Two particles move in the xy-plane. At time t, the position of particle A is given by x(t)=5t−5 and y(t)=2t−k, and the position of particle B is given by x(t)=4t and y(t)=t2−2t−1.

(a) If k=−6, do the particles ever collide?

(b) Find k so that the two particles are certain to collide.

k=

(c) At the time the particle collide in (b), which is moving faster?
A. particle A
B. particle B
C. neither particle (they are moving at the same speed)

Answer: Our required probability is [tex]\dfrac{31}{35}[/tex]

Step-by-step explanation:

Since we have given that

Number of red envelopes = 3

Number of blue envelopes = 3

Number of green envelopes = 1

We need to select 3 envelopes in such a way that at least one envelope is a red.

So, it becomes,

[tex]\dfrac{^3C_1\times ^4C_2}{^7C_3}+\dfrac{^3C_2\times ^4C_1}{^7C_3}+\dfrac{^3C_3}{7C_3}\\\\=\dfrac{18}{35}+\dfrac{12}{35}+\dfrac{1}{35}\\\\=\dfrac{18+12+1}{35}\\\\=\dfrac{31}{35}[/tex]

Hence, our required probability is [tex]\dfrac{31}{35}[/tex]

Final answer:

The probability of not selecting any red envelope is determined by choosing 3 envelopes without any red ones. The probability of selecting at least one red envelope is 17/24.

Explanation:

To find the probability of selecting at least one red envelope, we need to consider the complement of selecting no red envelopes.

The total number of ways to select 3 envelopes out of the given collection is (3+3+1+3)C3 = 10C3 = 120.

The number of ways to select 3 envelopes without selecting any red envelope is (3+1+3)C3 = 7C3 = 35.

Therefore, the probability of selecting at least one red envelope is 1 - P(no red envelope) = 1 - (35/120) = 85/120 = 17/24.

The proportion of Greenville County adult residents who are 65 years or older is approximately 13%.

To find the proportion of Greenville County adult residents who are 65 years or older, we need to divide the number of residents who are 65 years or older by the total number of adult residents in the county.

The proportion can be found using the formula: Proportion = Number of residents who are 65 years or older / Total number of adult residents.

Substituting the given values, we have:

Proportion = 59,969 / 461,299 = 0.1299

Therefore, the proportion of Greenville County adult residents who are 65 years or older is approximately 0.13, or 13%.

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Answer:

The most that a meal can cost before tip = $43.47

Step-by-step explanation:

Kristin's maximum budget for a birthday dinner = $50 inclusive of 15% tip.

Let the cost of meal that Kristin orders in dollars be [tex]=x[/tex]

15% of the cost of meal is tip which would be in dollars = 15% of [tex]x=0.15\ x[/tex]

So total cost of dinner would be [tex]=x+0.15x[/tex]

We know that the total should be no more that $50. So, we have

[tex]x+0.15x\leq50[/tex]

⇒ [tex]1.15x\leq50[/tex]

Dividing both sides by 1.15.

⇒ [tex]\frac{1.15x}{1.15}\leq\frac{50}{1.15}[/tex]

∴ [tex]x\leq43.47[/tex]

So, the most that a meal can cost before tip = $43.47

Answer:

y = -2(x - 4)^2 + 4.

Step-by-step explanation:

Vertex form:

y = a(x - 4)^2 + 4    (because the vertex is at (4, 4)).

To find the value of a we substitute the point (2, -4):

-4 = a(2-4)^2 + 4

4a = -8

a = -2.

Answer:

The expected value of the safe bet equal $0

Step-by-step explanation:

If  

[tex]S=\left\{s_1,s_2,...,s_n\right\}[/tex]

is a finite numeric sample space and

[tex]P(X=s_k)=p_k[/tex] for k=1, 2,..., n

is its probability distribution, then the expected value of the distribution is defined as

[tex]E(X)=s_1P(X=s_1)+s_2P(X=s_2)+...+s_nP(X=s_n)X) [/tex]

What is the expected value of the safe bet?

In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is

S = {100,-100}

Since the coin is supposed to be fair,  

P(X=100)=0.5

P(X=-100)=0.5

and the expected value is

E(X) = 100*0.5 - 100*0.5 = 0

The expected value of the safe bet is $0, which means that you would neither gain nor lose money on average if you played this game repeatedly.

The expected value of the safe bet can be calculated by multiplying the possible outcomes by their respective probabilities and summing them up. In this case, the safe bet has two possible outcomes: winning $100 with a 50% probability and losing $100 with a 50% probability. So, the expected value can be calculated as follows:

Expected value = (100 * 0.5) + (-100 * 0.5) = $0

Therefore, the expected value of the safe bet is $0, which means that you would neither gain nor lose money on average if you played this game repeatedly.

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the limits of the 99% confidence interval are approximately [tex]\( (230.068, 233.799) \).[/tex]

To find the limits of the 99% confidence interval (CI) for the true average natural frequency (Hz), we'll use the formula for confidence intervals for the mean:

[tex]\[ \text{CI} = \bar{x} \pm z \frac{s}{\sqrt{n}} \][/tex]

Where:

- [tex]\( \bar{x} \)[/tex] is the sample mean,

- [tex]\( z \)[/tex] is the z-score corresponding to the desired confidence level,

- [tex]\( s \)[/tex] is the sample standard deviation,

- [tex]\( n \)[/tex] is the sample size.

Given:

- Sample size (n ) = 5,

- 95% CI: Lower limit = 230.061, Upper limit = 233.807.

1. Calculate the sample mean [tex](\( \bar{x} \)):[/tex]

[tex]\[ \bar{x} = \frac{\text{Lower limit} + \text{Upper limit}}{2} \][/tex]

[tex]\[ \bar{x} = \frac{230.061 + 233.807}{2} \][/tex]

[tex]\[ \bar{x} = \frac{463.868}{2} \][/tex]

[tex]\[ \bar{x} = 231.934 \][/tex]

2. Calculate the sample standard deviation [tex](\( s \)):[/tex]

To calculate the sample standard deviation, we need to know the z-score corresponding to the desired confidence level.

3. Find the z-score for a 99% confidence level:

At 99% confidence level, the critical z-value is approximately 2.576 (you can find this value using a standard normal distribution table or a calculator).

4. Substitute the values into the confidence interval formula:

[tex]\[ 231.934 \pm 2.576 \frac{s}{\sqrt{5}} \][/tex]

We know that the width of the interval at 95% confidence level is[tex]\( 233.807 - 230.061 = 3.746 \).[/tex]

So, we can set up an equation to find \( s \):

[tex]\[ 2.576 \frac{s}{\sqrt{5}} = \frac{3.746}{2} \][/tex]

[tex]\[ \frac{s}{\sqrt{5}} = \frac{3.746}{2.576 \times 2} \][/tex]

[tex]\[ \frac{s}{\sqrt{5}} \approx \frac{3.746}{5.152} \][/tex]

[tex]\[ s \approx \frac{3.746 \times \sqrt{5}}{5.152} \][/tex]

[tex]\[ s \approx \frac{3.746 \times 2.236}{5.152} \][/tex]

[tex]\[ s \approx \frac{8.3753}{5.152} \][/tex]

[tex]\[ s \approx 1.6247 \][/tex]

Now, we can substitute [tex]\( s = 1.6247 \)[/tex] into the CI formula to find the limits of the 99% interval:

[tex]\[ 231.934 \pm 2.576 \frac{1.6247}{\sqrt{5}} \][/tex]

[tex]\[ 231.934 \pm 2.576 \times \frac{1.6247}{\sqrt{5}} \][/tex]

[tex]\[ 231.934 \pm 2.576 \times 0.7241 \][/tex]

[tex]\[ 231.934 \pm 1.8658 \][/tex]

Therefore, the limits of the 99% confidence interval are approximately [tex]\( (230.068, 233.799) \).[/tex]

Answer:

We accept H₀, we dont have evidence to say that the family size container

of ketchup has smaller quantity

Step-by-step explanation:

Population mean  μ₀ =  20 ounces

sample size  n  =  10   df = n -1  df = 10-1   df= 9

n < 30   use of t-student distribution

sample mean  μ  = 19.86

sample standard deviation  s  =  0,22

One tail-test ( left tail)

1.-Test Hypothesis

H₀     null hypothesis                μ₀ =  20

Hₐ  Alternative hypothesis       μ₀ < 20

2.- α  =  0,01   and one test tail

3.- Compute

t(s)  =  [ ( μ  -   μ₀ ) ] / s/√n           t(s)  = [( 19.86  -  20 )* √10 ] / 0.22

t(s)  =  - ( 0,14 * 3.16) / 0,22  

t(s)  =  - 2.01

4.- We go to  t-student table  t(c) for df = 9  and 0,01 = α  

and find t(c) =  - 2.821

5.-Compare   t (s)   and  t (c)

t (c)  <   t (s)      -  2.821  < - 2.01

6. t(s)  is inside de acceptance region  we accept H₀

Answer:

250

Step-by-step explanation:

36% are rabbits, and 28% are dogs.  That means the percent that are hamsters is:

100% − 36% − 28% = 36%

There are 160 dogs and hamsters, which are 28% + 36% = 64% of the pets.  Therefore, we can write and solve a proportion:

160 / 64% = x / 100%

x = 250

There are a total of 250 pets.

Answer: 250 pets

Step-by-step explanation:

Let the total number of pets in the pet shop be x

36% of them are rabbits. This means that the number of pets that are rabbit 36/100 × x = 0.36×x = 0.36x

28% of them are dogs. This means that the number of pets that are dogs are 28/100 × x = 0.28×x = 0.28x

The rest are hamsters. This means that total number of hamsters is total number of pets minus sum of rabbits and dogs.

Number of hamsters = x -(0.28x + 0.36x) = 0.36x

If there are 160 dogs and hamsters,it means that

0.28x + 0.36x = 160

0.64x =160

x = 160/0.64

x = 250

Answer:

The skitties are not evenly distributed by colour

Step-by-step explanation:

Given that Mr. T  aylor's 4th grade class uses Skittles to learn about probability. They open several randomly selected bags of Skittles and sort and count the different colors and want to determine if Skittles are evenly distributed by color.

[tex]H_0: Skitties are equally distributed\\H_a: atleast two are not equally distributed[/tex]

(Two tailed chi square test)

If all are equally distributed then expected values would be equal to 420/5 =104

Observed                       Red   Orange  Yellow   Purple   Green  Total

                                       107          101      87           115           10     420

Expected                         104          104     104          104         104    420

Chi square 0.0865 0.0865 2.7788 1.16345 84.9615 89.0769

Chi square is calculated as (obs-exp)^2/exp

Total chi square = 89.0769

df = 4

p value = <0.00001

Reject null hypothesis

The skitties are not evenly distributed by colour

Answer:

Option b - not significantly greater than 75%.

Step-by-step explanation:

A random sample of 100 people was taken i.e. n=100

Eighty of the people in the sample favored Candidate i.e. x=80

We have used single sample proportion test,

[tex]p=\frac{x}{n}[/tex]

[tex]p=\frac{80}{100}[/tex]

[tex]p=0.8[/tex]

Now we define hypothesis,

Null hypothesis [tex]H_0[/tex] : candidate A is significantly greater than 75%.

Alternative hypothesis [tex]H_1[/tex] : candidate A is not significantly greater than 75%.

Level of significance [tex]\alpha=0.05[/tex]

Applying test statistic Z -proportion,

[tex]Z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}[/tex]

Where, [tex]\widehat{p}=80\%=0.80[/tex] and [tex]p=75%=0.75[/tex]

Substitute the values,

[tex]Z=\frac{0.80-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}[/tex]

[tex]Z=\frac{0.80-0.75}{\sqrt{\frac{0.1875}{100}}}[/tex]

[tex]Z=\frac{0.05}{0.0433}[/tex]

[tex]Z=1.1547[/tex]

The p-value is

[tex]P(Z>1.1547)=1-P(Z<1.1547)[/tex]

[tex]P(Z>1.1547)=1-0.8789[/tex]

[tex]P(Z>1.1547)=0.1241[/tex]

Now, the p-value is greater than the 0.05.

So we fail to reject the null hypothesis and conclude that the A is not significantly greater than 75%.

Therefore, Option b is correct.

To answer if the proportion of the population in favor of Candidate A is significantly more than 75% at a .05 level of significance, we'd need to perform a statistical test. If the p-value from this test is less than .05, we can say the proportion is significantly more than 75%. However, we haven't been given a statistical result so we can't definitively select between options a. and b.

The problem is about determining the significance of a proportion in a population. In this case, the proportion represents the people who favor Candidate A. The question is whether this proportion is significantly more than 75% at a .05 level of significance. Eighty people out of the sample of hundred favor Candidate A, which is 80% of the population sample.

The next step is to set up the null hypothesis (H0) and the alternative hypothesis (Ha). The null hypothesis claims that the proportion of people in favor of Candidate A is 75%. The alternative hypothesis states that the proportion of people in favor of Candidate A is significantly more than 75%.

Next, we test the hypothesis using statistical analysis. As we are using a .05 level of significance, if the p-value is less than .05, we reject the null hypothesis in favor of the alternative hypothesis.

In this case, since our percentage in the sample (80%) is greater than the claim that we are testing against (75%), we can suggest that the proportion of the population in favor of Candidate A is significantly greater than 75% if our p-value is less than .05. However, without performing the statistical test or being given the resultant p-value we cannot choose between options a. and b.

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Answer:

0.546 is the probability that a randomly selected smartphone users in the age range 11 to 55+ is between 30 and 54 years old.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  34.8 years

Standard Deviation, σ = 14.1 years

We are given that the distribution of ages of smartphone is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P( age range is between 30 and 54 years old)

[tex]P(30 \leq x \leq 54) = P(\displaystyle\frac{30 - 34.8}{14.1} \leq z \leq \displaystyle\frac{54-34.8}{14.1}) = P(-0.3404 \leq z \leq 1.3617)\\\\= P(z \leq 1.3617) - P(z < -0.3404)\\= 0.913 - 0.367 = 0.546 = 54.6\%[/tex]

[tex]P(30 \leq x \leq 54) = 54.6\%[/tex]

0.546 is the probability that a randomly selected smartphone users in the age range 11 to 55+ is between 30 and 54 years old.

Answer:

L=0

Step-by-step explanation:

[tex]L=\lim\limits_{x \rightarrow \frac{\pi}{2}}3secx-3tanx[/tex]

Replacing the value of x we get ∞ - ∞ which is an indetermined expression

We must transform the limit so it can be shown as a fraction and the L'Hopital's rule can be applied:

[tex]L=\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{3-3sinx}{cosx}=\frac{0}{0}[/tex]

Now we can take the derivative in both parts of the fraction

[tex]L=\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{-3cosx}{-sinx}=3\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{cosx}{sinx}=3\times 0=0[/tex]

Answer:No, Diet and exercise does not reduce hypertension.

Step-by-step explanation:

Since we have given that

p = 0.23

n = 75

x = 18

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{18}{75}=0.24[/tex]

So, hypothesis would be

[tex]H_0:p=\hat{p}\\\\H_a:\hat{p}<p[/tex]

So, the test statistic value would be

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.24-0.23}{\sqrt{\dfrac{0.23\times 0.77}{75}}}\\\\z=\dfrac{0.01}{0.049}\\\\z=0.204[/tex]

At α = 0.05 level of significance, we get

critical value = 1.96

and 1.96>0.204.

so, we will accept the null hypothesis.

Hence, No, Diet and exercise does not reduce hypertension.

Answer:

The assumption that must be made for this study to follow the probabilities of a binomial experiment is that there must be only two outcomes of each trail in this study (meaning that it is either they prefer grey suits or they do not prefer grey suits). There must be no other option apart from those two options and each of the independent trails must be mutually exclusive, meaning that the two required options cannot occur together. It is either the first option (prefer grey suits) or the second option (do not prefer grey suits).

Step-by-step explanation:

Final answer:

The sampling distribution of the sample mean has the same mean as the original distribution (80) and a calculated standard error of 0.5. The z-value for x = 81 is 2, corresponding to a probability of P(x < 81) = 0.9772. It would not be unusual for a sample mean to be less than 81 based on this z-score.

Explanation:

If we have a variable x that follows a normal distribution with a known mean (μ) and standard deviation (σ), and we draw random samples of size n from it, we can describe the distribution of the sample means. The distribution of sample means will also be normally distributed, known as the sampling distribution of the sample mean, thanks to the Central Limit Theorem. For a sufficiently large sample size, this holds true regardless of the shape of the original distribution.

(a) Description of x Distribution and Calculation of Mean and Standard Deviation

The variable x has a distribution with a mean (μx) of 80 and a standard deviation (σx) of 3. When taking a sample size n = 36, by the Central Limit Theorem, the mean of the sampling distribution will remain the same (μx = 80), but the standard deviation will be the original standard deviation divided by the square root of the sample size (n), known as the standard error (SE). Hence, the standard error will be σx/√n = 3/√36 = 3/6 = 0.5.

(b) Finding the z-value for x = 81

To find the z-value for x = 81, we use the formula:
z = (x - μx) / SE
So, z = (81 - 80) / 0.5 = 1/0.5 = 2.

(c) Computing P(x < 81) - Probability Calculation

To find the probability P(x < 81), we would look up the z-value we just calculated in a standard normal distribution table. Let's assume it corresponds to a probability of 0.9772. Thus, P(x < 81) = 0.9772.

(d) Unusualness of a Sample Mean Less Than 81

To determine if it would be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 81, we consider the z-value and the empirical rule. Since our z-score of 2 corresponds to a percentage greater than 5% of the tail (assuming the previously stated probability is correct), it is not unusual for a sample mean to be less than 81 because more than 5% of samples would have means less than this.

Answer:

[tex]$ \sqrt{-1} = i $[/tex]

Step-by-step explanation:

(1) [tex]$ \sqrt{-100} $[/tex]

[tex]$ = \sqrt{100}.i = 10i $[/tex]

(1) - 10i

(2) [tex]$ -2\sqrt{50} $[/tex]

[tex]$ = -2\sqrt{50}  = -2(5) \sqrt{2} = -10\sqrt{2} $[/tex]

(2) - -10√2

(3) [tex]$ 5 + \sqrt{-13} $[/tex]

[tex]$ \implies 5 + \sqrt{13}i $[/tex]

(3) - 5 + √13i

(4) [tex]$ 5 + \sqrt{-120} $[/tex]

[tex]$ = 5 + \sqrt{120}i = 5 + 2\sqrt{30}i$[/tex]

(4) - 5 + 2√30i

(5) [tex]$ -10 +  \sqrt{-50} $[/tex]

[tex]$ = -10 + 5\sqrt{2} i $[/tex]

(5) - -10 + 5√2i

(6) - [tex]$ \sqrt{-45} $[/tex]

[tex]$ 3\sqrt{5}i $[/tex]

(6) - 3√5i

Answer:

for 1 ) Normal (rectangular) coordinates

for 2) Polar coordinates

for 3) Cylindrical coordinates

for 4) Spherical coordinates

for 5) Normal (rectangular) coordinates

Step-by-step explanation:

1. ∫10∫y20 1x dx dy 2. → Normal (rectangular) coordinates x=x , y=y → integration limits       ∫ [20,1]  and  ∫ [10,2]

2. ∫∫D 1x2+y2 dA. , D is: x2+y2≤4 → Polar coordinates x=rcosθ  , y=rsinθ  → integration limits  ∫ [2,0] for dr  and  ∫ [2π,0] for dθ

3. ∫∫∫E z2 dV , E is: −2≤z≤2, 1≤x2+y2≤2  → Cylindrical coordinates x=rcosθ  , y=rsinθ , z=z  → integration limits  ∫ [2,-2] for dz  , ∫ [√2,1] for dr and  ∫ [2π,0] for dθ

4. ∫∫∫E dV where E is: x2+y2+z2≤4, x≥0, y≥0, z≥0 → Spherical coordinates x=rcosθcosФ y=rsinθcosФ , z=rsinФ → integration limits  ∫ [2,0] for dr  ,∫ [-π/2,π/2] for dθ , ∫ [π/2,0] for dθ

5. ∫∫∫E z dV where E is: 1≤x≤2, 3≤y≤4, 5≤z≤6 → Normal (rectangular) coordinates x=x , y=y , z=z → integration limits ∫ [2,1] for dx ,∫ [4,3] for dy and ∫ [6,5] for dz

The type of coordinates easiest to use when solving integrals depends on the nature of the integral and its bounds. In cases where circular symmetry is present, Polar, Cylindrical, or Spherical coordinates may be used instead of Cartesian Coordinates.

The integrals listed in your question can be best solved depending on the coordinate system expressed in their limits of integration, or the area or volume they represent. Let's match them below:

Please note, the type of coordinates that are the easiest to use often depends heavily on the specific integrals and their bounds. For instance, in situations where circular symmetry is present, it's preferable to use Polar, Cylindrical or Spherical coordinates as compared to Cartesian coordinates.

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Answer:

-6-(-2) is equivalent to

-6 +2

and 2-6

a) and b) are correct options

Answer:

a and b.

Step-by-step explanation:

-6-(-2) = -6 + 2 = -4.

2 - 6 = -4.

Answer:

d = √(k/R)

Step-by-step explanation:

Eliminate fractions by multiplying by the denominator, then divide by the coefficient of d². Finally, take the square root.

[tex]R=\dfrac{k}{d^2}\\\\Rd^2=k \quad\text{multiply by $d^2$}\\\\d^2=\dfrac{k}{R} \quad\text{divide by $R$}\\\\d=\sqrt{\dfrac{k}{R}} \quad\text{take the square root}[/tex]

Answer:

The total length can be calculated by doing thesum of all sides

Step-by-step explanation:

The total length of planting around each stepping stone is 15x - 2.

To find the total length of planting around each stepping stone, we need to add up the lengths of all the sides of the irregular polygon.

The sides of the polygon are (3x-1), (2x+1), (4x-2), (4x-4), and (2x+4).

To find the total length, we can add up the lengths of all the sides:

(3x-1) + (2x+1) + (4x-2) + (4x-4) + (2x+4)

Simplifying the expression, we get:

15x - 2

Therefore, The total length of planting around each stepping stone is 15x - 2.

The probable question may be:

Belinda placed stepping stones in the shape of the irregular polygon shown. She will plant thyme around the edge of each stepping stone. What is the total length of planting around each stepping stone ?​

The sides of the irregular polygon are (3x-1), (2x+1), (4x-2), (4x-4), (2x+4)

Answer: the error is he placed the decimal point incorrectly after he multiplied. the answer is actually 2.7, as 0.3 times 9 is 2.7. Hope this helps.

Answer:

the answer is a

Step-by-step explanation:

i just took the test

Answer:

C) (-5%.15%)

Step-by-step explanation:

1) 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 =10\%=0.1 [/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma[/tex]=5% =0.05 represent the population standard deviation  

2) Confidence interval

We assume that the random variable X who represent The Ishares Bond Index fund (TLT) follows this distribution:

[tex]X \sim N(\mu, \sigma=10\%=0.1)[/tex]

The confidence interval for the returns on TLT is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\sigma[/tex]   (1)

In order to calculate the critical value [tex]z_{\alpha/2}[/tex]. Since the Confidence is 0.66 or 66%, the value of [tex]\alpha=0.34[/tex] and [tex]\alpha/2 =0.17[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.17,0,1)".And we see that [tex]z_{\alpha/2}=0.95[/tex]

Now we have everything in order to replace into formula (1):

[tex]0.05-0.95(0.1)=-0.05[/tex]    

[tex]0.05+0.95(0.1)=0.15[/tex]

So on this case the 66% confidence interval would be given by (-0.05;0.15) and we can convert this into % and wr got (-5%; 15%).    

Answer: the system of equations are

x + y = 35

3x + 2y = 100

Step-by-step explanation:

Let x= the number of short answer questions.

Let y= the number of multiple choice questions.

Noah wants 35 questions on the exam. This means that

x + y = 35

He plans to mix short answer questions, worth 3 points, with multiple choice questions worth 2 points. This means that x short answer questions will give 3x points and y multiple choice questions will give 2y points

Since the exam is worth 100 points, then,

3x + 2y = 100 - - - - - - - -1

Substituting x = 35 - y into equation 1, it becomes

3(35 - y) + 2y = 100

105 - 3y + 2y = 100

y = 105 - 100 = 5

x = 35 - y = 35 - 5

x = 30

The system of equation telling the number of each question Noah can have on the test is [tex]x+y = 35\\3x + 2y = 100[/tex]. The value of variables evaluates to:

You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on methods can be used to convert description to mathematical expressions.

For this case, we are given these facts:

Let there are:

Then, we get:

[tex]x+y = 35[/tex] (as total number of question is 35)

Thus, we get second equation as:

[tex]3x + 2y = 100[/tex] (as total maximum score = 100 points)

Therefore, the system of equation obtained for this condition is:

[tex]x+y = 35\\3x + 2y = 100[/tex]

From first equation, getting value of x in terms of  y, we get:

[tex]x = 35 -y[/tex]

Putting this in second equation, we get:

[tex]3x + 2y = 100\\3(35-y) + 2y = 100\\105 -3y + 2y = 100\\105-100 = y\\y = 5[/tex]

Putting this value of y in equation for x, we get;

[tex]x = 35 -y = 35 - 5 = 30[/tex]

Thus, the system of equation telling the number of each question Noah can have on the test is [tex]x+y = 35\\3x + 2y = 100[/tex]. The value of variables evaluates to:

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Answer:

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours.

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.

Step-by-step explanation:

1) Data given and notation    

[tex]\bar X=19.5[/tex] represent the battery life sample mean  

[tex]s=1.9[/tex] represent the sample standard deviation    

[tex]n=33[/tex] sample size    

[tex]\mu_o =20[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/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)    

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is less than 20 :    

Null hypothesis:[tex]\mu \geq 20[/tex]    

Alternative hypothesis:[tex]\mu < 20[/tex]    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic    

We can replace in formula (1) the info given like this:    

[tex]t=\frac{19.5-20}{\frac{1.9}{\sqrt{33}}}=-1.51[/tex]    

4) P-value    

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=33-1=32[/tex]

Since is a one-side lower test the p value would be:    

[tex]p_v =P(t_{(32)}<-1.51)=0.0704[/tex]    

5) Conclusion    

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the average battery life it's not significantly different less than 20 hours at 5% of signficance. If we analyze the options given we have:

(A) Reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. FALSE, we FAIL to reject the null hypothesis.

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. TRUE, we fail to reject the null hypothesis that the mean would be 20 or higher .

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.  TRUE, we FAIL to reject the null hypothesis that the mean is greater or equal to 20 hours, so we reject the alternative hypothesis that the mean is less than 20 hours.

(D) There is enough evidence at the α=0.05 level of significance to support the claim that the true population mean battery life of the smartphone is not equal to 20 hours. FALSE the claim is not that the mean is different from 20. The real claim is: "Javier believes the mean battery life is less than 20 hours".

Javier's t-test with a test statistic of −1.51 and 32 degrees of freedom has a p-value greater than 0.05, hence we fail to reject the null hypothesis, indicating not enough evidence to suggest the true mean is less than the claimed 20 hours. Thus, the correct answer is option (C).

This problem involves conducting a one-sample t-test to determine whether the true population mean battery life is different from the claimed 20 hours. The null hypothesis here is H0: [tex](mu = 20)[/tex] hours, and the alternative hypothesis is Ha: [tex](mu < 20)[/tex] hours. With 32 degrees of freedom, Javier's calculated test statistic is t ≈ −1.51. To decide whether to reject or fail to reject the null hypothesis, we must compare the p-value to the significance level, [tex]\(\alpha = 0.05\)[/tex].

If the p-value is less than [tex]\(\alpha = 0.05\)[/tex], then we reject the null hypothesis. If the p-value is higher, we fail to reject. In this exercise, the p-value associated with Javier's test statistic of t ≈ −1.51 for a one-tailed test is greater than 0.05, thus we should fail to reject the null hypothesis (Option B). This implies that there is not enough evidence at the 0.05 level of significance to support the claim that the true population mean battery life of the smartphone is less than 20 hours (Option C).

Answer:

a)  (50.30 , 52.50)

b) (50.85 , 51.95)

c) (50.68 , 52.12)

d)  (51.02 , 51.78)

e) 209

Step-by-step explanation:

(a)  Sample Mean = 51.4

σ = 2.8

Sample Size, n = 25

Standard Error, E = [tex]\frac{\sigma}{\sqrt{n}}[/tex] = 0.56

z critical value for 95% confidence interval

z = 1.96

Margin of Error (ME) = z × E = 1.0976

95% confidence interval is given as

⇒ Mean ± ME

= 51.4 ± 1.0976

or

= (50.30 , 52.50)

b) Sample Mean = 51.4

σ = 2.8

Sample Size, n = 100

Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28

z critical value for 95% confidence interval

z = 1.96

Margin of Error (ME) = z × E = 0.5488

95% confidence interval is given as

⇒ Mean ± ME

= 51.4 ± 0.5488

or

= (50.85 , 51.95)

c) Sample Mean = 51.4

σ = 2.8

Sample Size, n = 100

Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28

z critical value for 99% confidence interval

z = 2.5758

Margin of Error (ME) = z × E = 0.7212

99% confidence interval is given as

⇒ Mean ± ME

= 51.4 ± 0.7212

or

= (50.68 , 52.12)

d) Sample Mean = 51.4

σ = 2.8

Sample Size, n = 100

Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28

z critical value for 82% confidence interval

z = 1.3408

Margin of Error (ME) = z × E = 0.3754

82% confidence interval is given as

⇒ Mean ± ME

= 51.4 ± 0.3754

or

= (51.02 , 51.78)

e) Margin of Error, ME = (width of interval) ÷ 2 = 0.5

Now,

σ = 2.8

as ME = z × Standard Error,

z = 2.5758  for 99% confidence level

For ME = 0.5,

i,e

[tex]\frac{z\times\sigma}{\sqrt{n}}[/tex] = 0.5

or

[tex]\frac{2.5758 \times2.8}{\sqrt{n}}[/tex] = 0.5

or

n = [tex](\frac{2.5758 \times2.8}{0.5})^2[/tex]

or

n = 208.06

or

n ≈ 209

The question involves calculating different confidence intervals for an unknown population parameter. The computations require using the given standard deviation, sample size, and sample mean, along with appropriate Z-scores. The calculated intervals range between about 49.896 and 52.904 Watts for a 95% CI with a sample size of 25, and as narrow as between 50.626 and 52.174 Watts for an 82% CI with a sample size of 100. Using a desire for a 99% CI width of 1.0, a necessary sample size of about 43 is computed.

In statistics, confidence intervals (CI) provide an estimated range of values which is likely to include an unknown population parameter. Given the parameters of standard deviation (σ), sample size (n), and the sample mean (x), we can compute the confidence intervals. This involves finding the standard error of the mean (σ/√n), and using Z-scores depending on the percentage of the confidence interval. For 95%, 99%, and 82% CIs, the Z-scores are approximately 1.96, 2.58, and 1.34 respectively.

For part (e), we want the width of the 99% interval to be 1.0. This involves setting the equation for the interval to 1.0 and solving for n. This results in n being approximately 42.64, but since we can't have a fractional part of an individual, we round up to 43.

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Answer:

Step-by-step explanation:

The null hypothesis is that the mean weekly food budget for all households in the community is equal to the national average, while the alternative hypothesis is that it is higher. The test statistic is calculated using the sample mean and the standard deviation of the population. To determine if the results are significant, we compare the test statistic to the critical value.

The null hypothesis for this test is that the mean weekly food budget for all households in the community is equal to the national average, which is $99. The alternative hypothesis is that the mean weekly food budget for all households in the community is higher than the national average.

The test statistic in this case is calculated using the sample mean and the standard deviation of the population. It is equal to (sample mean - population mean) / (standard deviation / sqrt(sample size)).

To determine if the results are significant at the 5% level, we compare the test statistic to the critical value for a one-tailed test with a significance level of 0.05. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the mean weekly food budget for all households in the community is higher than the national average.

Answer: -204.8

Step-by-step explanation: if there are and more questions like this use m   a   t    h   w    a  y

Answer:

-cos(6)+sin(3)+cos(5)

=-0.53538809312 (using calculator)

Step-by-step explanation:

here, F = sin(x)dx + cos(y)dy

then f = -cos(x) + sin(y)

( because, [tex]\nabla f[/tex] should be F. by applying [tex]\nabla[/tex] operator on f we must obtain F. so to satisfy this condition f must be -cosx + siny.

where, [tex]\nabla g(x,y) =[/tex] partial derive of g(x,y) with respect to x +  partial derive of g(x,y) with respect to y )

={-cos(-6)}+sin(3) -[{-cos(-5)}+sin(0)]

=-cos(6)+sin(3)-[-cos(5)]    [since, sin(0)=0, cos(-a)=cos(a) where, a>0]

= -cos(6)+sin(3)+cos(5)

=-0.53538809312 (using calculator)

The evaluation of the line integral is: -0.53538809312 (using a calculator) or -cos(6)+sin(3)+cos(5)

If F is a vector field and if it is defined on a domain D and F

= ∇f for any scalar function f on D

Then f is known as the potential function of F

where, F = sin(x)dx + cos(y)dy

then f = -cos(x) + sin(y)

( because ∇f should be F. by applying ∇ operator on f we must obtain F. so to satisfy this condition f must be -cosx + siny.

where, ∇g(x,y) partial derive of g(x,y) with respect to x +  partial derive of g(x,y) with respect to y )

Therefore, the line integral of F over the given curve C

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Answer:

The error in the calculated volume is about [tex]0.0242\pi \approx 0.07602 \:cm^3[/tex]

Step-by-step explanation:

Given a function y=f(x) we call dy and dx differentials and the relationship between them is given by,

[tex]dy=f'(x)dx[/tex]

If the error in the measured value of the radius is denoted by [tex]dr=\Delta r[/tex], then the corresponding error in the calculated value of the volume is [tex]\Delta V[/tex], which can be approximated by the differential

[tex]dV=4\pi r^2dr[/tex]

When r = 1.1 cm and dr = 0.005 cm, we get

[tex]dV=4\pi (1.1)^2(0.005)=0.0242\pi[/tex]

The error in the calculated volume is about [tex]0.0242\pi \approx 0.07602 \:cm^3[/tex]

Answer:

The p-value is 0.175

Step-by-step explanation:

We have the null hypothesis [tex]H_{0}: \mu = 18[/tex] and the alternative hypothesis [tex]H_{1}: \mu \neq 18[/tex] (two-tailed alternative). Because the sample size is n = 17 and the test statistic is t=-1.421, we know that this last value comes from a t distribution with n-1=17-1=16 degrees of freedom. Therefore, the p-value is given by 2P(T < -1.421) because the p-value is the probability of getting a value as extreme as the observed value and because of the simmetry of the t distribution. Here, T has a t distribution with 16 df and we are using the t distribution because the sample size is small. So, 2P(T < -1.421) = 0.1745

The p-value is approximately 0.173, indicating that we do not have enough evidence to reject the null hypothesis. The mean data speed at airports is not significantly different from the claimed value of 18.00 Mbps.

Certainly! To find the p-value for the given hypothesis test, we can follow these steps:

1. **Set up Hypotheses:**

  - Null Hypothesis (H0): The mean [tex](\(\mu\))[/tex] data speed is equal to 18.00 Mbps.

  - Alternative Hypothesis (Ha): The mean [tex](\(\mu\))[/tex] data speed is different from 18.00 Mbps.

2. **Identify Test and Parameters:**

  - We are conducting a one-sample t-test.

  - Sample size (n) is 17.

  - Test statistic (t) is -1.421.

3. **Degrees of Freedom:**

  - Degrees of freedom (df) is calculated as n - 1, which is 16.

4. **Determine the Critical Region:**

  - Since it's a two-tailed test (not equal), we need to find the critical values for a significance level [tex](\(\alpha\))[/tex] of 0.05.

5. **Calculate the p-value:**

  - Using the t-distribution table or statistical software, we find the p-value associated with t = -1.421 and df = 16.

6. **Make a Decision:**

  - If the p-value is less than the significance level (commonly 0.05), we reject the null hypothesis.

Now, let's perform the calculation. Using statistical software or a t-distribution table, the p-value for t = -1.421 and df = 16 is approximately 0.173.

**Conclusion:**

Since the p-value (0.173) is greater than 0.05, we fail to reject the null hypothesis. There is not enough evidence to suggest that the mean data speed at airports is different from 18.00 Mbps.

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