After checking to ensure that the sample data follows the necessary condition for using the t-procedure, we use Minitab to get a 90% confidence interval to estimate mu. We find it to be (118.51 ounces, 124.29 ounces). What does this interval tell us?

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:

E.

simple random sample

Step-by-step explanation:

Simple random sample: Every member and set of members has an equal chance of being included in the sample. Random samples are usually fairly representative since they don't favor certain members.

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:

f''(0.3) > 0 therefore,

x = 0.3 is point of minima

and,

f''(0) = 0

thus,

x = 0 is point of neither maxima nor minima

Step-by-step explanation:

Given function:

f(x) = 5x⁴ − 2x³

Now,

To find the points of maxima or minima, put f'(x) = 0

thus,

f'(x) = (4)5x³ - (3)2x² = 0

or

20x³ - 6x² = 0

or

x(20x² - 6x) = 0

or

x = 0      and       20x² - 6x = 0

or

x = 0       and       2x(10x - 3) = 0

or

x = 0       and       2x = 0      and   (10x - 3) = 0

or

x = 0    and     x = 0        and     x = [tex]\frac{3}{10}[/tex] = 0.3

thus,

condition for maxima or minima

f''(x) = (3 × 4)5x² - (2 × 3)2x

or

f''(x) = 60x² - 12x

at

x = 0

f''(0) = 60(0)² - 12(0) = 0

at x = 0.3

f''(0.3) = 60(0.3)² - 12(0.3)

= 5.4 - 3.6

= 1.8

since,

f''(0.3) > 0 therefore,

x = 0.3 is point of minima

and,

f''(0) = 0

thus,

x = 0 is point of neither maxima nor minima

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

Read more about line segments here:

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Answer:

a. $134.66

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 =128[/tex] represent the sample mean for the sample  

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

s=38 represent the sample standard deviation

n=100 represent the sample size  

2) Calculate the confidence interval

Since the sample size is large enough n>30. The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

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

[tex]128-1.64\frac{38}{\sqrt{100}}=121.768[/tex]    

[tex]128+1.64\frac{38}{\sqrt{100}}=134.232[/tex]

The closest value would be $134.66 and that would be the answer for this case.

The upper bound of a 90% confidence interval for the mean monthly gasoline expenditure per household, based on the sample provided, is approximately $134.66. This is derived using the z-score for a 90% confidence level and the given sample mean and standard deviation.

To calculate the upper bound of a 90% confidence interval for the mean monthly expenditure on gasoline per household, we can use the formula for the confidence interval of the mean, which is sample mean ± (critical value * (sample standard deviation / sqrt(sample size))). Since the sample size is 100, the sample mean is $128, and the sample standard deviation is $38, we need to find the critical value for a 90% confidence level. For a 90% confidence interval and a sample size of 100, which results in a degrees of freedom of 99, we could use a t-table to find the critical value; however, given the large sample size, the critical value will approximate the z-score, which is about 1.645 for a 90% confidence level.

The calculation is:

$128 + (1.645 * ($38 / sqrt(100)))

$128 + (1.645 * ($38 / 10))

$128 + (1.645 * 3.8)

$128 + 6.251

$134.251

The closest answer choice to our calculation is a. $134.66, which we can reasonably conclude to be a slight rounding difference in the critical value used.

Answer:

90% of people getting married for the first time get married before 32 years.

Step-by-step explanation:

The average age for a person getting married for the first time is 26 years i.e. [tex]\mu=26[/tex]

The ages for the first marriages have a normal distribution with a standard deviation of about 4 years i.e. [tex]\sigma = 4[/tex]

90% of people getting married for the first time.

The z-value at 90% confidence interval is z=1.64.

The z-score formula is given by,

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Where, x is the required sample mean or age.

Substitute the values,

[tex]1.64=\frac{x-26}{4}[/tex]

[tex]1.64\times 4=x-26[/tex]

[tex]6.56=x-26[/tex]

[tex]x=6.56+26[/tex]

[tex]x=32.56[/tex]

Therefore, 90% of people getting married for the first time get married before 32 years.

Answer:

Step-by-step explanation:

The standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean(μ)  and standard deviation(σ)

The general formula for the sample size is given below:

[tex]n=p^{'}(1-p^{'})(\frac{Z_{\frac{a}{3} } }{E} )^{2}[/tex]

The formular for finding sample size is given as:

[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]

a.)

it is given that [tex]E=±0.02, p^{'}_{1}=0.216, p^{'}_{2}=0.192[/tex]

The confidence level is 0.90

For (1 - ∝) = 0.90

∝=0.10; ∝/2 = 0.05

frm the standard normal table, the required [tex]Z_{0.05}[/tex] value for 90% confidence is 1.645. The sample size is as shown:

[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]

=[tex]n=(\frac{1.645}{0.05} )^{2} * (0.216(1-0.216)+0.192(1-0.192))\\=351.22≅352[/tex]

The required sample size is 352 (nearest whole number)

b.)

it is given that [tex]E=±0.02, p^{'}_{1}=0.5, p^{'}_{2}=0.5[/tex]

The confidence level is 0.90

For (1 - ∝) = 0.90

∝=0.10; ∝/2 = 0.05

frm the standard normal table, the required [tex]Z_{0.05}[/tex] value for 90% confidence is 1.645. The sample size is as shown:

[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]

=[tex]n=(\frac{1.645}{0.05} )^{2} * (0.5(1-0.5)+0.5(1-0.5))\\=541.205≅542[/tex]

The required sample size is 542 (nearest whole number)

Using the estimates from a previous year (21.6% male and 19.2% female), a sample size of 811 should be obtained. If no prior estimates are used, a sample size of 848 is needed.

To determine the sample size needed for estimating the difference in the proportion of men and women who participate in regular sustained physical activity, we can use the formula:

[tex]n = (Z^2 * p * (1-p)) / E^2[/tex]

Where:

(a) If the therapist uses the estimates of 21.6% male and 19.2% female from a previous year, we can assume that the sample proportion for both genders is the same (20.4%).

Plugging in the values:

[tex]n = ((1.645^2) * 0.204 * (1-0.204)) / (0.02^2)[/tex]

= 810.611

Therefore, a sample size of 811 should be obtained.

(b) If the therapist does not use any prior estimates, we can assume that the sample proportion for both genders is 0.5 (maximum variability).

Plugging in the values:

[tex]n = ((1.645^2) * 0.5 * (1-0.5)) / (0.02^2)[/tex]

= 847.075

Therefore, a sample size of 848 should be obtained.

Answer:

[tex]\frac{67}{292} = 0.229[/tex]

Step-by-step explanation:

Out of 292 drawn Americans over 21, 225 of them don't smoke. That means the rest of them are smokers:

So there are 292 - 225 = 67 smokers here.

We can then calculate the portion of Americans over 21 who smoke:

[tex]\frac{67}{292} = 0.229[/tex]

The probability that a sample mean is 12 or larger for a sample from the horse population is approximately 0.0264, which matches the correct answer.

Given:

Population mean (μ): 10 years

Sample mean ([tex]\bar{x}[/tex]): 12 years

Sample size (n): 60 horses

Population standard deviation (σ): 8 years

First, let's calculate the standard error of the sample mean (SE) using the formula:

SE = σ / √n

SE = 8 / √60 ≈ 1.032

Now, let's find the z-score using the formula:

z = ([tex]\bar{x}[/tex] - μ) / SE

z = (12 - 10) / 1.032 ≈ 1.938

Next, we find the probability that a sample mean is 12 or larger by finding the area under the standard normal curve to the right of z = 1.938.

Using a standard normal distribution table or calculator, we find that P(Z > 1.938) ≈ 0.0264.

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.

a. The expression for the varying number of radians Cody has swept out since the ride started is 6.5t radians.

b. Cody's height above the center of the Ferris wheel is[tex]\(30\cos(6.5t)\)[/tex] feet.

c. Cody's height above the ground is [tex]\(36 + 30\cos(6.5t)\)[/tex] feet.

a. To represent the varying number of radians Cody has swept out since the ride started, we use the formula for angular distance:

[tex]\[ \text{Angular distance} = \text{angular speed} \times \text{time} \][/tex]

Given that the Ferris wheel rotates at a constant angular speed of 6.5 radians per minute, the expression in terms of t is:

[tex]\[ \text{Angular distance} = 6.5t \][/tex]

b. To represent Cody's height above the center of the Ferris wheel, we use the relationship between angular displacement and height:

[tex]\[ \text{Height} = \text{radius} \times \cos(\text{angular distance}) \][/tex]

Substituting the expression for angular distance from part (a), we get:

[tex]\[ \text{Height} = 30 \times \cos(6.5t) \][/tex]

c. To represent Cody's height above the ground, we add the height of the center of the Ferris wheel to Cody's height above the center:

[tex]\[ \text{Total height} = \text{Center height} + \text{Height above center} \][/tex]

Given that the center of the Ferris wheel is 36 feet above the ground, the expression in terms of t becomes:

[tex]\[ \text{Total height} = 36 + 30 \times \cos(6.5t) \][/tex]

These expressions represent Cody's angular distance, height above the center of the Ferris wheel, and height above the ground as functions of time t.

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:

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:

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:

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 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]

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:

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:

We reject H₀

we accept Hₐ seeds in the packet would germinate smaller than 93%

Step-by-step explanation:

Test of proportions

One tail-test  (left side)

93 %   =  0.93

p₀  =  0,93

1.- Hypothesis

Answer:

P(1st health insurance)*p(2nd health insurance) = 0.1296  

Step-by-step explanation:

p(health insurance) =    =25/100 =   0.25

p(life insurance) =   =36/100 =   0.36

p(other) =   =24/100 =   0.24

p(car or home) =   =1-0.25-0.36-0.24 =   0.15

a)       P(1st health insurance)*p(2nd health insurance)      

= 0.36*0.36      

= 0.1296      

Answer: The length of the rectangle is 15 inches

Step-by-step explanation:

Let L represent the length of the rectangle.

Let W represent the width of the rectangle.

The perimeter of a rectangle is expressed as 2 length + 2 width

Perimeter = 2(L + W)

The rectangle has a perimeter of 52 inches. It means that

2(L + W) = 52

L + W = 52/2 = 26 - - - - - - - 1

The length of the rectangle is 4 inches more than its width. It means that

L = W + 4

Substituting L = W + 4 into equation 1, it becomes

W + 4 + W = 26

2W = 26 - 4 = 22

W = 22/2 = 11

L = W + 4 = 11 + 4 = 15

The Midpoint Rule is used to approximate the value of an integral. In this case, we are approximating the integral ∫ sin(x) dx on the interval [0, 64] using the Midpoint Rule with n = 4. The approximate value of the integral is approximately 39.6007.

The Midpoint Rule is used to approximate the value of an integral by dividing the interval into equal subintervals and evaluating the function at the midpoint of each subinterval. In this case, the integral is ∫ sin(x) dx on the interval [0, 64] and n = 4.

We can calculate the width of each subinterval by dividing the total interval length by the number of subintervals: (64-0)/4 = 16.

Next, we evaluate the function sin(x) at the midpoint of each subinterval and sum up the results, multiplying by the width of each subinterval:

Approximation = 16 * [sin(8) + sin(24) + sin(40) + sin(56)]

Calculating the values of sin(8), sin(24), sin(40), and sin(56) using a calculator, we get:

Approximation ≈ 16 * [0.1392 + 0.4121 + 0.7451 + 0.9309] ≈ 39.6007

Therefore, the approximate value of the integral ∫ sin(x) dx on the interval [0, 64] using the Midpoint Rule with n = 4 is approximately 39.6007.

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To approximate the integral, we divide the interval into subintervals, find the midpoints, evaluate the function at the midpoints, and multiply by the width of the subintervals.

To approximate the integral using the Midpoint Rule, we divide the interval [0, 64] into n subintervals of equal width. In this case, n = 4, so each subinterval has width (64-0)/4 = 16.

The midpoint of each subinterval is used to estimate the value of the function sin(x) within that subinterval.

The midpoints of the subintervals are: 8, 24, 40, and 56.

We evaluate sin(x) at these midpoints and multiply each value by the width of the subintervals (16).

Finally, we sum these values to approximate the integral.

Using a calculator or computer, we find that sin(8) = 0.989, sin(24) = -0.905, sin(40) = 0.745, and sin(56) = -0.521.

Therefore, the approximation of the integral is (0.989 + (-0.905) + 0.745 + (-0.521)) * 16 = 3.296.

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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.

Final answer:

The volume of the solid with a triangular base and semicircular cross sections is found by integrating the area of the semicircles along the height of the triangle. This approach involves calculating the area of semicircles using their radius which changes with y, and then integrating from 0 to 15 along the y-axis. The total volume is (56.25)π cubic units.

Explanation:

To find the volume of the solid with a triangular base and semicircular cross sections perpendicular to the base and parallel to the y-axis, we use the general slicing method. The base of the solid is a triangle with vertices at (0,0), (15,0), and (0,15), implying that it is a right-angled isosceles triangle on the xy-plane. The length of the legs are both 15 units.

For a slice at a particular y-value, the length of the base of the semicircle (which is also the diameter) is equal to the x-value at that y (since the triangle's equation is x + y = 15, x = 15 - y). The radius of the semicircle is thus ½(15 - y). The area of a semicircle is given by ½[tex]πr^{2}[/tex], substituting the expression for the radius we get ½π(½(15 - y))2.

To find the volume, we integrate this area from y = 0 to y = 15 along the y-axis. The integral of ½π(½(15 - y))2 dy from 0 to 15 is ½π × ½2 × ∫015 (15 - y)2 dy, which simplifies to ∑ volume = ⅔π(152) (⅓) = ½π(225)(⅓) = (56.25)π cubic units.

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: 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: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.

Step-by-step explanation:

The answer is A because since the median is resistant to extreme values but the mean is not, the mean tends to move toward extreme values in the distribution.

Sorry if I'm wrong but i tried my best sorry :) :(

The statement which is generally true about a distribution that is highly skewed to the right is A) The median is less than the mean.

Mean of a set of data is defined as the average of all the values. It gives the exact middle point of the data set.

In a right skewed distribution, the data values are more concentrated on the right side.

So the values of the data will be larger, which in turn makes the value of the mean to be higher.

Since median is the exact middle value in the set, it does not depend on the right values.

So mean will be larger than median.

Hence the correct option is A.

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Answer:

We have the line parametrized by

[tex]x=8+6t\\y=3+2t[/tex]

Solving for t in each equation we have that

[tex]t=\frac{x-8}{6}\\t=\frac{y-3}{2}[/tex]

The point (a,b) lies in the line if when we replace a in the first equation and b in the second equation, the values of t coincide.

a)

1. (-4,-1)

[tex]t=\frac{-4-8}{6}=-2\\t=\frac{-1-3}{2}=-2[/tex]

Then, (-4,-1) lies in the line but no lies in the section of the line obtained by restricting t to nonnegative numbers.

2. (26,9)

[tex]t=\frac{26-8}{6}=3\\t=\frac{9-3}{2}=3[/tex]

Since t is positive then (26,9) lies in the line and lies in the section of the line obtained by restricting t to nonnegative numbers.

3. (32,11)

[tex]t=\frac{32-8}{6}=4\\t=\frac{11-3}{2}=4[/tex]

Since t is positive then (32,11) lies in the line and lies in the section of the line obtained by restricting t to nonnegative numbers.

4. If we take t=2 we obtain the point

[tex]x=8+6(2)=20\\y=3+2(2)=7[/tex]

(20,7) that lies in the section of the line obtained by restricting t to nonnegative numbers.

b)

When t=-5,

[tex]x=8+6(-5)=-22\\y=3+2(-5)=-7[/tex]

correspond to the point (-22,-7).

when t=-2

[tex]x=8+6(-2)=-4\\y=3+2(-2)=-1[/tex]

correspond to the point (-4,-1).

-22<-4 and -7<-1

then the left endpoint (-22,-7) and right endpoint (-4,-1)