A mass that weighs 8 lb stretches a spring 6 in. The system isacted on by an external force of 8 sin 8t lb. If the mass is pulleddown 3 in and then released, determine the position of the mass atany time. Determine the first four times at which the velocity ofthe mass is zero.

Answer: G=21

Step-by-step explanation:

Solve for G by simplifying both sides of the equation, then isolating the variable.

Answer:

10.0

Step-by-step explanation:

4.0+8.0=12.0-2.0=10.0

The probability that both students will not wear blue is 0.36, obtained by multiplying the probability of one student not wearing blue, 0.60, by itself.

To find the probability that both students will not be wearing blue, we first need to determine the probability that one student is not wearing blue, and then multiply that probability by itself for two students.

Step 1: Calculate the probability that one student is not wearing blue.

Given that 40% of students wore blue, the probability that one student is not wearing blue is 1 - 0.40 = 0.60.

Step 2: Multiply the probability for one student by itself for two students.

The probability that both students will not be wearing blue is (0.60) ×(0.60) = 0.36.

So, the probability that both students will not be wearing blue is 0.36.

Answer:

Step-by-step explanation:

FIRST FOR ORANGES SHE BOUGHT 10 SO 10 TIMES 3 IS 30 AND THEN 11 TIMES 2.40 IS 26.40. 30 PLUS 26.40 IS 56.40 THATS YOUR AWNSER

Answer: A

Step-by-step explanation:

Answer:

the first opition is the correct answer

Answer:

4n-3=5

Step-by-step explanation:

If you read it step by step the product of and a number is basically 4* any variable.( In this case I use n.) Next the beginning part is 3 less than the product part so it is 4n-3. And finally the whole equation is equal to 5 so itis 4n-3=5.

Answer:

The final integration in the given limits will be 89.876

Given:

Given that the radius of the merry - go - round is 5 feet.

The arc length of AB is 4.5 feet.

We need to determine the measure of the minor arc AB.

Measure of the minor arc AB:

The measure of the minor arc AB can be determined using the formula,

[tex]Arc \ length=(\frac{\theta}{360})2 \pi r[/tex]

Substituting arc length = 4.5 and r = 5, we get;

[tex]4.5=(\frac{\theta}{360})2 (3.14)(5)[/tex]

Multiplying the terms, we get;

[tex]4.5=(\frac{\theta}{360})31.4[/tex]

Dividing, we get;

[tex]4.5=0.087 \theta[/tex]

Dividing both sides of the equation by 0.087, we get;

[tex]51.7=\theta[/tex]

Rounding off to the nearest degree, we have;

[tex]52=\theta[/tex]

Thus, the measure of the minor arc AB is 52°

Answer:

52°

Step-by-step explanation:

Arc length = (theta/360) × 2pi × r

4.5 = (theta/360) × 2 × 3.14 × 5

theta/360 = 45/314

Theta = 51.59235669

Answer:

We conclude that % of DC area adults who have traveled outside of the United States is different from 40%.

Step-by-step explanation:

We are given that 40% of DC area adults have traveled outside of the United States. Nardole wants to know if his customers are typical in this respect. He surveys 40 customers and finds 60% have traveled outside of the U.S.

We have to test is this result a statistically significant difference.

Let p = % of DC area adults who have traveled outside of the United States

SO, Null Hypothesis, [tex]H_0[/tex] : p = 40%  {means that 40% of DC area adults have traveled outside of the United States}

Alternate Hypothesis, [tex]H_a[/tex] : p [tex]\neq[/tex] 40%  {means that % of DC area adults who have traveled outside of the United States is different from 40%}

The test statistics that will be used here is One-sample z proportion statistics;

                  T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = % of customers who have traveled outside of the United States

                  in a survey of 40 customers = 60%

          n  = sample of customers = 40

So, test statistics = [tex]\frac{0.60-0.40}{\sqrt{\frac{0.60(1-0.60)}{40} } }[/tex]

                             = 2.582

Since in the question we are not given with the significance level so we assume it to be 5%. So, at 0.05 level of significance, the z table gives critical value of 1.96 for two-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that % of DC area adults who have traveled outside of the United States is different from 40%.

Answer:

434.98

Step-by-step explanation:

To calculate the probability that the sample mean would differ from the true mean by greater than $38, if a sample of 208 persons is randomly selected, we need to use the Central Limit Theorem. First, we determine the standard error of the mean (SEM) using the formula SEM = standard deviation / square root of sample size. Then, we calculate the Z-score using the formula Z = (sample mean - true mean) / SEM. Finally, we find the probability associated with the Z-score using a Z-table or calculator.

To calculate the probability that the sample mean would differ from the true mean by greater than $38, if a sample of 208 persons is randomly selected, we need to use the Central Limit Theorem.

According to the Central Limit Theorem, the distribution of sample means will be approximately normal regardless of the shape of the population distribution, as long as the sample size is large enough.

Since the sample size is greater than 30, we can assume that the distribution of sample means will be approximately normal.

To calculate the probability, we first need to determine the standard error of the mean (SEM), which is the standard deviation divided by the square root of the sample size. In this case, the SEM = $397 / √208.

Next, we calculate the Z-score using the formula Z = (sample mean - true mean) / SEM = ($38 - 0) / ($397 / √208). Finally, we can use a Z-table or calculator to find the probability associated with the Z-score.

In this case, it is the probability that Z is greater than the calculated Z-score. Hence, the probability that the sample mean would differ from the true mean by greater than $38 is the probability that Z is greater than the calculated Z-score.

Answer:

The truck will skid for 151 ft before stopping when the brakes are applied

Step-by-step explanation:

From the equations of motion, we will use

[tex]v^{2} = u^{2}-2aS[/tex]

We have to make sure that the parameters we are working with are in the same unit of length. Here, we will be converting from ft to miles

When the truck is travelling at 10 mph.

S = distance the truck skids = [tex]\frac{5ft}{5286ft/mile}= 0.00094697miles[/tex]

Final velocity of truck, v = 0 m/s (this is because the truck decelerates to a halt)

Initial velocity of truck u = 10 mph

Hence, we have

[tex]0^{2}=10^{2}-2a\times 0.00094697[/tex]

[tex]a= 52799.9miles/hr^{2}[/tex]

This is the deceleration of the truck

We will work based on the assumption that the car decelerates at the same rate each time the brakes are fully applied.

When the truck is travelling at u= 55 mph.

We will need to use the deceleration of the car to find the distance traveled when it skids.

[tex]0^{2}=55^{2}-2\times52799.98\times S[/tex]

[tex]S= 0.0286 miles\approx 151 ft[/tex]

∴The car skids for about 151 ft when it is travelling at 55 mph and the brakes are applied.

Answer: 3(3x−4) is the answer

Answer:

3(3x-4) is the answer

Step-by-step explanation:

The center of mass of the lamina is located at the point (0, (39/12) (1 / ln(4))).

To find the center of mass of the given lamina, we need to calculate the moment about the x-axis and the y-axis, and then divide them by the total mass of the lamina.

Given information:

- The boundary of the lamina consists of the semicircles y = sqrt(1 - x^2) and y = sqrt(16 - x^2), and the portions of the x-axis that join them.

- The density at any point is inversely proportional to its distance from the origin.

Find the total mass of the lamina.

Let the density function be [tex]\[ \rho(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \][/tex], where k is a constant.

The total mass, M, is given by the double integral of the density function over the region of the lamina.

M = ∫∫ ρ(x, y) dA

To evaluate this integral, we need to express the lamina in polar coordinates.

The semicircles can be represented as:

0 ≤ r ≤ 1, 0 ≤ θ ≤ π

0 ≤ r ≤ 4, π ≤ θ ≤ 2π

The total mass can be calculated as:

[tex]\[ M = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r \, dr \, d\theta \]\[ M = k \left( \pi \ln(1) + 2\pi \ln(4) \right) \]\[ M = 2\pi k \ln(4) \][/tex]

Calculate the moment about the x-axis.

The moment about the x-axis, Mx, is given by:

[tex]\[ M_x = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r^2 \sin(\theta) \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r^2 \sin(\theta) \, dr \, d\theta \]\[ M_x = k \left( \frac{\pi}{2} + \frac{32\pi}{3} \right) \]\[ M_x = \frac{39\pi}{6} k \][/tex]

Calculate the moment about the y-axis.

The moment about the y-axis, My, is given by:

My = ∫∫ x ρ(x, y) dA

In polar coordinates:

[tex]\[ M_y = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r^2 \cos(\theta) \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r^2 \cos(\theta) \, dr \, d\theta \][/tex]

My = 0 (due to symmetry)

Find the coordinates of the center of mass.

The coordinates of the center of mass (x_cm, y_cm) are given by:

x_cm = My / M

y_cm = Mx / M

Substituting the values, we get:

x_cm = 0 / (2πk ln(4)) = 0

y_cm = (39π/6) k / (2πk ln(4)) = (39/12) (1 / ln(4))

Therefore, the center of mass of the lamina is located at the point (0, (39/12) (1 / ln(4))).

Complete question:

The boundary of a lamina consists of the semicircles y=sqrt(1 − x^2) and y= sqrt(16 − x^2) together with the portions of the x-axis that join them. Find the center of mass of the lamina if the density at any point is inversely proportional to its distance from the origin.

The center of mass of the lamina is at the origin (0, 0)

To find the center of mass of the lamina, we first need to find the mass and the moments about the  x- and  y-axes.

The mass  M of the lamina can be calculated by integrating the density function over the lamina. Since the density at any point is inversely proportional to its distance from the origin, we can express the density[tex]\( \delta \) as \( \delta(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \)[/tex], where  k is a constant.

Let's denote [tex]\( \delta(x, y) \) as \( \frac{k}{\sqrt{x^2 + y^2}} \)[/tex]. Then the mass M  is given by the double integral of [tex]\( \delta(x, y) \)[/tex] over the region  R bounded by the semicircles and the portions of the x-axis:

[tex]\[ M = \iint_R \delta(x, y) \, dA \][/tex]

Where  dA  represents the differential area element.

To find the moments about the  x- and  y -axes, we calculate:

[tex]\[ M_x = \iint_R y \delta(x, y) \, dA \]\[ M_y = \iint_R x \delta(x, y) \, dA \][/tex]

Then, the coordinates [tex]\( (\bar{x}, \bar{y}) \)[/tex] of the center of mass are given by:

[tex]\[ \bar{x} = \frac{M_y}{M} \]\[ \bar{y} = \frac{M_x}{M} \][/tex]

Now, let's proceed to find [tex]\( M \), \( M_x \), and \( M_y \)[/tex]

First, let's express the density [tex]\( \delta(x, y) \)[/tex] in terms of k:

[tex]\[ \delta(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \][/tex]

Now, we'll find the mass  M by integrating [tex]\( \delta(x, y) \)[/tex] over the region  R :

[tex]\[ M = \iint_R \frac{k}{\sqrt{x^2 + y^2}} \, dA \][/tex]

Since the region  R  is symmetric about the  x-axis, we can integrate over the upper half and double the result:

[tex]\[ M = 2 \iint_{R_1} \frac{k}{\sqrt{x^2 + y^2}} \, dA \][/tex]

Now, we'll switch to polar coordinates [tex]\( (r, \theta) \)[/tex]. In polar coordinates, the region [tex]\( R_1 \)[/tex]is described by [tex]\( 0 \leq \theta \leq \pi \) and \( 1 \leq r \leq 4 \).[/tex]

So, the integral becomes:

[tex]\[ M = 2 \int_{0}^{\pi} \int_{1}^{4} \frac{k}{r} \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \int_{1}^{4} 1 \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} (4 - 1) \, d\theta \]\[ = 2k \int_{0}^{\pi} 3 \, d\theta \]\[ = 6k \pi \][/tex]

For [tex]\( M_x \)[/tex], we integrate [tex]\( x \delta(x, y) \)[/tex] over the region R :

[tex]\[ M_x = \iint_R x \cdot \frac{k}{\sqrt{x^2 + y^2}} \, dA \]\[ = 2 \int_{0}^{\pi} \int_{1}^{4} r \cos(\theta) \cdot \frac{k}{r} \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \int_{1}^{4} \cos(\theta) \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \left[ \frac{1}{2} r^2 \cos(\theta) \right]_{1}^{4} \, d\theta \][/tex]

[tex]\[ = 2k \int_{0}^{\pi} \left( 8 \cos(\theta) - \frac{1}{2} \cos(\theta) \right) \, d\theta \]\[ = 2k \int_{0}^{\pi} \left( \frac{15}{2} \cos(\theta) \right) \, d\theta \]\[ = 2k \left[ \frac{15}{2} \sin(\theta) \right]_{0}^{\pi} \]\[ = 2k \cdot 0 \]\[ = 0 \][/tex]

Now, for[tex]\( M_y \)[/tex], we integrate [tex]\( y \delta(x, y) \)[/tex]over the region  R :

[tex]\[ M_y = \iint_R y \cdot \frac{k}{\sqrt{x^2 + y^2}} \, dA \\\[ = 2 \int_{0}^{\pi} \int_{1}^{4} r \sin(\theta) \cdot \frac{k}{r} \cdot r \, dr \, d\theta \\\[ = 2k \int_{0}^{\pi} \int_{1}^{4} \sin(\theta) \cdot r \, dr \, d\theta \\\[ = 2k \int_{0}^{\pi} \left[ \frac{1}{2} r^2 \sin(\theta) \right]_{1}^{4} \, d\theta \\[/tex]

[tex]\[ = 2k \int_{0}^{\pi} \left( 8 \sin(\theta) - \frac{1}{2} \sin(\theta) \right) \, d\theta \\\[ = 2k \int_{0}^{\pi} \left( \frac{15}{2} \sin(\theta) \right) \, d\theta \\\[ = 2k \left[ -\frac{15}{2} \cos(\theta) \right]_{0}^{\pi} \\\[ = 2k \cdot 0 \\\[ = 0 \][/tex]

Now, we have [tex]\( M = 6k \pi \), \( M_x = 0 \), and \( M_y = 0 \).[/tex]

Finally, we can find the coordinates of the center of mass [tex]\( (\bar{x}, \bar{y}) \):[/tex]

[tex]\[ \bar{x} = \frac{M_y}{M} = \frac{0}{6k \pi} = 0 \]\[ \bar{y} = \frac{M_x}{M} = \frac{0}{6k \pi} = 0 \][/tex]

So, the center of mass of the lamina is at the origin (0, 0) .

Answer:

a) Type 1 Error: 1.5%

b) Type 2 Error: 5.5%

Step-by-step explanation:

Probability of positive reaction when infact the person has disease = 94.5%

This means, the probability of negative reaction when infact the person has disease = 100- 94.5% = 5.5%

Probability of positive reaction when the person does not have the disease = 1.5%

This means,

Probability of negative reaction when the person does not have disease = 100% - 1.5% = 98.5%

Our Null Hypothesis is:

"The individuals does not have the disease"

Part a) Probability of Type 1 Error:

Type 1 error is defined as: Rejecting the null hypothesis when infact it is true. Therefore, in this case the Type 1 error will be:

Saying that the individual have the disease(positive reaction) when infact the individual does not have the disease. This means giving a positive reaction when the person does not have the disease.

From the above data, we can see that the probability of this event is 1.5%. Therefore, the probability of Type 1 error is 1.5%

Part b) Probability of Type 2 Error:

Type 2 error is defined as: Accepting the null hypothesis when infact it is false. Therefore, in this case the Type 2 error will be:

Saying that the individual does not have the disease(negative reaction) when infact the individual have the disease.

From the above data we can see that the probability of this event is 5.5%. Therefore, the probability of Type 2 error is 5.5%

Answer:

0.0668

Step-by-step explanation:

Assuming the distribution is normally distributed with a mean of $75,

with a standard deviation of $12.

We can find the z-score of 78 using;

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

[tex]\implies z=\frac{78-75}{\frac{12}{36} } =1.5[/tex]

Using our normal distribution table, we obtain the area that corresponds to 0.25 to be 0.9332

This is the area corresponding to the probability that, the average is less or equal to 78.

Subtract from 1 to get the complement.

P(x>78)=1-0.9332=0.0668

The probability that the average sales per customer from a sample of 36 customers, taken at random from this population, exceeds $78 is 0.0668.

Since The average sales per customer at a home improvement store during the past year is $75 with a standard deviation of $12.

Here  we need to find out the z score

= [tex]78-75\div 12\div 36[/tex]

= 1.5

Here we considered normal distribution table, we obtain the area that corresponds to 0.25 to be 0.9332

So,  the average is less or equal to 78.

Now

Subtract from 1 to get the complement.

So,

P(x>78)=1-0.9332

=0.0668

Learn more about probability here: https://brainly.com/question/24613748

Answer:

$15.6

Step-by-step explanation:

one bag= $1.95

8 bags = 8×1.95=15.6

so answer is $15.6.

Answer:

840

Step-by-step explanation:

the volume of a rectangular prism is the base(width)height

Answer:

The old version awards 50 more points for each level

Step-by-step explanation:

Function Modeling

Two models are being compared: The old model relates the total number of points awarded in a computer game (y) with the number of levels completed (x) as

[tex]y=175x+150[/tex]

The new model is given as a graph. To find the equation of the line we must locate a couple of 'good' points. The graph is very clear, so we select the extreme points (1,250) (9,1250)

Find the equation of the line with the point-point formula

[tex]\displaystyle y-250=\frac{1250-250}{9-1}(x-1)[/tex]

[tex]\displaystyle y-250=\frac{1000}{8}(x-1)[/tex]

[tex]y=250+125x-125[/tex]

[tex]y=125x+125[/tex]

Comparing the new function with the old function we can note the coefficient of x (the slope of the line) is 50 points more in the old version than the points in the new version.

Thus the answer is

The old version awards 50 more points for each level

Answer: the answer is B

Step-by-step explanation:

Answer: b. A 90% confidence level and a sample of 24 observations. 1.714 1.714 Correct

Step-by-step explanation:

Without sufficient data, we cannot calculate the probabilities or the cost for the 3% highest domestic airfares. The calculations require details about the total number of observed airfares and the number that falls in the specified price ranges.

I regret to inform you that I cannot provide a factual answer to your question regarding domestic airfares as you have not provided sufficient data. Probability depends on the sample space and given conditions which are not provided in your question. For instance:

https://brainly.com/question/22962752

#SPJ11

Answer:

The median of the following set of data is 9 since the question is implying, which is the center of the data distribution.

Step-by-step explanation:

Answer:

m=2/3

Step-by-step explanation:

your answer would be A.

Answer: It would make the meal $40.00 and with tax it would be $42.80

Step-by-step explanation: 50 - 10 = 40

                                              40 / 0.07 = 2.8

                                              40 + 2.8 = 42.8

Answer:

Step-by-step explanation:

Answer:

50

Step-by-step explanation:

add 5 to -25 to get -2/5x by itself.

Answer:

Step-by-step explanation:

Table(2) shows the relative frequency opted from the table(1).

It is defined as the number of waves that crosses a fixed point in one second known as frequency. The unit of frequency is per second.

We have a table in which data has shown:

To find the relative frequency:

For the first cell:

[tex]=\frac{67}{92} \times100 \approx 73\%[/tex]

For the second cell:

= 100 - 73 ⇒ 27%

For the third cell:  

[tex]\rm = \frac{46}{66} \times 100 \approx 70\%[/tex]

For the fourth cell:

= 100 - 70 = 30%

Thus, table(2) shows the relative frequency opted from table(1).

Learn more about the frequency here:

brainly.com/question/27063800

Answer:

28.57% probability that the oldest is a girl and the youngest is a boy​, given that there is at least one boy and at least one girl.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

These are all the possible outcomes: from youngest to oldest, b is boy and g is girl

b - b - b - g

b - b - g - b

b - b - g - g

b - g - b - b

b - g - b - g

b - g - g - b

b - g - g - g

g - b - b - b

g - b - b - g

g - b - g - b

g - b - g - g

g - g - b - b

g - g - b - g

g - g - g - b

The nuber of total outcomes is 14.

Desired outcomes:

Oldest(last) girl, youngest(first) boy

b - b - b - g

b - b - g - g

b - g - b - g

b - g - g - g

So 4 desired outcomes

Probability:

4/14 = 0.2857

28.57% probability that the oldest is a girl and the youngest is a boy​, given that there is at least one boy and at least one girl.

Final answer:

The probability that a family with four children will have the oldest being a girl and the youngest being a boy, given that there is at least one boy and one girl, is 1/4.

Explanation:

To solve this probability question, we need to consider all possible combinations of children while adhering to the given conditions: the oldest child must be a girl (G), the youngest must be a boy (B), and there must be at least one boy and at least one girl in the family.

The possible genders for the four children can be represented as a sequence of G (girl) and B (boy) like this: G---B. There are two positions in the middle that can be either a boy or a girl. Since each position can be filled independently with either a boy or a girl, there are 2 options for each of the middle children, giving us 2 x 2 = 4 combinations: GBGB, GBBB, GGBB, GGBG.

To calculate the probability of any single one of these combinations occurring, we need to remember that the probability of giving birth to a boy or a girl is equal, which means each event (birth of a child) has a probability of 1/2. Thus, the probability of each combination is (1/2)^4 since there are four independent events (births). However, since we have 4 combinations that meet the criteria, we multiply this probability by 4. So, the probability is 4 * (1/2)^4 = 1/4.

Therefore, the probability that a family with four children will have the oldest being a girl and the youngest being a boy, given that there is at least one boy and at least one girl, is 1/4.

Answer:

Answer : D

Step-by-step explanation:

A school has two kindergarten classes. There are 21 children in Ms. Toodle's kindergarten class. Of these, 17 are "pre-readers" children on the verge of reading. There are 19 children in Mr. Grimace's kindergarten class. Of these, 13 are pre-readers. Using the plus four confidence interval method, a 90% confidence interval for the difference in proportions of children in these classes that are pre-readers is â0.104 to 0.336.  

Which of the following statements is correct?

A) This confidence interval is not reliable because the samples are so small.  

B)This confidence interval is of no use because it contains 0, the value of no difference between classes.  

C)This confidence interval is reasonable because the sample sizes are both at least 5.  

D) This confidence interval is not reliable because these samples cannot be viewed as simple random samples taken from a larger population.

The Answer is D - This confidence interval is not reliable because these samples cannot be viewed as simple random samples taken from a larger population.

In this setup, all the students are already involved in the data. This is not a sample from a larger population, but probably, the population itself.