEM)
By carrying out these calculations, we will obtain the sample mean that is greater than 95% of possible sample means.
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The sample mean that would be greater than exactly 95% of possible sample means is approximately 3.4164.
Given that the population mean [tex](\(\mu\))[/tex] is 3.25 and the population standard deviation [tex](\(\sigma\))[/tex] is 1.75, we can use the z-score formula to find the sample mean [tex](\(\bar{x}\))[/tex] that corresponds to the 95th percentile:
[tex]\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \][/tex]
We know that for a standard normal distribution, the z-score that corresponds to the 95th percentile is approximately 1.645 (since we want the value that is greater than exactly 95%, we use the one-tailed z-score). We can rearrange the formula to solve for [tex]\(\bar{x}\)[/tex]:
[tex]\[ \bar{x} = \mu + (z \times \frac{\sigma}{\sqrt{n}}) \][/tex]
Now we plug in the values:
[tex]\[ \bar{x} = 3.25 + (1.645 \times \frac{1.75}{\sqrt{300}}) \] \[ \bar{x} = 3.25 + (1.645 \times \frac{1.75}{\sqrt{300}}) \] \[ \bar{x} = 3.25 + (1.645 \times \frac{1.75}{17.3205080756888}) \] \[ \bar{x} = 3.25 + (1.645 \times 0.1010101010101) \] \[ \bar{x} = 3.25 + 0.16641664166417 \] \[ \bar{x} = 3.4164166416642 \][/tex]
A weight loss company wanted to predict how much weight a client would lose if they followed a prescribed exercise program in addition to the company's diet program. Volunteers were randomly divided into two groups, one group dieted but didn’t exercise, and the other group dieted and followed the exercise program. For the exercise group, they used linear regression with percent compliance with the exercise program as the explanatory variable and pounds lost in three months as the response variable. One of the clients was told that his residual was 5.5 pounds. What does this mean?
Options:
a. His predicted weight loss was 5.5 pounds higher than his actual weight loss.
b. His actual weight loss was 5.5 pounds higher than his predicted weight loss.
c. His actual weight loss was 5.5 pounds higher than it would have been if he didn’t exercise.
d. His predicted weight loss was 5.5 pounds higher than it would have been if he didn’t exercise.
Answer:
Option B. His actual weight loss was 5.5 pounds higher than his predicted weight loss.
Step-by-step explanation:
In regression analysis, the difference between the observed value of the dependent variable (y) and the predicted value (ŷ) is called the residual (e). Residual value= Observed value - Predicted value.i.e. e = y - ŷ
Since the residual weight of the client was 5.5 pounds, this means that His actual weight after compliance with the exercise program his 5.5 pounds higher than what he predicted.
Final answer:
The client's residual of 5.5 pounds means they lost 5.5 pounds more than predicted by the exercise program's linear regression model, indicating a discrepancy between the predicted and actual weight loss.
Explanation:
When a client has been told that his residual was 5.5 pounds in the context of a linear regression, it refers to the difference between the actual weight the client lost and the weight the regression model predicted they would lose based on their percent compliance with the exercise program. A residual of 5.5 pounds indicates that the client lost 5.5 pounds more than what the model predicted. Residuals are used to measure the accuracy of regression models; if they are small, it suggests that the model is accurately predicting outcomes. However, large or systematic residuals can indicate a problem with the model's fit to the data.
A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of 0.81 second and a sample standard deviation of 0.34 second. (a) Calculate a 99% confidence interval for the true average echo duration. (b) This sample data is used as a pilot study, and now the investigator would like to design a new study to construct a 99% confidence interval with width 0.1. What is the necessary sample size?
Answer:
a) [tex]0.81-2.62\frac{0.34}{\sqrt{110}}=0.725[/tex]
[tex]0.81+2.62\frac{0.34}{\sqrt{110}}=0.895[/tex]
So on this case the 99% confidence interval would be given by (0.725;0.895)
b) [tex]n=(\frac{2.58(0.34)}{0.1})^2 =76.95 \approx 77[/tex]
So the answer for this case would be n=77 rounded up to the nearest integer
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".
[tex]\bar X=0.81[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=0.34 represent the sample standard deviation
n=110 represent the sample size
Part a
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=110-1=109[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,119)".And we see that [tex]t_{\alpha/2}=2.62[/tex]
Now we have everything in order to replace into formula (1):
[tex]0.81-2.62\frac{0.34}{\sqrt{110}}=0.725[/tex]
[tex]0.81+2.62\frac{0.34}{\sqrt{110}}=0.895[/tex]
So on this case the 99% confidence interval would be given by (0.725;0.895)
Part b
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigmas}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =0.1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
We can assume the following estimator for the population deviation [tex]\hat \sigma =s =0.34[/tex]
The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got [tex]z_{\alpha/2}=2.58/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.58(0.34)}{0.1})^2 =76.95 \approx 77[/tex]
So the answer for this case would be n=77 rounded up to the nearest integer
A certain television is advertised as a 40-inch TV (the diagonal length). If the width of the TV is 24 inches, how many inches tall is the TV?
Answer:
Answer is 32 if you are doing DeltaMath
Step-by-step explanation: