Answer:
There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.
Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]
P-value = 0.1290
As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.
Step-by-step explanation:
We have to test the hypothesis of the difference between means.
The claim is that Portland has more average yearly rainfall than Seattle.
Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]
The significance level is 0.10.
The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.
The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.
The difference between means is:
[tex]M_d= M_1-M_2=37.50-37.07=0.43[/tex]
The standard error for the difference between means is:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774[/tex]
We can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393[/tex]
The degrees of freedom are:
[tex]df=n1+n2-2=45+35-2=78[/tex]
Then, the p-value for this one-tailed test with 78 degrees of freedom is:
[tex]P-value=P(t>1.1393)=0.1290[/tex]
As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.
There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.
A model of a soccer ball is made up of regular pentagons and hexagons.
A soccer ball is made up of regular pentagons and hexagons.
The side length of one of the pentagons measures 2 inches and the apothem measures about 1.38 inches. What is the area of one of the pentagons? State your answer to the nearest tenth.
square inches
The side length of one of the hexagons measures 2 inches and the apothem measures about 1.73 inches. What is the area of one of the hexagons? State your answer to the nearest tenth.
square inches
Answer:
- pentagon area= 6.9
-hexagon area=10.4
Answer:
Pentagons: 6.9 square inches
Hexagons: 10.4 square inches
Step-by-step explanation: i got it right on edge
Bryce reads in the latest issue of Pigskin Roundup that the average number of rushing yards per game by NCAA Division II starting running backs is 50 with a standard deviation of 8 yards. If the number of yards per game (X) is normally distributed, what is the probability that a randomly selected running back has 64 or fewer rushing yards
Answer:
0.9599 is the probability that a randomly selected running back has 64 or fewer rushing yards.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 50
Standard Deviation, σ = 8
We are given that the distribution of number of rushing yards per game is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(running back has 64 or fewer rushing yards)
[tex]P( x \leq 64) = P( z \leq \displaystyle\frac{64 - 50}{8}) = P(z \leq 1.75)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x \leq 64) = 0.9599[/tex]
0.9599 is the probability that a randomly selected running back has 64 or fewer rushing yards.
please help, im so confused!
Which point is on the graph of the function
f(x) = One-half(2)x?
(0, 1)
(0, 2)
(1, One-half)
(1, 1)
Answer:
D (1,1)
Step-by-step explanation:
Answer:
D. (1,1)
Hope it works!
Step-by-step explanation:
Lucille has a collection of more than 500 songs on her phone that have a mean duration of 215 seconds and a standard deviation of 35 seconds. Suppose that every week she makes a playlist by taking an SRS of 49 of these songs, and we calculate the sample mean duration ë of the songs in each sample. Calculate the mean and standard deviation of the sampling distribution of ________. seconds L = seconds
Answer:
The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.
Step-by-step explanation:
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.
All songs
Mean 215 seconds, standard deviation 35 seconds
Sample
49
Mean 215, standard deviation [tex]s = \frac{35}{\sqrt{49}} = 5[/tex]
The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.
Answer:
The sample mean would be:
[tex]\mu_{\bar X} = 215 seconds[/tex]
And the deviation:
[tex]\sigma_{\bar X} = \frac{35}{\sqrt{49}}= 5 seconds[/tex]
Step-by-step explanation:
Previous concepts
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".
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
We know the following info for the random variable X who represent the duration
[tex]\mu = 215, \sigma=35[/tex]
For this case we select a sampel size of n =49>30. So we can apply the central limit theorem. 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]
The sample mean would be:
[tex]\mu_{\bar X} = 215 seconds[/tex]
And the deviation:
[tex]\sigma_{\bar X} = \frac{35}{\sqrt{49}}= 5 seconds[/tex]
Evaluate the following expression. Round your answer to two decimal places.
Answer: 2.18
Step-by-step explanation:
log(11)/log(3)=2.18
Answer:
2.18
Step-by-step explanation:
I just did it and got it right
Suppose Julio is a veterinarian who is doing research into the weight of domestic cats in his city. He collects information on 188 cats and finds the mean weight for cats in his sample is 10.97 lb with a standard deviation of 4.41 lb. What is the estimate of the standard error of the mean (SE)
Answer:
The standard error of the mean is 0.3216
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 188
Sample mean =
[tex]\bar{x}= 10.97\text{ lb}[/tex]
Sample standard deviation =
[tex]s = 4.41\text{ lb}[/tex]
We have to estimate the standard error of the mean.
Formula for standard error:
[tex]S.E = \dfrac{s}{\sqrt{n}}[/tex]
Putting values, we get,
[tex]S.E =\dfrac{4.41}{\sqrt{188}} = 0.3216[/tex]
Thus, the standard error of the mean is 0.3216
i need to find the length
Answer: I am pretty sure it is 7
Step-by-step explanation:
the smaller one is to smaller than the bigger one so you would just add two to the 5
Answer:
7.5
Step-by-step explanation:
The triangles are proportional, so 6/4=x/5
1.5=x/5
x=1.5*5=7.5
i need to know number 7
Answer:
it would be A. -25
Step-by-step explanation:
`for example if she sold 0 dollars worth of necklaces and she spent 25 dollars on supplies she would have lost 25 dollars so it would be -25.
Answer:A,-$25
Step-by-step explanation:
She bought the necklace. When she sold it, she sold it for $25 less than what she bought it for. Therefore, she lost $25, the same as -$25.
The table shows information about the numbers of hours 30 children spent on their tablets one evening. a) find the class interval that contains the median. B)work out an estimate for the mean number of hours
(a) The class interval that contains the median is [tex]\( 1 < h \leq 2 \).[/tex] (b)The estimated mean number of hours is 1.8 hours.
To analyze the data and answer the questions about the time 30 children spent on their tablets one evening, let's break it down step by step.
(a) The median is the value that separates the higher half from the lower half of the data. For 30 children, the median position will be the average of the 15th and 16th values in the ordered data set.
Given the frequencies:
- [tex]\( 0 < h \leq 1 \): 6[/tex] children
- [tex]\( 1 < h \leq 2 \):[/tex] 13 children
- [tex]\( 2 < h \leq 3 \):[/tex] 7 children
- [tex]\( 3 < h \leq 4 \): 4[/tex] children
To find the median class interval, we need to determine where the 15th and 16th values fall.
Cumulative frequency:
[tex]- \( 0 < h \leq 1 \): 6\\ - \( 1 < h \leq 2 \): 6 + 13 = 19\\ - \( 2 < h \leq 3 \): 19 + 7 = 26\\ - \( 3 < h \leq 4 \): 26 + 4 = 30[/tex]
From the cumulative frequency:
The 15th and 16th values fall within the second class interval [tex]\( 1 < h \leq 2 \),[/tex] because the cumulative frequency reaches 19 in this interval.
The class interval that contains the median is [tex]\( 1 < h \leq 2 \).[/tex]
(b) To estimate the mean, we'll use the midpoint of each class interval and multiply by the frequency of that interval, then divide by the total number of children.
Determine the midpoints of each class interval:
[tex]- \( 0 < h \leq 1 \): midpoint = \( \frac{0 + 1}{2} = 0.5 \)\\ - \( 1 < h \leq 2 \): midpoint = \( \frac{1 + 2}{2} = 1.5 \)\\ - \( 2 < h \leq 3 \): midpoint = \( \frac{2 + 3}{2} = 2.5 \)\\ - \( 3 < h \leq 4 \): midpoint = \( \frac{3 + 4}{2} = 3.5 \)[/tex]
Multiply each midpoint by its respective frequency:
[tex]- \( 0.5 \times 6 = 3 \)\\ - \( 1.5 \times 13 = 19.5 \)\\ - \( 2.5 \times 7 = 17.5 \)\\ - \( 3.5 \times 4 = 14 \)\\[/tex]
Sum these products:
[tex]\[ 3 + 19.5 + 17.5 + 14 = 54 \][/tex]
Divide by the total number of children to find the mean:
[tex]\[ \text{Mean} = \frac{54}{30} = 1.8 \][/tex]
The estimated mean number of hours is 1.8 hours.
The complete question is:
The table shows information about the numbers of hours 30 children spent on their tablets one evening.
Number of hours (m) Frequency
D<h<1 6
1<h<2 13
2<h<3 7
3<h<4 4
a) Find the class interval that contains the median.
b) Work out an estimate for the mean number of hours.
If tan (k•90)=0 then k is an even integer true of false
Answer:
True
Step-by-step explanation:
Use the unit circle
tan(x) = 0 only when x is 0 or a multiple of 180 (in degrees)
Answer:
true
Step-by-step explanation:
Height of rectangular prism if the volume is 334.8 meters
Only the volume is given, so you can already assume the rectangular prism is a cube, which in that case you can divide by 3 to get the answer.
Write an inequality for each statement.
1. Lacrosse practice will not be more than 45 minutes.
2. Mario measures more than 60 inches in height
Write an inequality for each statement.
1. Lacrosse practice will not be more than 45 minutes.
2. Mario measures more than 60 inches in height
3. More than 8000 fans attended the first football game of the Wizard amd Arrowhead Stadium in Kansas City, Missouri. Write an inequality to describe attendance.
3. More than 8000 fans attended the first football game of the Wizard amd Arrowhead Stadium in Kansas City, Missouri. Write an inequality to describe attendance.
Answer:
1. [tex]l \leq 45[/tex] ( in this inequality, the time can be less than or equal to 45, but no more than 45)
2. [tex]m > 60[/tex] (Mario's height is more than 60.)
3. [tex]f > 8000[/tex] (More than 8000 fans attended.)
Two tanks of brine are connected via two tubes. Tank 1 contains x(t) pounds of salt in 500 gallons of brine, and tank 2 contains y(t) pounds of salt in 500 gallons of brine. Brine is pumped from tank 1 to tank 2 at a rate of 50 gallons/min and brine is pumped from tank 2 to tank 1 at the same rate of 50 gallons/min, ensuring that the total volume of brine in the tanks remains constant over time.
a) Show that x(t) and y(t) satisfy the differential equations x'= − (1/10)x + (1/10)y ; y' = (1/10)x − (1/10)y.
b) If initially tank 1 contains no salt, and tank 2 contains 10 pounds of salt, use the method of elimination to find x(t) and y(t).
a. Salt flows into tank A at a rate of
(y(t)/500 lb/gal) * (50 gal/min) = y(t)/10 lb/gal
and out at a rate of
(x(t)/500 lb/gal) * (50 gal/min) = x(t)/10 lb/gal
so the net rate of change of the amount of salt in tank A is
[tex]x'(t)=\dfrac{y(t)}{10}-\dfrac{x(t)}{10}[/tex]
Similarly, you would find
[tex]y'(t)=\dfrac{x(t)}{10}-\dfrac{y(t)}{10}[/tex]
b. We have
[tex]x'=-\dfrac x{10}+\dfrac y{10}\implies x''=-\dfrac{x'}{10}+\dfrac{y'}{10}[/tex]
Notice that x' = -y', so
[tex]x''=-{2x'}{10}\implies 5x''+x'=0[/tex]
Solve for x: the characteristic equation
[tex]5r^2+r=r(5r+1)=0[/tex]
has roots [tex]r=0[/tex] and [tex]r=-\frac15[/tex], so
[tex]x(t)=C_1+C_2e^{-t/5}[/tex]
Again using the fact that y' = -x', we then find
[tex]x'(t)=-\dfrac{C_2}5e^{-t/5}\implies y'(t)=\dfrac{C_2}5e^{-t/5}\implies y(t)=C_1-C_2e^{-t/5}[/tex]
Given that x(0) = 0 and y(0) = 10, we find
[tex]\begin{cases}0=C_1+C_2\\10=C_1-C_2\end{cases}\implies C_1=5,C_2=-5[/tex]
A company needs 150,000 items per year. It costs the company $640 to prepare a production run of these items and $7 to produce each item. If it also costs the company $0.75 per year for each item stored, find the number of items that should be produced in each run so that total costs of production and storage are minimized.
The requried x represents the number of items produced in each run, we should produce approximately 127 items in each run to minimize the total costs of production and storage.
To find the number of items that should be produced in each run to minimize the total costs of production and storage, we can use the following steps:
Let x be the number of items produced in each run. The total cost of production and storage (C) can be expressed as:
C(x) = (640/150,000) + (7x) + (0.75 * 150,000 / x)
The first term (640/150,000) represents the cost of preparing a production run, the second term (7x) represents the cost of producing each item, and the third term (0.75 * 150,000 / x) represents the cost of storing each item.
To minimize the total cost, we need to find the value of x that minimizes the function C(x). We can do this by taking the derivative of C(x) with respect to x and setting it equal to zero:
C'(x) = 0
Now, let's find the derivative of C(x):
C'(x) = d/dx [(640/150,000) + (7x) + (0.75 * 150,000 / x)]
C'(x) = 0 + 7 - (0.75 * 150,000 / x²)
Now, set C'(x) equal to zero and solve for x:
7 - (0.75 * 150,000 / x²) = 0
0.75 * 150,000 / x² = 7
150,000 / x² = 7 / 0.75
150,000 / x² = 9.333...
Now, solve for x:
x² = 150,000 / 9.333...
x² ≈ 16,071.43
x ≈ sqrt(16,071.43)
x ≈ 126.77
Since x represents the number of items produced in each run, we should produce approximately 127 items in each run to minimize the total costs of production and storage.
Learn more about minimizing the total costs of production here:
https://brainly.com/question/24107867
#SPJ4
Aiden needs to find the percent of shaded squares in the bar diagram. Which model should he use? plz help
Answer: C
Step-by-step explanation:
Answer:
It might be C
Step-by-step explanation:
I am so sorry if it is the wrong answer. I didn't have a test about this.... :(
2) 5, 28, 16, 32,5, 16, 48, 29, 5, 35
Mean:?
Median:?
Mode:?
Range:?
The mean is 21.9, the median is 22, the mode is 5 and the range is 43.
Important information:
The given data values are 5, 28, 16, 32,5, 16, 48, 29, 5, 35.Mean, Median, Mode, Range:Mean of the data set is:
[tex]Mean=\dfrac{5+28+16+32+5+16+48+29+5+35}{10}[/tex]
[tex]Mean=\dfrac{219}{10}[/tex]
[tex]Mean=21.9[/tex]
Arrange the data set in asccending order.
5, 5, 5, 16, 16, 28, 29, 32, 35, 48
The number of observation is 10, which is an even number. So, the median is average of [tex]\dfrac{10}{2}=5th[/tex] term and [tex]\dfrac{10}{2}+1=6th[/tex].
[tex]Median=\dfrac{16+28}{2}[/tex]
[tex]Median=\dfrac{44}{2}[/tex]
[tex]Median=22[/tex]
Mode is the most frequent value.
In the given data set 5 has the highest frequency 3. So, the mode of the data is 5.
Range is the data set is:
[tex]Range=Maximum-Minimum[/tex]
[tex]Range=48-5[/tex]
[tex]Range=43[/tex]
Therefore, the mean is 21.9, the median is 22, the mode is 5 and the range is 43.
Find out more about 'Mean, Median, Mode, Range' here:
https://brainly.com/question/361615
Find the slope of the line that passes through the two points below (-4,9) and (8,7)
Answer:
-1/6
Step-by-step explanation:
The slope of the line can be found by
m = (y2-y1)/(x2-x1)
= (7-9)/(8 - -4)
-2 /(8+4)
-2/(12)
-1/6
a student says the prime factors of 17 are 1 and 17. Is the student correct? explain.
Answer:
Correct
Step-by-step explanation:
1. Factors
Are whole numbers that can divide the number completely.
2. Prime number
Number of factors that can be express in form of prime factorisation.
Prime factorisation of a number can be done by repeated division by prime numbers (number is written as the product of its prime factors)
For example, 17
List of factors, 1 and 17
Prime number, 17
Prime factorisation, 1×17=17
So why prime numbers cannot be express in form of prime factorisation?
Because, a prime number can only be divided by 1 or itself, so it cannot be factored any further. Not like the other number.
Brainliest would be appreciated :)
Final answer:
The student is incorrect; 17 is a prime number, and the only prime factor of 17 is 17 itself. The number 1 is not considered a prime factor as it is not a prime number.
Explanation:
No, the student is not correct in stating that the prime factors of 17 are 1 and 17. Prime factors are numbers that are both prime and divide the original number without leaving a remainder. The definition of a prime number is a number greater than 1 that has no positive divisors other than 1 and itself. In the case of 17, it is indeed a prime number, so it cannot have any prime factors other than itself. Therefore, the only prime factor of 17 is 17.
The number 1 is not considered a prime factor, as it is not a prime number because it does not meet the basic criterion of having exactly two distinct natural number divisors: 1 and itself. Examples of prime factors can be seen in other numbers, such as 15, whose prime factors are 3 and 5, both of which are prime numbers that divide 15 without leaving a remainder. In contrast, 17 can only be divided evenly by 1 and 17, and since 1 is not a prime number, 17 is the sole prime factor of 17.
A researcher claims that the mean annual cost of raising a child (age 2 and under) by husband-wife families in the U.S. is $13,960. In a random sample of husband-wife families in the U.S. the mean annual cost of raising a child (age 2 and under) is $13,725. The sample consists of 500 children and the population standard deviation is $2,345. At the α = 0.10, is there enough evidence to reject the claim? Use the p-value approach.
Answer:
[tex]z=\frac{13725-13960}{\frac{2345}{\sqrt{500}}}=-2.24[/tex]
[tex]p_v =2*P(z<-2.24)=0.0251[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 13960 at 10% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=13275[/tex] represent the sample mean
[tex]\sigma=2345[/tex] represent the sample standard deviation
[tex]n=500[/tex] sample size
[tex]\mu_o =68[/tex] represent the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is 13960, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 13690[/tex]
Alternative hypothesis:[tex]\mu \neq 13690[/tex]
If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-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".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{13725-13960}{\frac{2345}{\sqrt{500}}}=-2.24[/tex]
P-value
Since is a two sided test the p value would be:
[tex]p_v =2*P(z<-2.24)=0.0251[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 13960 at 10% of signficance.
Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last-minute" filers who mail their returns during the last five days of the income tax period (typically April 10 to April 15).
a. A researcher suggests that one of the reasons that individuals wait until the last five days to file their returns is that on average those individuals have a lower refund than early filers. Develop appropriate hypotheses such that rejection of H0 will support the researcher’s contention.
b. For a sample of 600 individuals who filed a return between April 10 and April 15, the sample mean refund was $1050 and the standard deviation was $500. Compute the p-value.
c. Using α =.05, what is your conclusion?
d. Test the hypotheses using the critical value approach (α = 0.025).
Answer:
a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.
[tex]H_0: \mu=1102\\\\H_a:\mu < 1102[/tex]
b) P-value = 0.0055
c) The null hypothesis is rejected.
There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.
d) Critical value tc=-1.96.
As t=-2.55, the null hypothesis is rejected.
Step-by-step explanation:
We have to perform a hypothesis test on the mean.
The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).
a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.
[tex]H_0: \mu=1102\\\\H_a:\mu < 1102[/tex]
b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.
We can calculate the z-statistic as:
[tex]t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55[/tex]
The degrees of freedom are df=599
[tex]df=n-1=600-1=599[/tex]
The P-value for this test statistic is:
[tex]P-value=P(t<-2.55)=0.0055[/tex]
c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.
There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.
d) If the significance level is α=0.025, the critical value for the test statistic is t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.
This is the case, as the test statistic is t=-2.55 and falls in the rejection region.
Identify if there is a relationship between the variables.
No, there is no relationship because the points are
all up and down.
Yes, the relationship displayed shows arm span
increasing as height increases.
O Yes, the relationship displayed shows arm span
decreasing as height increases.
Answer:
increases
Step-by-step explanation:
Answer:
the answer is:B
Step-by-step explanation:
Find the product.
-1(5) =
Answer:
-5
Step-by-step explanation:
when you multiply a number by -1 it becomes its opposite
The regression equation Credits = 17.9 − 0.09 Work was estimated from a sample of 21 statistics students. Credits is the number of college credits taken and Work is the number of hours worked per week at an outside job. (Round your answers to 1 decimal place.) (a) Choose the correct statement. An increase in the number of hours worked per week increases the expected number of credits. A decrease in the number of hours worked per week decreases the expected number of credits. An increase in the number of hours worked per week decreases the expected number of credits. (b) Choose the right option. The intercept is not meaningful as a student who works outside may not earn any credits. The intercept is meaningful as a student may not have a job outside of school. (c) If Work = 0, then Credits = . If Work = 60, then Credits = .
Answer:
Step-by-step explanation:
Hello!
Given the regression equation
Credit= 17.9 - 0.09 Work
That was estimated from a sample of n= 21 statistic students.
Y: number of college credits taken
X: number of hours worked per week at an outside job
a)
The estimation for the slope is -0.09
As you see the estimated slope is negative, this means that the association between the two variables is indirect or inverse. Meaning, when the number of Work increases, the number of Credit decreases. (negative regression)
The correct option is: "An increase in the number of hours worked per week decreases the expected number of credits."
b)
The estimation for the intercept is 17.9
In general, the intercept is the estimated average of Y when X=0. You can interpret the intercept of this equation as "the value of Credits earned by a student that does not work outside of school"
The correct option is: " The intercept is meaningful as a student may not have a job outside of school. "
c)
If Work=0, then Credit= 17.9 (As explained in item b)
By replacing the value in the equation: Credit= 17.9 - 0.09 * 0= 17.9
If Work=60, you replace it in the equation and:
Credit= 17.9 - 0.09 Work= 17.9-0.09*60= 12.5
If the student works 60 hs outside of school it is expected that he takes 12.5 college credits.
I hope this helps!
The correct answers are that an increase in work hours decreases the number of expected credits, the intercept is meaningful as it represents the credits for a student not working, and credits equal to 17.9 when work hours are zero and 12.5 when work hours are 60.
The regression equation given is Credits = 17.9 - 0.09 Work, where Credits is the number of college credits taken and Work is the number of hours worked per week at an outside job.
(a) From this equation, we can see that for an increase in the number of hours worked (Work), the number of expected credits (Credits) decreases because the coefficient of Work is negative (-0.09). So, the correct statement is: An increase in the number of hours worked per week decreases the expected number of credits.
(b) Regarding the intercept, it is meaningful since it represents the expected number of credits a student would take if they did not work at all (Work = 0). Therefore, the correct option is: The intercept is meaningful as a student may not have a job outside of school.
(c) If Work = 0, then Credits = 17.9. If Work = 60, then Credits = 17.9 - 0.09(60) = 17.9 - 5.4 = 12.5.
Please help!!
Which expression is equivalent to 10 - 8?
Choose 1 answer:
Answer:
Option A is the right answer choice.
Step-by-step explanation:
10 plus a negative number is the same as subtracting it as a positive
5.Richard Miyashiro purchased a condominium and obtained a 30-year loan of $196,000 at an annual interest rate of 8.20%. (Round your answers to the nearest cent.)
(a) What is the mortgage payment?
$
(b) What is the total of the payments over the life of the loan?
$
(c) Find the amount of interest paid on the mortgage loan over the 30 years.
$
6.
Marcel Thiessen purchased a home for $205,700 and obtained a 15-year, fixed-rate mortgage at 7% after paying a down payment of 10%. Of the first month's mortgage payment, how much is interest and how much is applied to the principal? (Round your answer to the nearest cent.)
interest $
applied to the principal $
Answer:
5 a) PMT=$1,465.60
b) Total Payments=$527,616
c) Total Interest=$331,616
6a) Interest=$1,079.93
b) Principal=$584.07
Step-by-step explanation:
a. Given the loan amount is $196,000, annual rate is 8.2% and the loan term is 30 years.
-The monthly mortgage payment can be calculated as follows:
[tex]PMT=A(\frac{(r/n)}{1-(1+\frac{r}{n})^{-nt}})[/tex]
Where:
PMT is the monthly mortgage paymentr is the annual interest raten,t is the number of annual payments and time in years respectively-We substitute to solve for PMT:
[tex]PMT=A(\frac{(r/n)}{1-(1+\frac{r}{n})^{-nt}})\\\\=196000[\frac{(0.082/12)}{1-(1+\frac{0.082}{12})^{-12\times30}}]\\\\=\$1,465.60[/tex]
Hence, the monthly mortgage payment is $1,465.60
b. The total number of payments is obtained by multiplying the total number of payments by the amount of each payment:
[tex]\sum(payments)=PMT\times nt\\\\=1465.60\times 12\times 30\\\\=\$527,616.00[/tex]
Hence, the total amount of payments is $527,616
c. The amount of interest paid over the loan's term is obtained by subtracting the principal loan amount from the total payments made:
[tex]Interest=Payments-Principal\\\\=527,616.00-196,000.00\\\\=\$331,616[/tex]
Hence, an interest amount of $331,616 is paid over the loan's term.
6 a) We first obtain the effective loan amount by subtracting the down-payment:
[tex]Loan \ Amount= Regular \ Price -Downpayment\\\\=205700-0.1(205700)\\\\=\$185,130[/tex]
The interest paid on the first mortgage payment is calculated as below:
[tex]I=\frac{r}{n}\times P\\\\I=Interest\\r=interest \ rate\\n=Payments \ per \ year\\P=Outstanding \ loan \ balance\\\\\therefore I=\frac{0.07}{12}\times 185130\\\\=\$1,079.93[/tex]
Hence, the amount of interest in the first payment is $1,079.93
b. The amount of principal repaid is obtained by subtracting the interest amount from the monthly mortgage payments;
[tex]Principal \ Paid=PMT-Interest\\\\PMT=A[\frac{(r/n)}{1-(1+\frac{r}{n})^{-nt}}]\\\\=185130[\frac{(0.07/12)}{1-(1+\frac{0.07}{12})^{-180}}\\\\=1664.00\\\\\\Principal \ Paid=1664.00-1079.93\\\\=\$584.07[/tex]
Hence, the amount of principal applied is $584.07
A jogger runs along a straight track. The jogger’s position is given by the function p(t), where t is measured in minutes since the start of the run. During the first minute of the run, the jogger’s acceleration is proportional to the square root of the time since the start of the run. Write a differential equation that describes this relationship, where k is a positive constant?
Answer:
[tex]\frac{d^{2}p}{dt^{2}}=k*\sqrt{t}[/tex]
Step-by-step explanation:
Given
The jogger’s position: p(t)
We can express the acceleration a as follows
[tex]a=\frac{d^{2}p}{dt^{2}}[/tex]
then
[tex]\frac{d^{2}p}{dt^{2}}=k*\sqrt{t}[/tex]
only if
[tex]0 min\leq t\leq 1 min[/tex]
The required differential equation will be [tex]\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex] or [tex]\dfrac{d^2P(t)}{dt^2}-k\sqrt t=0[/tex].
Given information:
A jogger runs along a straight track. The jogger’s position is given by the function p(t), where t is measured in minutes since the start of the run.
During the first minute of the run, the jogger’s acceleration is proportional to the square root of the time.
Let a be the acceleration of the jogger.
So, the expression for acceleration can be written as,
[tex]a=\dfrac{d}{dt}(\dfrac{dP(t)}{dt})\\a=\dfrac{d^2P(t)}{dt^2}[/tex]
Now, the acceleration is proportional to square root of time.
So,
[tex]a\propto \sqrt t\\a=k\sqrt t\\\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex]
Therefore, the required differential equation will be [tex]\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex] or [tex]\dfrac{d^2P(t)}{dt^2}-k\sqrt t=0[/tex].
For more details about differential equations, refer to the link:
https://brainly.com/question/1164377
please help :wildlife society wants to investigate the effects of a chemical spill on the growth of various types of fish. Scientists will start by measuring the lengths and weights of a sample of fish in the area. Which of the following samples should be selected for this study?
A.
fish in local lakes
B.
fish in oceans worldwide
C.
fish at the local zoo
D.
pet fish owned by local citizens
The weights of six animals at the zoom are shown in the table below
5
24
38
52
66
85
What is the mean absolute deviation? Round to the nearest tenth
Answer:
22.7
Step-by-step explanation:
To find the mean absolute deviation (MAD), first find the mean (or average).
μ = (5 + 24 + 38 + 52 + 66 + 85) / 6
μ = 45
Next, subtract the mean from each value and take the absolute value.
|5 − 45| = 40
|24 − 45| = 21
|38 − 45| = 7
|52 − 45| = 7
|66 − 45| = 21
|85 − 45| = 40
Sum the results and divide by the number of animals.
MAD = (40 + 21 + 7 + 7 + 21 + 40) / 6
MAD ≈ 22.7
In a study of the accuracy of fast food drive-through orders, one restaurant had 36 orders that were not accurate among 324 orders observed. Use a 0.01 significance level to test the claim that the rate of inaccurate orders is equal to 10%. Does the accuracy rate appear to be acceptable?
Identify the null and alternative hypotheses for this test. Choose the correct answer below.
Answer:
We do not have sufficient evidence to reject the claim that ,the rate of inaccurate orders is equal to 10%.
Step-by-step explanation:
We want to use a 0.01 significance level to test the claim that the rate of inaccurate orders is equal to 10%.
We set up our hypothesis to get:
[tex]H_0:p=0.10[/tex]------->null hypothesis
[tex]H_1:p\ne0.10[/tex]------>alternate hypothesis
This means that: [tex]p_0=0.10[/tex]
Also, we have that, one restaurant had 36 orders that were not accurate among 324 orders observed.
This implies that: [tex]\hat p=\frac{36}{324}=0.11[/tex]
The test statistics is given by:
[tex]z=\frac{\hat p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n} } }[/tex]
We substitute to obtain:
[tex]z=\frac{0.11-0.1}{\sqrt{\frac{0.1(1-0.1)}{324} } }[/tex]
This simplifies to:
[tex]z=0.6[/tex]
We need to calculate our p-value.
P(z>0.6)=0.2743
Since this is a two tailed test, we multiply the probability by:
The p-value is 2(0.2723)=0.5486
Since the significance level is less than the p-value, we fail to reject the null hypothesis.
We do not have sufficient evidence to reject the claim that ,the rate of inaccurate orders is equal to 10%.
A researcher collected data of systolic blood pressure and weight for 5 patients, as are shown in the table below.
Patient Systolic Blood Pressure (in mmHg) Weight (in lbs)
1 145 210
2 155 245
3 160 260
4 156 230
5 150 219
1. Draw a scatter plot of systolic blood pressure (response) versus weight (regressor).2. What is the direction of the association?
Answer:
We can see the details in the pic.
Step-by-step explanation:
We can see the details in the pic shown.