Hi to whoever is reading this. I really need help on breaking down the expression. I am doing this without a calculator and I don’t Know what steps to take to get the answer.

Answer:

(A) Type I error is convicting an innocent person.

Step-by-step explanation:

When someone is on trial for suspicion of committing a crime, the hypotheses are:

[tex]H_0[/tex] : innocent

[tex]H_A[/tex] : guilty

From the given options, A Type I error is convicting an innocent person. If the null hypothesis holds (i.e. a person is innocent) nut we still go ahead to convict the person, we have rejected a true null hypothesis.

Answer:

0.0879 is the probability that out of 5 randomly selected consumers, three are comfortable with delivery by drones.        

Step-by-step explanation:

We are given the following information:

We treat drone deliveries as a success.

P(consumers comfortable having drones deliver) = 25% = 0.25

Then the number of consumers follows a binomial distribution, where

[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]

where n is the total number of observations, x is the number of success, p is the probability of success.

We have to evaluate:

P(Exactly 3 customers out of 5 are comfortable with delivery by drones)

Here,

[tex]n = 5\\x = 3\\p = 0.25\\q = 1 - p = 1-0.25=0.75[/tex]

Putting values, we get,

[tex]P(x =3)\\\\= \binom{5}{3}(0.25)^3(1-0.25)^2\\\\= 0.0879[/tex]

0.0879 is the probability that out of 5 randomly selected consumers, three are comfortable with delivery by drones.

Answer:

A) Confidence Interval will become narrower. B) Confidence Interval will become narrower. C) Confidence Interval will become broader.

Step-by-step explanation:

Confidence Interval is the probable range around sample statistic, in which the population parameter is expected to lie.

Confidence Level shows the average percentage level of confidence interval, expected to contain population parameter. Lower confidence level implies narrower Confidence Interval

Bigger sample size reduces margin error (sample statistic, population parameter difference). Parameter-statistic proximity implies: narrower confidence interval around statistic, expected to contain parameter.

Standard Deviation is a measure of dispersion, spread. So, higher standard deviation implies more spread & broader confidence interval.

Final answer:

To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Using the formula New Population = Current Population x 2^(Number of Doubling Cycles), the new population will be approximately 4.8 million. Option A .

Explanation:

To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Since the city doubles in size every 35 years, in 20 years it will go through 20/35 of a doubling cycle.

To calculate how many people there will be, we use the formula:

New Population = Current Population x 2^(Number of Doubling Cycles)

Plugging in the values, we have:

New Population = 2.4 million x 2^(20/35)

Using a calculator, we find that the new population will be approximately 4.8 million. Therefore, the answer is 4.8 million.

Answer:

28x7=196 (Pls give Brainliest)

Answer:

Step-by-step explanation:

The volume of a cuboid is the product of its dimensions:

  V = LWH = (3 ft)(3 ft)(2 ft)

  V = 18 ft³

The area is the sum of the areas of the faces. Since opposite faces have the same area, we can figure the area from ...

  A = 2(LW +H(L+W)) = 2((3 ft)(3 ft) +(2 ft)(3 ft +3 ft)) = 2(9 ft² +12 ft²)

  A = 42 ft²

Answer:

50

Step-by-step explanation:

Answer:

the answer is 200

Step-by-step explanation:

Answer:

D.

Step-by-step explanation:

Rearrange

[tex]0.95x - 5.00 < 33.95[/tex]

[tex]0.95x<38.95[/tex]

[tex]x<\frac{38.95}{0.95}[/tex]

[tex]x<41[/tex]

Answer:

95

Step-by-step explanation:

66-66=0

0-29=-29

66+29=95

The range of the temperatures during this period is 95.

Given the following data:

To determine the range of the temperatures during this period:

Range is simply calculated by taking the difference between the highest number and the lowest number in a sample.

Mathematically, range is given by the formula;

[tex]Range = highest \;number -lowest \;number[/tex]

Substituting the given parameters into the formula, we have;

[tex]Range = 66-(-29)\\\\Range =66+29[/tex]

Range = 95

Therefore, the range of the temperatures during this period is 95.

Read more: https://brainly.com/question/16205323

Answer:

Check the explanation

Step-by-step explanation:

Ans=

A: For m = 5: P(³≥1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134

M = 10: 1 – 0.9973^10 = 0.0267

M = 20: 1 – 0.9973^20 = 0.0526

M = 30: 1 – 0.9973^30 = 0.0779

M = 50: 1 – 0.9973^50 = 0.126

18)

Ans=

Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values

Answer:

y = sin(t) is a solution to the differential equation

(dy/dt)² = 1 - y²

Step-by-step explanation:

Given (dy/dt)² = 1 - y²

Suppose y = sin(t) is a solution, then it satisfies the differential equation.

That is

[d(sin(t))]² = 1 - y²

Let y = sin(t)

dy/dt = d(sin(t)) = cos(t)

(dy/dt)² = cos²t

But cos²t + sin²t = 1

=> 1 - sin²t = cos²t

So

(dy/dt)² = 1 - sin²t

Since sin²t = (sint)² = y²,

we have

(dy/dt)² = 1 - y²

as required.

The differential equation becomes [tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]

Given the function;

Take the differential of the function

Square both sides of the equation to have:

[tex](\frac{dy}{dx} )^2 = (cost)^2[/tex]

Recall from trigonometry identity that [tex]sin^2t + cos^2t = 1[/tex]

Hence, [tex]cos^2t = 1- sin^2t[/tex]

Replace into the differential expression to have:

[tex](\frac{dy}{dx} )^2 = 1-sin^2t[/tex]

Recall that y = sin(t). On replacing, the differential equation becomes:

[tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]

learn more on differential equation here: https://brainly.com/question/18760518

Answer:

A. The interval will be narrower if 15 men are used in the sample.

Step-by-step explanation:

Hello!

When all other things remain the same, which of the following statements about the width of the interval is correct?

A. The interval will be narrower if 15 men are used in the sample.

B. The interval will be wider if 15 men are used in the sample.

C. The interval will be narrower if 5 men are used in the sample.

D. The interval will be narrower if the level is increased to 99% confidence.

E. The interval will be wider if the level is decreased to 90% confidence.

Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.

The general structure for the CI is "point estimate"±" margin fo error"

Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)

The margin of error of the interval is:

d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)

1) The sample size changes, all other terms of the interval stay the same.

As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.

↓d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↑n)

↑d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↓n)

Correct option: A. The interval will be narrower if 15 men are used in the sample.

2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:

↓d= ↓[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)

↑d= ↑[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)

I hope it helps!

Answer:

The interval will be narrower if 15 men are used in the sample.

Step-by-step explanation:

Answer:

attached

Step-by-step explanation:

attached

Answer:

22

Step-by-step explanation:

PEMDAS

4^2 = 16

16 - 5 = 11

11 x 2 = 22

Answer:

22

Step-by-step explanation:

2 (c^2 -5)

Let c=4

2 (4^2 -5)

PEMDAS

Parentheses first,

(4^2 -5)

Exponents

16-5 =11

Replace into expression

2(11)

22

Answer:

[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

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

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

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

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=24.8[/tex]

The sample deviation calculated [tex]s=4.367[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=10-1=9[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that [tex]t_{\alpha/2}=1.833[/tex]

And the margin of error would be given by:

[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]

Answer:

It was hypothesized that more people would choose the number 7 as their 'lucky' number than any other number.

Step-by-step explanation:

Given that one variable chi-square is used to test whether a single categorical variable follows a hypothesized population distribution. The Chi Square statistic compares the tallies or counts of categorical responses between two (or more) independent groups

The null hypothesis (H0) for the test is that all proportions are equal.

The alternate hypothesis (H1) is given condition in the question.

A. It was hypothesized that more people would choose the number 7 as their 'lucky' number than any other number.

This is suited for testing with a one-variable chi-square test because we are testing if the proportion of people who choose number 7 is greater than the proportion of any other numbers. So, we are therefore comparing more than 2 proportions.

B. People who choose the number 7 as their 'lucky' number are significantly more superstitious than people who choose the number 13 as their 'lucky' number.

This is not suited for testing with a one-variable chi-square test. A z test is more preferable in this instance because we are testing just two proportions.

C. Choice of 'lucky' number is directly related to measures superstition.

This is not suited for testing with a one-variable chi-square test because chi square test is not used for showing relationship between variables.

D. All of these. Since option A is correct, this option can not be correct.

Answer:
The equivalent system of first-order differential equations is:

[tex]\[ \begin{cases}u' = v \\v' = 8\sin(3t) - 6.5v + 1.5u\end{cases} \][/tex]

with the initial conditions:

[tex]\[ \begin{cases}u(1) = -3 \\v(1) = -1.5\end{cases} \][/tex]


To convert the given second-order differential equation into an equivalent system of first-order equations, we follow these steps:

1. Identifying the original second-order differential equation:

  u'' + 6.5u' - 1.5u = 8sin(3t)

2. Then we introduce a new variable v to represent the first derivative of u:

  v = u'

3. Now ,we write the second-order equation as a system of first-order equations:

  The original equation is:

  u'' + 6.5u' - 1.5u = 8sin(3t)

  Using the new variable v = u' , the second derivative u'' can be written as v'. Therefore, we have:

  v' + 6.5v - 1.5u = 8sin(3t)

4. Formulating the system of first-order equations:

  The first equation comes directly from the definition of v:

  u' = v

  The second equation comes from the rewritten second-order equation:

  v' = 8sin(3t) - 6.5v + 1.5u

5. Let us include the initial conditions:

  The initial conditions provided are:

  u(1) = -3

  u'(1) = -1.5

  Since v = u', this translates to:

  v(1) = -1.5

6. Writing the system with initial conditions:

  The system of first-order equations is:

[tex]\[ \begin{cases} u' = v \\ v' = 8\sin(3t) - 6.5v + 1.5u \end{cases} \][/tex]

  With the initial conditions:

 [tex]\[ \begin{cases} u(1) = -3 \\ v(1) = -1.5 \end{cases} \][/tex]

Answer:

1 / 15

Step-by-step explanation:

1/3 / 5 =

1/ 15

Final answer:

Each person will get 1/15 of a pound of peanuts when a 1/3 pound bag is shared evenly among 5 people.

Explanation:

To figure out how much each of the 5 people should get from a 1/3 pound bag of peanuts, we need to divide the total weight of the peanuts by the number of people. This gives us:

1/3 pound ÷ 5 = 1/15 pound per person.

This means each person will get 1/15 of a pound of peanuts. In other calculations such as the candy survey, determining percent uncertainty, or unit conversions as in Michaela's party scenario, a similar process of division or unit conversion is applied to find the answer.

Answer:

Do the data from this shipment indicate statistical control: No

Step-by-step explanation:

Calculating the mean of the sample, we have;

Mean (x-bar) = sum of individual sample/number of sample

                     = (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10

                     = 0.044/10

                    = 0.0044

Calculating the lower control limit (LCL) using the formula;

LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n

      = 0.0044 - 3*√(0.0044(1-0.0044))

       = 0.0044- (3*0.0042)

        = 0.0044 - 0.01256

        = -0.00816 ∠ 0

Calculating the upper control limit (UCL) using the formula;

UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n

      = 0.0044 + 3*√(0.0044(1-0.0044))

       = 0.0044+ (3*0.0042)

        = 0.0044 + 0.01256

       =0.01696∠ 0

Do the data from this shipment indicate statistical control: No

Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that  the data from this shipment do not indicate statistical control.

The data from this shipment does not indicate statistical control.

Calculating the mean of the sample, we have;

Mean (x-bar) = sum of individual sample/number of sample

[tex]\frac{(0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)}{10}\\=\frac{0.044}{10}\\=0.0044[/tex]

Calculating the lower control limit (LCL) using the formula;

LCL

= (x-bar) - 3*√(x-bar(1-x-bar))/n

[tex]= 0.0044 - 3*\sqrt{(0.0044(1-0.0044))}\\= 0.0044- (3*0.0042)\\= 0.0044 - 0.01256\\= -0.00816[/tex]

Calculating the upper control limit (UCL) using the formula;

UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n

[tex]= 0.0044 + 3*\sqrt{(0.0044(1-0.0044))}\\= 0.0044+ (3*0.0042)\\= 0.0044 + 0.01256\\=0.01696[/tex]

Do the data from this shipment indicate statistical control:

Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that  the data from this shipment does not indicate statistical control.

Learn more about mean: https://brainly.com/question/12892403

Answer: The 95% confidence interval will be wider.

Step-by-step explanation:

Confidence interval for population proportion is written as

Sample proportion ± margin of error

margin of error = z score × √pq/n

The z score is determined by the confidence level. The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%

This means that with all other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.

The higher the margin of error, the wider the confidence interval. Therefore,

The 95% confidence interval will be wider.

The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'

Given that,

If construct a 90% confidence interval for the population proportion,

And a 95% confidence interval for the population proportion,

Tim: 'The 90% confidence interval will be wider.'

Trevor: 'The 95% confidence interval will be wider.

Tracy: 'There is not enough information to tell.

Either interval could be wider.'

We have to determine,

Which of these intervals will be wider.

According to the question,

Three students provide their answers as follows:

Tim: 'The 90% confidence interval will be wider.

'Trevor: 'The 95% confidence interval will be wider.

'Tracy: 'There is not enough information to tell. Either interval could be wider.'

Therefore, Confidence interval for population proportion is written as

Sample proportion ± margin of error

Margin of error  [tex]= z score \times \frac{ \sqrt{pq}}{n}[/tex]

The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%.

Other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.

The higher the margin of error, the wider the confidence interval.

Therefore, The 95% confidence interval will be wider.

Hence, The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'

For more information about Probability click the link given below.

https://brainly.com/question/15688515

Answer:

1. Mean is 1.75

2. The variance is 1.6042

3.

The distribution function is:

Z           Z/K

0           21/128

1             5/16

2         33/128

3          1/6

4           29/384

5             1/48

6          1/384

Step-by-step explanation:

The mean of Z is given as:

E(Z) =Σ6, k=0 Kp (Z = k)

Σ6,k=0 K 1/6 Σ6, n=k (n k) (1/2)^n

=( 0(21/128) + 1(5/16) + 2( 33/128) + 3 (1/6) + 4 (29/384) + 5 (1/48) + 6 (1/384))

=7/4

=1.75

Thus, the mean Z is 1.75

The variance of Z is given as:

Var (Z) = E (Z^2) - (E (Z)) ^2

Therefore,

E(Z^2) = Σ 6, k=0 K^2P ( Z=K)

= ( 0(21/128 + 1(5/16) + 4(33/128) + 9(1/6) + 16(29/384) + 25(1/48) + 36(1/384))

=14/3

Var (Z) = 14/7 - (7/4)^2

= 14/7 - 49/16

=77/48

=1.6042

Thus, the variance is 1.6042

The probability of mass function is given as:

P(Z=k) = 1/6 Σ 6, n=k (n  k) (1/2)^n

The distribution function is

Z           Z/K

0           21/128

1             5/16

2         33/128

3          1/6

4           29/384

5             1/48

6          1/384

Answer:

[tex] \huge \pi \: units [/tex]

Step-by-step explanation:

[tex]l = \frac{48 \degree}{360 \degree} \times 2\pi \: r \\ \\ = \frac{2}{24} \times 2 \times \pi \times 6 \\ \\ = \frac{24}{24} \times \pi \\ \\ = \pi \: units \\ [/tex]

Answer:

The slope is -1

Step-by-step explanation:

Let's find the slope between your two points.

(1,4);(6,−1)

(x1,y1)=(1,4)

(x2,y2)=(6,−1)

Use the slope formula:

m= y2−y1/x2−x1  = −1−4/6−1

= −5/5

= −1

Hope this is a good explanation :)

Answer:

-10; no

Step-by-step explanation:

Final answer:

The inner product of the vectors (-4,9,8) and (3,2,-2) is -10. Since the inner product is not zero, the vectors are not perpendicular. Therefore, the correct answer is: -10; no.

Explanation:

The inner product (also known as the dot product) of two vectors (-4,9,8) and (3,2,-2) is calculated by multiplying the corresponding components of the two vectors and summing the result:

Inner product = (-4)×3 + 9×2 + 8×(-2)
= -12 + 18 - 16
= -10

To determine if the vectors are perpendicular, we check if their inner product is zero. Since the inner product in this case is -10, not zero, the vectors are not perpendicular.

Step-by-step explanation:

x²=a²+b²

x=√6²+12²

x=√180

x=3√2v

y²=16²+12²

y=√400

y=20

Answer: the answer for rafter 1 is 13.4 and the answer for rafter 2 is 20

Step-by-step explanation: I just know

Answer:

5 faces

Step-by-step explanation:

4 triangular, 1 square

Answer:

Step-by-step explanation:

Hello!

You have the output:

Two-Sample T-Test and Cl

Sample N Mean StDev SE Mean

1(M) 112 7.38 6.95 0.66

2(F) 101 7.15 6.31 0.63

Difference = mu (1) - mu (2)

Estimate for difference: 0.230000

95% lower bound for difference: -1.271079

T-Test of difference: T-value = 0.25 P-Value = 0.400 DF= 210

This output summarizes the information of the two samples and indicates the order the populations where studied.

It also informs you of the value of the statistic under the null hypothesis and the p-value.

Unfortunately, there is no information on the type of hypotheses that were tested, i.e. if they where two-tailed or one-tailed, in the latter case, there is no information if it was left-tailed or right-tailed). Likewise is not specified if the test was done for a specific value of the parameter. (for example μ₁ - μ₂ = 0 or μ₁ - μ₂ = θ₀)

For these reasons, the data provided by the output isn't enough to conclude any hypothesis.

From all the provided answers the one more likely to be correct is:

(iii) The data do not provide sufficient evidence to conclude that the mean depression score of male teens is larger than that of female teens.

I hope this helps!

Answer:

iv

Step-by-step explanation:

Since the p-value (0.4) is greater than the significance level (0.05), we can conclude that the result is not significant. This means that there is no enough statistical evidence to reject the null hypothesis H0. Therefore, we must accept it and conclude that the mean depression score for male and female teens do not differ.

Answer:

10% probability that a given class period runs between 51.25 and 51.75 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability of finding a value X between c and d, d greater than c, is given by the following formula:

[tex]P(c \leq X \leq d) = \frac{d-c}{b-a}[/tex]

Uniformly distributed between 49 and 54 minutes

This means that [tex]b = 54, a = 49[/tex]

Find the probability that a given class period runs between 51.25 and 51.75 minutes.

[tex]P(51.25 \leq X \leq 51.75) = \frac{51.75 - 51.25}{54 - 49} = 0.1[/tex]

10% probability that a given class period runs between 51.25 and 51.75 minutes.