Answer: 1.41
Step-by-step explanation:
Test statistic(z) for proportion is given by :-
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
, where p=population proportion.
[tex]\hat{p}[/tex]= sample proportion
n= sample size.
As per given , we have
[tex]H_0:\mu=0.68\\\\ H_a: \mu\neq0.68[/tex]
n= 102
[tex]\hat{p}=0.745[/tex]
Then, the test statistic (z) for this hypothesis test will be :-
[tex]z=\dfrac{0.745-0.68}{\sqrt{\dfrac{0.68(1-0.68)}{102}}}\\\\=\dfrac{0.065}{\sqrt{\dfrac{0.2176}{102}}}\\\\=\dfrac{0.065}{\sqrt{0.0021333}}\\\\=\dfrac{0.065}{0.04618802}=1.40729132792\approx1.41[/tex]
[Rounded to the two decimal places]
Hence, the test statistic (z) for this hypothesis test = 1.41
Question 1 (40 pt). You are given designs of 3 caches for a 16-bit address machine: D1: Direct-mapped cache. Each cache line is 1 byte. 10-bit index, 6-bit tag. 1 cycle hit time. D2: 2-way set associative cache. Each cache line is 1 word (4 bytes). 7-bit index, 7-bit tag. 2 cycle hit time. D3: fully associative cache with 256 cache lines. Each cache line is 1 word. 14-bit tag. 5 cycle hit time. Answer the following set of questions: a) What is the size of each cache? b) How much space does each cache need to store tags? c) Which cache design has the most conflict misses? Which has the least? d) The following information is given to you: hit rate for the 3 caches is 50%, 70% and 90% but did not tell you which hit rate corresponds to which cache, which cache would you guess corresponded to which hit rate? Why? e) Assuming the miss time for each is 20 cycles, what is the average service time for each? (Service Time = (hit rate)*(hit time) + (miss rate)*(miss time)). Question 2 (30 pt). Assume we have a computer where the CPI is 1.0 when all memory accesses (including data and instruction accesses) hit in the cache. The cache is a unified (data + instruction) cache of size 256 KB, 4-way set associative, with a block size of 64 bytes. The data accesses (loads and stores) constitute 50% of the instructions. The unified cache has a miss penalty of 25 clock cycles and a miss rate of 2%. Assume 32-bit instruction and data addresses. Now, answer the following questions:
Answer:
Step-by-step explanation:
a) What is the size of each cache?
Direct mapped cache= 2^index * size of cache line= 2^10 * 1B lines = 1KB.
2-way set associative cache= 2^index * size of cache line * 2 ways=2^7 * 4 words *2ways= 128 4B lines * 2 ways = 1KB
Fully associative cache= number of cache lines* size of each line= 256 * 4B lines = 1KB
b) How much space does each cache need to store tags?
Direct mapped cache= 1024 * 6-bit tags = 6Kb
2-way set associative cache= 256 * 7-bit tags = 1792 bits
Fully associative cache= 256 * 14-bit tags = 3584 bits
c) Which cache design has the most conflict misses? Which has the least?
Direct mapped cache has likely the most conflict misses, because it is direct mapped. Fully associative cache has the least since it is fully associative so it can never have conflict misses.
d) The following information is given to you: hit rate for the 3 caches is 50%, 70% and 90% but did not tell you which hit rate corresponds to which cache, which cache would you guess corresponded to which hit rate? Why?
Since the size of all three caches is same size and as we said in the previous answer that direct mapped cache has more conflict misses and fully associative has the least so direct mapped will have 50%, 2-way set associative 70%, and Fully associative will have 90% hit rate.
e) Assuming the miss time for each is 20 cycles, what is the average service time for each? (Service Time = (hit rate)*(hit time) + (miss rate)*(miss time)
We are given hit rates and miss rates. Also miss time=2o cycles for each cache and hit time= 1, 2, 5 for direct mapped, 2-way set associative and fully associative cache respectively.
Direct mapped= 0.5*1 + 0.5*20 = 10.5 cycles
2-way set associative= 0.7*2 + 0.3*20 = 7.4 cycles
Fully associative cache= 0.9*5 + 0.1*20 = 6.5 cycles.
The size of each cache, the space needed to store tags, the cache design with the most and least conflict misses, guessing which cache corresponds to each hit rate, and calculating the average service time for each cache.
Explanation:D1: Direct-mapped cache:Size: Each cache line is 1 byte, so the total cache size is 2^10 * 1 byte = 1024 bytesTag space: Each cache line has a 6-bit tag, so the total tag space required is 2^6 * 1 byte = 64 bytesD2: 2-way set associative cache:D3: Fully associative cache:Size: Each cache line is 1 word, so the total cache size is 256 lines * 2^2 bytes = 1024 bytesTag space: Each cache line has a 14-bit tag, so the total tag space required is 2^14 * 1 byte = 16384 bytesEach cache design has different levels of conflict misses. The direct-mapped cache (D1) has the most conflict misses because multiple memory locations map to the same cache line. The 2-way set associative cache (D2) has fewer conflict misses because each set can hold two cache lines, reducing the chance of multiple memory locations mapping to the same set. The fully associative cache (D3) has the least conflict misses because any memory location can be stored in any cache line, reducing conflicts.
Based on the given hit rates, we can guess which cache corresponds to each hit rate. The cache with the highest hit rate (90%) is likely to be D3 (fully associative cache), as it has the lowest miss rate and therefore the highest hit rate. The cache with the lowest hit rate (50%) is likely to be D1 (direct-mapped cache), as it has the highest miss rate and therefore the lowest hit rate. The cache with the intermediate hit rate (70%) is likely to be D2 (2-way set associative cache).To calculate the average service time for each cache, we use the formula: Average service time = (hit rate) * (hit time) + (miss rate) * (miss time). Given a miss time of 20 cycles for each cache, we can calculate the average service time for each cache using their respective hit rates:D1: Average service time = (0.5) * (1 cycle) + (0.5) * (20 cycles) = 10.5 cyclesD2: Average service time = (0.7) * (2 cycles) + (0.3) * (20 cycles) = 4.6 cyclesD3: Average service time = (0.9) * (5 cycles) + (0.1) * (20 cycles) = 5.5 cyclesCalculate the standard deviation σ of X for the probability distribution. (Round your answer to two decimal places.)σ =x 1 2 3 4P(X = x)0.2 0.2 0.2 0.4Calculate the standard deviation σ of X for the probability distribution. (Round your answer to two decimal places.)σ =x −20 −10 0 10 20 30P(X = x)0.1 0.2 0.4 0.1 0 0.2
Answer:
a) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{1.36}=1.166[/tex]
b) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{149}=12.21[/tex]
Step-by-step explanation:
Part a
So then the random variable is given by this table
X | 1 | 2 | 3 | 4 |
P(X) | 0.2 | 0.2 | 0.2 | 0.4 |
First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.
In order to calculate the expected value we can use the following formula:
[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]
And if we use the values obtained we got:
[tex]E(X)=1*0.2 +2*0.2 +3*0.2 +4*0.4=2.8[/tex]
In order to find the standard deviation we need to find first the second moment, given by :
[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]
And using the formula we got:
[tex]E(X^2)=(1^2 *0.2)+(2^2 *0.2)+(3^2 *0.2)+(4^2 *0.4)=9.2[/tex]
Then we can find the variance with the following formula:
[tex]Var(X)=E(X^2)-[E(X)]^2 =9.2-(2.8)^2 =1.36[/tex]
And then the standard deviation would be given by:
[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{1.36}=1.17[/tex]
Part b
So then the random variable is given by this table
X | -20 | -10 | 0 | 10 |20 |
P(X) | 0.1 | 0.2 | 0.4 | 0.1 | 0.2 |
First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.
In order to calculate the expected value we can use the following formula:
[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]
And if we use the values obtained we got:
[tex]E(X)=(-20*0.1) +(-10*0.2) +(0*0.4) +(10*0.1)+(20*0.2)=1[/tex]
In order to find the standard deviation we need to find first the second moment, given by :
[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]
And using the formula we got:
[tex]E(X^2)=((-20)^2 *0.1)+((-10)^2 *0.2)+(0^2 *0.4)+(10^2 *0.1)+(20^2 *0.2)=150[/tex]
Then we can find the variance with the following formula:
[tex]Var(X)=E(X^2)-[E(X)]^2 =150-(1)^2 =149[/tex]
And then the standard deviation would be given by:
[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{149}=12.21[/tex]
The standard deviation of the first probability distribution is 1.10 and of the second distribution is 15.23.
Explanation:The standard deviation σ of a probability distribution is calculated by first finding its mean μ, then using the formula:
σ = √[Σ(x-μ)^2 * P(X = x)]
For the first distribution, the mean μ = (1*0.2) + (2*0.2) + (3*0.2) + (4*0.4) = 3.2. The standard deviation σ = √[(1-3.2)^2 * 0.2 + (2-3.2)^2 * 0.2 + (3-3.2)^2 * 0.2 + (4-3.2)^2 * 0.4] = 1.10.
For the second distribution, the mean μ = (-20*0.1) + (-10*0.2) + (0*0.4) + (10*0.1) + (20*0) + (30*0.2) = -2. The standard deviation σ = √[(-20+2)^2 * 0.1 + (-10+2)^2 * 0.2 + (0+2)^2 * 0.4 + (10+2)^2 * 0.1 + (20+2)^2 * 0 + (30 + 2)^2 * 0.2] = 15.23.
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Calculate ∫C(7(x2−y)i⃗ +3(y2+x)j⃗ )⋅dr⃗ if: (a) C is the circle (x−2)2+(y−3)2=9 oriented counterclockwise. ∫C(7(x2−y)i⃗ +3(y2+x)j⃗ )⋅dr⃗ = (b) C is the circle (x−a)2+(y−b)2=R2 in the xy-plane oriented counterclockwise. ∫C(7(x2−y)i⃗ +3(y2+x)j⃗ )⋅dr⃗ =
By Green's theorem,
[tex]\displaystyle\int_C(7(x^2-y)\,\vec\imath+3(y^2+x)\,\vec\jmath)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial3(y^2+x)}{\partial x}-\frac{\partial7(x^2-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]
[tex]\displaystyle=10\iint_D\mathrm dx\,\mathrm dy[/tex]
where [tex]D[/tex] is the region bounded by the closed curve [tex]C[/tex]. The remaining integral is 10 times the area of [tex]D[/tex].
Since [tex]D[/tex] is a circle in both cases, and we're given the equations for them right away, it's just a matter of determining the radius of each one and plugging it into the well-known formula for the area of a circle with radius [tex]r[/tex], [tex]\pi r^2[/tex].
(a) [tex]C[/tex] is a circle with radius 3, so the line integral is [tex]10\pi(3^2)=\boxed{90\pi}[/tex].
(b) [tex]C[/tex] is a circle with radius [tex]R[/tex], so the line integral is [tex]\boxed{10\pi R^2}[/tex].
Let ,
First calculate the value for given equation by using green's theorem,
Since, the formula for greens theorem is:
[tex]\int\ CFds=\int\ \int\ CurlFkdA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ CurlFkdA[/tex]...(1)The given equation is,
[tex]\int\ C [7(x^2-y)i +3(y^2+x)j]dr[/tex]
Here,
[tex]F=[7(x^2-y)i +3(y^2+x)j][/tex]
Now to calculate the value of [tex]Curl F[/tex],
[tex]Curl F=\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial k} \\7(x^2-y)&(y^2+x)&0\end{array}\right] \\\\CurlF=[\frac{\partial}{ \partial y} (0)-\frac{\partial}{\partial k} (y^2+x)]+[\frac{\partial}{ \partial x} (0)-\frac{\partial}{\partial k} (y^2+x)]+[\frac{\partial}{ \partial x} (y^2+x)-\frac{\partial}{\partial y} (7x^2-y)]\\\\Curl F=0+0+10k\\\\CurlF=10k[/tex]
Substitute in equation (1),
[tex]\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ 10k *kdA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=\int\ \int\ 10dA\\\\\int\ C(7(x^2-y)i +3(y^2+x)j=10\int\ \int\ dA[/tex]
The remaining integral is [tex]10[/tex] times the area of region .
The general equation is,
[tex]x^2+y^2=r^2[/tex]
The area of a circle is [tex]\pi r^2[/tex] .
Hence, area of region of circle is [tex]10\pi r^2[/tex].
Now,
(a) The given equation is [tex](x-2)^2+(y-3)^2=3^2[/tex],
C is a circle with radius 3, so the line integral is
[tex]10\pi (3)^2=90\pi[/tex] .
(b) The given equation is [tex](x-a)^2+(y-b)^2=R^2[/tex],
C is a circle with radius [tex]R[/tex] , so the line integral is,
[tex]10\pi (R)^2=10\pi R^2[/tex].
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The following price quotations are for exchange-listed options on Primo Corporation common stock. Company Strike Expiration Call Put Primo 61.12 55 Feb 7.25 0.48 With transaction costs ignored, how much would a buyer have to pay for one call option contract. Assume each contract is for 100
Answer:
$ 725
Step-by-step explanation:
Price of call option = 7.25
buyer have to pay for one call option contract. Assume each contract is for 100 = 100 * 7.25 = $ 725
A major software company is arranging a job fair with the intention of hiring 6 recent graduates. The 6 jobs are different, and numbered 1 through 6. No candidate can receive more than one offer. In response to the company's invitation, 136 candidates have appeared at the fair. a. How many ways are there to extend the 6 offers to 6 of the 136 candidates? {1 point} b. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer, but we do not know which? {1 point} C. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer for job number 2? {1 point} d. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is not getting any offers? {1 point} e. How many ways are there for 3 interviewers to select 3 resumes (one resume for each interviewer) from the pile of 136 resumes for the first interview round?
Answer:
a) 7,858,539,612
b) 2,080,201,662
c) 346,700,277
d) 7,511,839,335
e) 410,040
Step-by-step explanation:
a. How many ways are there to extend the 6 offers to 6 of the 136 candidates?
Combinations of 136 (candidates) taken 6 (offers) at a time without repetition:
[tex]\large \binom{136}{6}=\frac{136!}{6!(136-6)!}=\frac{136!}{6!130!}=7,858,539,612[/tex]
b. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer, but we do not know which?
There are 6 ways Computer Joe can get an offer. Now there are left 5 offers and 135 candidates. So there are
6 times combinations of 135 taken 5 at a time without repetition:
[tex]\large 6*\binom{135}{5}=6*\frac{135!}{5!(135-5)!}=6*\frac{135!}{5!130!}=2,080,201,662[/tex]
c. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is getting an offer for job number 2?
Now, we only have 5 offers and 135 candidates. So there are combinations of 135 taken 5 at a time without repetition:
[tex]\large \binom{135}{5}=\frac{135!}{5!(135-5)!}=\frac{135!}{5!130!}=346,700,277[/tex]
d. How many ways are there to extend the 6 offers to 6 of the 136 candidates, if we already know that Computer Joe is not getting any offers?
Here we have 6 offers and 135 candidates, given that Computer Joe is out. So there are combinations of 135 taken 6 at a time without repetition:
[tex]\large \binom{135}{6}=\frac{135!}{6!(135-6)!}=\frac{135!}{6!129!}=7,511,839,335[/tex]
e. How many ways are there for 3 interviewers to select 3 resumes (one resume for each interviewer) from the pile of 136 resumes for the first interview round?
There are combinations of 136 taken 3 at a time without repetition:
[tex]\large \binom{136}{3}=\frac{136!}{3!(136-3)!}=\frac{136!}{3!133!}=410,040[/tex]
To determine whether the means of two populations are equal,
A. a t test must be performed.
B. an analysis of variance must be performed.
C. either a t test or an analysis of variance can be performed.
D. a chi-square test must be performed.
Answer:
The correct option is C. either a t test or an analysis of variance can be performed.
Step-by-step explanation:
Consider the provided information.
The t-test, is used for whether the means of two groups are equal or not. The assumption for the test is that both groups are sampled from normal distributions with equal variances. Analysis of Variance (ANOVA) is a statistical method evaluating variations between two or more methods. ANOVA is used in a study to analyze the gaps between group methods.ANOVA is used not for specific differences between means, but for general testing.The chi-squared test is often used to evaluate whether there was a significant difference in one or more groups between the predicted frequencies and the observed frequencies.Hence, Either a t test or an analysis of variance can be performed to determine whether the means of two population are equal.
Therefore, the correct option is C. either a t test or an analysis of variance can be performed.
To determine the means of two populations are equal or not: C. either a t-test or an analysis of variance can be performed.
What is the t-test and ANOVA used for?The t-test is a statistical test that is used to compare the means of two groups, to determine if there is any significant difference between the two groups or populations.
The ANOVA, like the t-test is also used to compare means, however, it is used when the groups or populations involved are more than two.
Therefore, to determine the means of two populations are equal or not: C. either a t-test or an analysis of variance can be performed.
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You wish to test the claim that p > 33 at a level of significance of a = 0.05 and are given sample 19) statistics n = 5O x = 33.3. Assume the population standard deviation is 12. Compute the value of the standardized test statistic. Round your answer to two decimal places.
Answer:
test statistic is 0.176
Step-by-step explanation:
Given Data
p>33
a=0.05
n=50
x=33.3
d(population deviation)=12
Test statistics=?
Solution
Test statistic z=(p-x)\(d/sqrt(50))
z=(33.3-30)\(12\sqrt(50))
z=0.176
use the functions f(x)=2x and g(x)=x^2+1 to find the value of each expression
1. f(g(3))
2. f(3)+g(4)
3. f(5)-2×g(1)
Answer:
1. 20
2. 23
3. 6
Step-by-step explanation:
We have that:
f(x) = 2x
g(x) = x² + 1
f(g(x)) is the composite function of f and g. So
f(g(x)) = f(x²-1) = 2(x²+1) = 2x² + 2
1. f(g(3))
f(g(x)) = 2x² - 2 = 2(3)² + 2 = 18 + 2 = 20
2. f(3)+g(4)
f(3) = 2(3) = 6
g(4) = 4² + 1 = 17
f(3) + g(4) = 6 + 17 = 23
3. f(5) - 2g(1)
f(5) = 2(5) = 10
g(1) = (1)² + 1 = 2
f(5) - 2g(1) = 10 - 2*2 = 10 - 4 = 6
find the equation of a line that is perpendicular to the given line and passes through the given point. enter your answer in slope intercept form y=mx+b with the values of m and b given as decimals, rounded to the nearest hundredth. y =10x ; (8,-2)
Answer:
m= -0.1
b= -1.2
Step-by-step explanation:
The equation of the given line is
y=10x, slope of the given line is 10.
As the line is perpendicular to this line, the slope is [tex]-\frac{1}{10}[/tex]
m= -0.1
( product of slope of two perpendicular lines is -1)
thus the equation is
[tex]y = (-\frac{1}{10})x + b[/tex]
[tex]y = (-0.1)x + b[/tex]
now to find the value of b, input the point coordinates (8, -2) in the above equation.
[tex]-2 = (-0.1)(8) + b[/tex]
[tex]b = -2 + (0.8)[/tex]
b = -1.2
A car dealer is interested in comparing the average gas mileages of four different car models. The dealer believes that the average gas mileage of a particular car will vary depending on the person who is driving the car due to different driving styles. Because of this, he decides to use a randomized block design. He randomly selects five drivers and asks them to drive each of the cars. He then determines the average gas mileage for each car and each driver. Can the dealer conclude that there is a significant difference in average gas mileages of the four car models? The results of the study are as follows. Average Gas Mileage Driver Car A Car B Car C Car D Driver 1 29 31 20 34 Driver 2 27 37 35 39 Driver 3 24 23 31 23 Driver 4 38 24 22 38 Driver 5 20 33 37 36 ANOVA Source of Variation SS df MS Rows 190.2000 4 47.5500 Columns 114.5500 3 38.1833 Error 534.2000 12 44.5167 Total 838.9500 19 Step 1 of 3: Find the value of the test statistic for testing whether the average gas mileage is the same for the four car models. Round your answer to two decimal places, if necessary.
Answer:
Step-by-step explanation:
The F statistic, calculated through one-way ANOVA, for this problem is 1.07, which aims to examine the differences in the averages of multiple groups. However, without details such as the significance level and degrees of freedom, this task cannot determine if there's a significant difference between the average gas mileages of the four car models.
Explanation:The average gas mileage comparison across four car models represented by the car dealer is an example of a problem solved by the One-Way ANOVA statistical approach. This test aims to determine if there is a statistically significant difference between the means of multiple groups, in this case, the average mileage of four different car models.
To find the test statistic, we consider the between-group mean square (MS Between) and the within-group mean square (MS Within). In ANOVA, the F statistic is used which performs the test of two variances, and is calculated as the ratio of MS Between to MS Within. In this case, MS Between is represented by 'MS Rows' (47.5500) and MS Within by 'MS Error' (44.5167). So, the F statistic = MS Between / MS Within = 47.5500 / 44.5167 = 1.07 (rounded to two decimal places).
However, the value of the F statistic alone is not enough to conclude the test. The conclusion depends on the significance level, degrees of freedom, and the value from the F-distribution table. Without these details, we cannot conclude whether there's a significant difference in the average gas mileages of the four car models.
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For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,3] into n equal subintervals and using the right-hand endpoint for each c[Subscript]k. Then take a limit of this sum as n approaches infinity to calculate the area under the curve over [0,3].
f(x)=2x^2
We divide [0, 3] into [tex]n[/tex] subintervals,
[tex]\left[0,\dfrac3n\right]\cup\left[\dfrac3n,\dfrac6n\right]\cup\left[\dfrac6n,\dfrac9n\right]\cup\cdots\cup\left[\dfrac{3(n-1)}n,3\right][/tex]
so that the right endpoint of each subinterval is given according to the arithmetic sequence,
[tex]r_k=\dfrac{3k}n[/tex]
for [tex]1\le k\le n[/tex].
The Riemann sum is then
[tex]\displaystyle\sum_{k=1}^nf(r_k)\Delta x_k[/tex]
where
[tex]\Delta x_k=r_k-r_{k-1}=\dfrac{3k}n-\dfrac{3(k-1)}n=\dfrac3n[/tex]
With [tex]f(x)=2x^2[/tex], we have
[tex]\displaystyle\frac3n\sum_{k=1}^n2\left(\frac{3k}n\right)^2=\frac{54}{n^3}\sum_{k=1}^nk^2[/tex]
Recall that
[tex]\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6[/tex]
The area under the curve [tex]f(x)[/tex] over the interval [0, 3] is then
[tex]\displaystyle\int_0^32x^2\,\mathrm dx=\lim_{n\to\infty}\frac{54n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}9\left(2+\frac3n+\frac1{n^2}\right)=\boxed{18}[/tex]
Describe how the variability of the distribution changes as the sample size increases. As the sample size increases, the variability decreases. It cannot be determined. As the sample size increases, the variability stays the same. As the sample size increases, the variability increases.
Answer:
As the sample size increases, the variability decreases.
Step-by-step explanation:
Variability is the measure of actual entries from mean. The less the deviations the less would be the variance.
For a sample of size n, we have by central limit theorem the mean of sample follows a normal distribution for random samples of large size.
X bar will have std deviation as [tex]\frac{s}{\sqrt{n} }[/tex]
where s is the square root of variance of sample
Thus we find the variability denoted by std deviation is inversely proportion of square root of sample size.
Hence as sample size increases, std error decreases.
As the sample size increases, the variability decreases.
In statistics, as the sample size increases, the variability typically decreases because more data points allow a closer approximation of the true population mean. Therefore, larger sample sizes provide a narrower confidence interval, leading to less variability.
Explanation:The variability of a distribution is a measure of the differences from the mean that occur in the data points. The sample size refers to the number of data points collected in your sample from a population. In statistics, as the sample size increases, the variability or scatter of your dataset normally decreases, because a larger number of data points give a more accurate representation of the population you are studying.
Variability is affected by sample size in the following way: Increasing the sample size leads to a decrease in the error bound and makes a narrower confidence interval. This is because more data points enable a closer estimation of the true population mean. Thus, as your sample size grows larger, the variability decreases, and your data forms a tighter grouping around the mean.
For example, if you are conducting a survey, and you take four different samples of 50 people each from the same population, you might see differing outcomes due to sample variability. However, if you were to increase your sample size to perhaps 500 people, the results are likely to have less sample variability.
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A function y(t) satisfies the differential equation dy dt = y 4 − 6y 3 + 5y 2 . (a) What are the constant solutions of the equation? (Recall that these have the form y = C for some constant, C.) (b) For what values of y is y increasing? (c) For what values of y is y decreasing?
Answer:
Hence increasing in (-\infty,0) U (1,5)
c) Decreasing in (0,1)
Step-by-step explanation:
Given that y(t) satisfies the differential equation
[tex]\frac{dy}{dt} =y^4-6y^2+5y^2\\=y^2(y^2-6y+5)\\=y^2(y-1)(y-5)[/tex]
Separate the variables to have
[tex]\frac{dy}{y^2(y-1)(y-5)} =dt[/tex]
Left side we can resolve into partial fractions
Let [tex]\frac{1}{y^2(y-1)(y-5)} =\frac{A}{y} +\frac{B}{y^2}+\frac{C}{y-1} \frac{D}{y-5}[/tex]
Taking LCD we get
[tex]1= Ay(y-1)(Y-5) +B(y-1)(y-5)+Cy^2 (y-5)+Dy^2 (y-1)\\Put y =1\\1 = -4C\\Put y =5\\ 1 = 25(4)D\\Put y =0\\1=5B\\[/tex]
By equating coeff of y^3 we have
A+C+D=0
[tex]C=\frac{-1}{4} \\D=\frac{1}{100} \\B =\frac{1}{5} \\A = -C-D = \frac{6}{25}[/tex]
Hence left side =
[tex]\frac{6}{25y} +\frac{1}{5y^2}+\frac{-1}{4(y-1)}+ \frac{1}{100(y-5)}=dt\\\frac{6}{25}ln y -\frac{1}{5y}-\frac{1}{4}ln|(y-1)| +\frac{1}{100}ln|y-5| = t+C[/tex]
b) y is increasing whenever dy/dt>0
dy/dt =0 at points y =0, 1 and 5
dy/dt >0 in (-\infty,0) U (1,5)
Hence increasing in (-\infty,0) U (1,5)
c) Decreasing in (0,1)
Answer:
a) y = 0 , 5,1
b) y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)
Step-by-step explanation:
Given data:
differential equation is given as
[tex]\frac{dy}[dt} = y^4 -6y^3+ 5y^2[/tex]
a) constant solution
[tex] y^4 -6y^3+ 5y^2 = 0 [/tex]
taking y^2 from all part
[tex]y^2(y^2 - 6y -5) = 0[/tex]
solution of above equation is
y = 0 , 5,1
b) for which value y is increasing
[tex]\frac{dy}{dt} > 0[/tex]
y^2(y - 5) (y -1) > 0
y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)
The Hypotenuse of a right triangle is 0.5 units long. The longer leg is 0.1 units longer than the shorter leg. Find the lengths of the sides of the triangle.
Answer:
The lengths of the sides of a right triangle are
Longer leg = 0.4 units.
Shorter leg = 0.3 units.
Step-by-step explanation:
Given:
Hypotenuse = 0.5 units
Let the length of shorter leg of right triangle be x units then
According to the given condition,
length of longer leg will be (0.1 + x) units
Now,we know for a right triangle,by Pythagoras theorem we have
[tex](\textrm{Hypotenuse})^{2} = (\textrm{Longer leg})^{2}+(\textrm{Shorter leg})^{2}[/tex]
substituting the values we get
[tex]0.5^{2}= (x+0.1)^{2}+ x^{2}[/tex]
Applying [tex](a+b)^{2}= a^{2}+2ab+b^{2}[/tex] we get
[tex]0.25= x^{2} +2\times 0.1\times x+ 0.1^{2} + x^{2} \\2x^{2} +0.2x+0.01-0.25=0\\2x^{2} +0.2x-0.24=0\\[/tex]
which is a quadratic equation
dividing the equation throughout by two we get
[tex]x^{2} +0.1x-0.12=0\\\textrm{on factorizing we get}\\x^{2} +0.4x-0.3x-0.12=0\\(x+0.4)(x-0.3)=0[/tex]
[tex]\therefore (x-0.3)= 0\\\therefore x=0.3[/tex]
Since x cannot be negative we take
x = 0.3 units
∴ Longer leg = x + 0.1
= 0.3+0.1
=0.4 units
So, the lengths of the sides of a right triangle are
Longer leg = 0.4 units.
Shorter leg = 0.3 units.
Final answer:
To find the lengths of the sides of the right triangle with a hypotenuse of 0.5 units and one leg being 0.1 units longer than the other, you can use the Pythagorean theorem to set up an equation and solve for the lengths of the legs.
Explanation:
To find the lengths of the sides of the triangle, let's assume the length of the shorter leg is x units. Then, the length of the longer leg would be x + 0.1 units. Applying the Pythagorean theorem, where a and b are the legs and c is the hypotenuse:
a² + b² = c²
(x)² + (x + 0.1)² = (0.5)²
Simplifying the equation and solving for x, we get x ≈ 0.226 units for the shorter leg, and the longer leg would be x + 0.1 ≈ 0.326 units.
evaluate x(y+3)/(3+y)z for x=6 y=9 z=2
I'm struggling to solve this equation. Please help me. thank you!
Answer:3
Step-by-step explanation:plug in the numbers in the algebraic expression. Should look like this: 6(9+3)/(3+9)2. After you set it up like this you have to get rid of the parentheses by multiplying the outside number to the numbers in the parentheses. So do: 6×9=54, then 6×3=18 so now your problem should look like this 54+18/ (3+9)2 do the same with the other side so: 2×3=6, then 9×2=18 now your problem should look like this: 54+18/6+18 now you add each side up 54+18=72 and 6+18=24 . Then divide those 2 answers which looks like this 72/24 = 3.
Lucero wants to hang 3 paintings in her room. The widths of the paintings are 10 1/2 inches, 3 1/2 feet, 2 feet, and 2 3/4 inches. If she hangs them next to each other with 3 inches between them, what is the total width of the wall space she will need?
Answer:
7 ft 1 1/4 in
Step-by-step explanation:
The total width of 3 paintings and 2 spaces is ...
(10.5 in) + (3 in) + (3 ft 6 in) + (3 in) + (2 ft 2 3/4 in)
= (10.5 +3 +6 +3 +2.75) in + (3 +2) ft
= 25.25 in + 5 ft = 1.25 in + 24 in + 5 ft
= 7 ft 1 1/4 in
The total width of wall space needed for the paintings is 7 feet 1 1/4 inches.
_____
If two more 3-inch spaces are added, one on each end, then the total width is 7 feet 7 1/4 inches. The problem isn't clear about that, saying only that there are spaces between the paintings.
The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 50 and a standard deviation of 11. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 39 and 50?
Answer:
34% of lightbulb replacement requests numbering between 39 and 50.
Step-by-step explanation:
The 68-95-99.7 rule states that, for a normally distributed random variable:
68% are within 1 standard deviation of the mean(34% between one standard deviation below and the mean, 34% between the mean and one standard deviation above the mean).
95% are within 2 standard deviations of the mean.
99.7% are within 3 standard deviations of the mean.
In this problem, we have that:
The distribution of the number of daily requests is bell-shaped and has a mean of 50 and a standard deviation of 11.
Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 39 and 50?
50 is the mean
39 is one standard deviation below the mean.
This means that 34% of lightbulb replacement requests numbering between 39 and 50.
Given below are the number of successes and sample size for a simple random sample from a population. xequals6, nequals50, 90% level a. Determine the sample proportion. b. Decide whether using the one-proportion z-interval procedure is appropriate. c. If appropriate, use the one-proportion z-interval procedure to find the confidence interval at the specified confidence level. d. If appropriate, find the margin of error for the estimate of p and express the confidence interval in terms of the sample proportion and the margin of error.
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]
I hope you have a SUPER day!
A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 505π ft3, determine the radius and height of the silo that requires the least amount of material to build. Hint: The volume of the silo is πr2h + 2 3 πr3, and the surface area (including the floor) is π(3r2 + 2rh). (Round your answers to one decimal place.)
r = (3V/2π)^(1/3) and h = V/(πr^2), respectively.
Explanation:To determine the dimensions of the silo that requires the least amount of material to build, we need to find the values of the radius and height that minimize the surface area. Let's start by expressing the volume of the silo as a function of one variable. The volume is given by V = πr^2h + (2/3)πr^3, where r is the radius and h is the height. Since we're looking for the minimum surface area, we'll differentiate the surface area expression, equate it to zero, and solve the resulting equation to find the values of r and h that minimize the surface area.
Taking the derivative of the surface area function A(r, h) = π(3r^2 + 2rh) with respect to r and h, we get:
dA/dr = 6πr + 2πh = 0
dA/dh = 2πr = 0
Solving these equations simultaneously, we find that r = 0 and h = 0, which are not meaningful values in this context. Therefore, there are no critical points inside the domain. Instead, we need to examine the endpoints of the domain. Since r and h must be positive, we find that A(r, h) goes to infinity as r or h approaches zero. Therefore, the surface area does not have a minimum in the interior of the domain and we need to consider the endpoints.
First, we'll consider the case where h = 0. In this case, the silo is just a hemispherical dome with no height, so the surface area is equal to the curved surface area of the hemisphere, which is given by 2πr^2. We can rewrite the volume equation as V = πr^2h + (2/3)πr^3 = (2/3)πr^3, since h = 0. Solving this equation for r, we find that r = (3V/2π)^(1/3).Next, we'll consider the case where r = 0. In this case, the silo is just a right circular cylinder with no radius, so the surface area is equal to the curved surface area of the cylinder, which is given by 2πrh. We can rewrite the volume equation as V = πr^2h + (2/3)πr^3 = πr^2h, since r = 0. Solving this equation for h, we find that h = V/(πr^2).Now, we compare the surface areas of the two cases to determine which requires the least amount of material. A(h = 0) = 2πr^2 and A(r = 0) = 2πrh = 2V/r. Substituting the values of r and h from the previous calculations, we find that A(h = 0) = (2π/3)(3V)^(2/3) and A(r = 0) = 6V^(1/3)/(3π)^(2/3). Since the surface area of the dome is smaller than that of the cylinder, the silo with a hemispherical dome on top requires the least amount of material to build. Therefore, the radius and height of the silo are given by r = (3V/2π)^(1/3) and h = V/(πr^2), respectively.Learn more about Optimization here:https://brainly.com/question/33322941
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To determine the dimensions of the silo that requires the least material, set up an equation involving the volume and the surface area. Solve for the values of r and h that minimize the surface area. Use differentiation and the second derivative test to find the minimum point.
Explanation:To determine the radius and height of the silo that requires the least amount of material to build, we need to find the dimensions that minimize the surface area of the silo. The surface area of the silo is given by A = π(3r^2 + 2rh), and the volume of the silo is given by V = πr^2h + (2/3)πr^3.
We are given that the volume of the silo is 505π ft^3. We can use this information to set up an equation involving the volume and the dimensions of the silo. Solving this equation will give us the values of r and h that minimize the surface area.
By differentiating the surface area equation with respect to r and h, and setting the derivatives equal to zero, we can find the critical points. By analyzing the second derivative test, we can determine which point is the minimum.
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The test statistic of zequalsnegative 2.40 is obtained when testing the claim that less than 0.32. a. Using a significance level of alphaequals0.10, find the critical value(s). b. Should we reject Upper H 0 or should we fail to reject Upper H 0
Answer:
a) Critical value = -1.285
b) We should reject null hypothesis that the mean equals 0.32
Step-by-step explanation:
Given that the statistic of z equals negative 2.40 is obtained when testing the claim that less than 0.32
i.e. for hypotheses
[tex]H_0: \bar = 0.32\\H_a: \bar x <0.32\\[/tex]
(one tailed test at 10% significance level)
Z critical value for 90% one tailed = -1.285
Since our test statistic is less than -1.285 we reject null hypothesis
a) Critical value = -1.285
b) We should reject null hypothesis that the mean equals 0.32
As a general rule, the sampling distribution of the sample proportions can be approximated by a normal probability distribution whenever
a)np greater than or equal to 5
b) n(1-p) greater than or equal to 5
c) n greater than or equal to 30
d) both a and b are true
A continuous random variable is uniformly distributed between a and b. The probability density function between a and b is
a) zero.
b) (a - b).
c) (b - a).
d) 1/(b - a).
A population has a mean of 84 and a standard deviation of 12. A sample of 36 observations will be taken. The probability that the sample mean will be between 80.54 and 88.9 is
a) 0.0347.
b) 0.7200.
c) 0.9511.
d) None of the alternative answers is correct.
In probability, the sampling distribution of the sample proportions can be approximated by a normal probability distribution when D. Both a and b are true.
How to calculate the probability?The sampling distribution of the sample proportions can be approximated by a normal probability distribution whenever np is greater than or equal to 5 and when n(1-p) is greater than or equal to 5.
When the continuous random variable is uniformly distributed between a and b, the probability density function between a and b is 1/(b - a).
The probability that the sample mean will be between 80.54 and 88.9 will be:
= P[Z = 88.9 - 84)/(12/✓36)] - P(Z = 80.54 - 84)/(12/✓36)]
= P(Z = 2.45) - P(Z = -1.73)
= 0.9929 - 0.0418
= 0.9511
Therefore, the probability is 0.9511.
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Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 10pm and 3 am, find the probability that:
a) Exactly two will be drunken drivers.
b) Three or four will be drunken drivers.
c) At least 7 will be drunken drivers.
d) At most 5 will be drunken drivers.
Answer:
(a) 0.28347
(b) 0.36909
(c) 0.0039
(d) 0.9806
Step-by-step explanation:
Given information:
n=12
p = 20% = 0.2
q = 1-p = 1-0.2 = 0.8
Binomial formula:
[tex]P(x=r)=^nC_rp^rq^{n-r}[/tex]
(a) Exactly two will be drunken drivers.
[tex]P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}[/tex]
[tex]P(x=2)=66(0.2)^{2}(0.8)^{10}[/tex]
[tex]P(x=2)=\approx 0.28347[/tex]
Therefore, the probability that exactly two will be drunken drivers is 0.28347.
(b)Three or four will be drunken drivers.
[tex]P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)[/tex]
[tex]P(x=3\text{ or }x=4)=P(x=3)+P(x=4)[/tex]
Using binomial we get
[tex]P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}[/tex]
[tex]P(x=3\text{ or }x=4)=0.236223+0.132876[/tex]
[tex]P(x=3\text{ or }x=4)\approx 0.369099[/tex]
Therefore, the probability that three or four will be drunken drivers is 0.3691.
(c)
At least 7 will be drunken drivers.
[tex]P(x\geq 7)=1-P(x<7)[/tex]
[tex]P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)][/tex]
[tex]P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155][/tex]
[tex]P(x\leq 7)=1-[0.9961][/tex]
[tex]P(x\leq 7)=0.0039[/tex]
Therefore, the probability of at least 7 will be drunken drivers is 0.0039.
(d) At most 5 will be drunken drivers.
[tex]P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)[/tex]
[tex]P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315[/tex]
[tex]P(x\leq 5)=0.9806[/tex]
Therefore, the probability of at most 5 will be drunken drivers is 0.9806.
a) Exactly two will be drivers: 0.2835. b) Three or four will be drivers: 1.5622; c) At least 7 will be drivers: 32.5669 (rounded to four decimal places). d) At most 5 will be drivers: 0.8749
a) Exactly two will be drivers:
For this case, we'll use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
Where:
n is the number of trials (in this case, the number of drivers in the sample), which is 12.
k is the number of successful trials (in this case, the number of drivers), which is 2.
p is the probability of success in a single trial (in this case, the probability of a driver being), which is 0.20.
(n choose k) is the number of combinations of n items taken k at a time.
Calculating:
P(X = 2) = (12 choose 2) * (0.20)^2 * (0.80)^10
= 66 * 0.04 * 0.1073741824
= 0.2834678413
b) Three or four will be drivers:
For this case, we'll find P(X = 3) and P(X = 4) and then add them together.
For P(X = 3):
P(X = 3) = (12 choose 3) * (0.20)^3 * (0.80)^9
= 220 * 0.008 * 0.134456
= 0.23757696
For P(X = 4):
P(X = 4) = (12 choose 4) * (0.20)^4 * (0.80)^8
= 495 * 0.016 * 0.16777216
= 1.3245924272
Adding them together:
P(X = 3 or 4) = 0.23757696 + 1.3245924272
= 1.5621693872
c) At least 7 will be drivers:
To find this probability, we need to calculate P(X = 7) + P(X = 8) + ... + P(X = 12).
For P(X = 7):
P(X = 7) = (12 choose 7) * (0.20)^7 * (0.80)^5
= 792 * 0.128 * 0.32768
= 32.5669376
Similarly, find P(X = 8), P(X = 9), P(X = 10), P(X = 11), and P(X = 12) using the same method.
Finally, add all these probabilities together.
d) At most 5 will be drivers:
To find this probability, we need to calculate P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5).
For P(X = 0):
P(X = 0) = (12 choose 0) * (0.20)^0 * (0.80)^12
= 1 * 1 * 0.0687194767
= 0.0687194767
Similarly, find P(X = 1), P(X = 2), P(X = 3), P(X = 4), and P(X = 5) using the same method.
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I don't know how to approach it, I've been looking at my notes and I can't grasp it.
Answer:
8
Step-by-step explanation:
You can skip directly to the formula for the sum of an infinite sequence with first term a₁ and common ratio r:
S = a₁/(1-r)
Your values of the variables in this formula are a₁ = 6 and r = 2/8. Putting these into the formula gives ...
S = 6/(1 -2/8) = 6/(6/8) = 8
The sum of the infinite geometric sequence is 8.
_____
The above formula is the degenerate form of the formula for the sum of a finite sequence:
S = a₁((rⁿ -1)/(r -1))
When the common ratio r has a magnitude less than 1, the term rⁿ tends to zero as n gets very large. When that term is zero, the sum of the infinite sequence is ...
S = a₁(-1/(r-1)) = a₁/(1-r)
Define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and 8-bit strings with exactly five 1's. A subset X of S with five elements maps on to a string x so that j ∈ X if and only if the jth bit of x is 1. What string corresponds to the set {1, 3, 4, 5, 8}?
To define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and 8-bit strings with exactly five 1's, we can assign each element of the subset to a bit position in the string.
Explanation:In mathematics, a bijection is a one-to-one correspondence between two sets, such that each element in one set is paired with a unique element in the other, and vice versa. It implies both injective (no duplicates) and surjective (every element has a match) properties.
To define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and 8-bit strings with exactly five 1's, we can assign each element of the subset to a bit position in the string. For example, the subset {1, 3, 4, 5, 8} corresponds to the 8-bit string 11011001, where the 1st, 3rd, 4th, 5th, and 8th bits are set to 1, representing the elements in the subset.
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A kite flier wondered how high her kite was flying. She used a protractor to measure an angle of 33° from level ground to the kite string. If she used a full 90 yard spool of string, how high, in feet, was the kite? Round your answer to 3 decimal places. (Disregard the string sag and the height of the string reel above the ground.)
Answer: height of kite is 147.042 feets
Step-by-step explanation:
The diagram of the kite is shown in the attached photo
Triangle ABC is formed and it is a right angle triangle.
The kite string made an angle of 33 degrees with the ground. The string used was 90 yards We will convert the 90 yards to feets.
I yard = 3 feets
90 yards would become
90×3 = 270 feets
This 270 feets form the hypotenuse of the triangle.
To determine the height of the kite h, we will use trigonometric ratio
Sin# = opposite / hypotenuse
Where
# = 33 degrees
Hypotenuse = 270 feets
Opposite = h feets
Sin 33 = h/270
h = 270sin33
h = 270 × 0.5446 = 147.042 feets
One of the loudest sounds in recent history was that made by the explosion of Krakatoa on August 26-27, 1883. According to barometric measurements, the sound had a decibel level of 180 dB at a distance of 161 km. Assuming the intensity falls off as the inverse of the distance squared, what was the decibel level on Rodriguez Island, 4,800 km away?
Answer:
150.51 dB
Step-by-step explanation:
Data provided in the question:
decibel level of sound at 161 km distance = 180 dB
d₁ = 161 km
d₂ = 4800 km
I₁ = 180 db
The formula for intensity of sound is given as:
I = [tex]10\log(\frac{I_2}{I_1})[/tex]
and the relation between intensity and distance is given as:
I ∝ [tex]\frac{1}{d^2}[/tex]
or
Id² = constant
thus,
I₁d₁² = I₂d₂²
or
[tex]\frac{I_2}{I_1}=\frac{d_1}{d_2}[/tex]
therefore,
I = [tex]10\log(\frac{d_1}{d_2})^2[/tex]
or
I = [tex]10\times2\times\log(\frac{161}{4,800})[/tex]
or
I = 20 × (-1.474)
or
I = -29.49
Therefore,
the decibel level on Rodriguez Island, 4,800 km away
= 180 - 29.49
= 150.51 dB
To find the decibel level on Rodriguez Island 4,800 km away from the explosion of Krakatoa, you can use the inverse square law and the formula dB1 - dB2 = 20log10(d1/d2), where d1 and d2 are the distances from the explosion. Plugging in the values, you can calculate the decibel level on Rodriguez Island to be approximately 201.94 dB.
Explanation:To find the decibel level on Rodriguez Island 4,800 km away from the explosion of Krakatoa, we can use the inverse square law. According to the problem, the decibel level at a distance of 161 km is 180 dB. To find the decibel level at 4,800 km, we can use the formula: dB1 - dB2 = 20log10(d1/d2), where d1 and d2 are the distances from the explosion. Plugging in the values, we get: dB2 = dB1 - 20log10(d1/d2). Substituting dB1 = 180 dB, d1 = 161 km, and d2 = 4800 km, we can calculate the decibel level on Rodriguez Island.
dB2 = 180 - 20log10(161/4800) ≈ 180 - 20(-1.097) ≈ 180 + 21.94 ≈ 201.94 dB
Therefore, the decibel level on Rodriguez Island, 4,800 km away, is approximately 201.94 dB.
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The popularity of computer, video, online, and virtual reality games has raised concerns about their ability to negatively impact youth. The data in this exercise are based on a recent survey of 14 ‑ to 18 ‑year‑olds in Connecticut high schools. Assume the table displays the grade distributions of boys who have and have not played video games.
Grade Average
A's and B's C's D's and F's
Played games 730 444 190
Never played games 214 137 87
Give the conditional distribution of the grades of those who have played games. (Enter your answers rounded to two decimal places.)
Answer:
P(The grade of the boy is A| He has played video games)
is, [tex]\simeq 0.54[/tex]
P(The grade of the boy is B| He has played video games)
is [tex]\simeq 0.33[/tex]
P(The grade of the boy is C| He has played video games)
is [tex]\simeq 0.14[/tex]
Step-by-step explanation:
The total no. of boys who have played video games,
= (730 + 444 + 190)
=1360
Now, from the given data,
P(The grade of the boy is A| He has played video games)
= [tex]\frac {730}{1360}[/tex]
[tex]\simeq 0.54[/tex]
P(The grade of the boy is B| He has played video games)
= [tex]\frac {444}{1360}[/tex]
[tex]\simeq 0.33[/tex]
P(The grade of the boy is C| He has played video games)
= [tex]\frac {190}{1360}[/tex]
[tex]\simeq 0.14[/tex]
The conditional distribution of grades for boys who have played games shows that 53% received A's and B's, 33% received C's, and 14% received D's and F's. This is calculated by dividing each grade category by the total number of boys who played games and rounding to two decimal places.
To find the conditional distribution of grades for boys who have played games, we need to calculate the proportion of each grade category relative to the total number who played.
Steps to Calculate Conditional Distribution
Calculate the total number of boys who played games:Therefore, the conditional distribution of grades for boys who have played games is 0.53 for A's and B's, 0.33 for C's, and 0.14 for D's and F's.
The standard deviation of pulse rates of adult males is less than 12 bpm. For a random sample of 135 adult males, the pulse rates have a standard deviation of 11.5 bpm. Complete parts (a) and (b) below. a. Express the original claim in symbolic form.
Answer:
Step-by-step explanation:
Hello!
The population variance is symbolized σ² and the standard deviation σ (remember that for estimations and statistics test, the parameter of study is always the variance)
The sentence "The standard deviation of pulse rates of adult males is less than 12 bpm." is symbolized σ <12
The sample variance is symbolized S² and the standard deviation is S.
The sample standard deviation is S= 11.5.
I hope it helps!
The symbol σ is used to represent the standard deviation in statistics. So, the standard deviation of pulse rates of adult males being less than 12 bpm in symbolic form is σ < 12. This value tells us that the dispersion of pulse rates among adult males is quite small.
Explanation:The original claim is that the standard deviation of pulse rates of adult males is less than 12 bpm. This can be expressed in symbolic form as follows:
σ < 12
where σ represents the standard deviation.
In the context of this question, the standard deviation is a measure of the dispersion or spread in the pulse rates of adult males. The sample that was taken had a standard deviation of 11.5 bpm, which is less than 12 bpm, thus supporting the original claim.
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A county clerk wants to estimate the proportion of voters who will need special election facilities.
Suppose a sample of 400 voters was taken.
If 150 need special election facilities, what is the upper confidence limit (UCL) for the 90% confidence interval for the population proportion of voters who will need special election facilities?
Round your answer to 3 decimal places.
Answer: The upper confidence limit for the 90% confidence interval would be 0.415.
Step-by-step explanation:
Since we have given that
n = 400
x = 150
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{150}{400}=0.375[/tex]
At 90% confidence interval, z = 1.645
So, margin of error would be
[tex]z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.375\times 0.625}{400}}\\\\=0.0398[/tex]
So, the upper limit would be
[tex]\hat{p}+0.0398\\\\=0.375+0.0398\\\\=0.415[/tex]
Hence, the upper confidence limit for the 90% confidence interval would be 0.415.
Suppose y varies jointly as x and z. Find y when x = –13 and z = 7, if y = 205 when x = –5 and z = –8. Round your answer to the nearest hundredth, if necessary.
Answer: y = - 466.375
Step-by-step explanation:
y varies jointly as x and z.
This means that y varies directly as x and also varies directly as z.
In order to remove the proportionality symbol, we will introduce a constant of proportionality, k. Therefore,
y = kxz
The next step is to determine the value of k
if y = 205 when x = –5 and z = –8.
we will substitute these values into the equation to determine k.
205 = k × -5 × -8
205 = 40k
k = 205/40 = 5.125
Therefore, the equation becomes
y = 5.125xz
We want to determine y when x = - 13 and z = 7
y = 5.125 × - 13 × 7
y = 5.125 × - 91
y = - 466.375