The ratio of cats to dogs at the shelter is 3 to 2 if there are 15 cats at the shelter how many dogs are there

Answers

Answer 1
Answer:

Answer:

3/2=15/x

Step-by-step explanation:

Solve that and you’ll get x, which is your answer


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The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow. Click on the datafile logo to reference the data. 6 4 6 8 7 7 6 3 3 8 10 4 8 7 8 7 5 9 5 8 4 3 8 5 5 4 4 4 8 4 5 6 2 5 9 9 8 4 8 9 9 5 9 7 8 3 10 8 9 6 Develop a 95% confidence interval estimate of the population mean rating for Miami. If required, round your answers to two decimal places. Do not round intermediate calculations.

Solve for brainliest

Answers

Answer:

D

but not sure because I'm not that good

A survey of 300 parks showed the following. 15 had only camping. 20 had only hiking trails. 35 had only picnicking. 185 had camping. 140 had camping and hiking trails. 125 had camping and picnicking. 210 had hiking trails. Determine the number of parks that:

a. Had at least one of these features.
b. Had all three features.
c. Did not have any of these features.
d. Had exactly two of these features.

Answers

Answer:

A. 290 parks dad at least one of these features.

b. 95 parks had all three features.

c. 10 parks did not have any of these features.

d. 125 parks had exactly two of these features.

Step-by-step explanation:

This will be solved using set notation according to the venn diagram attached.

Let n(U) be the total number of parks surveyed

n(C) be those that had camping = 185

n(H) be those that had hiking trails = 210

n(C∩H) be those that had camping and hiking trails = 140

n(C∩P) be those that had camping and picnicking = 125

n(C∩P'∩H') be those that had only camping = 15

n(C'∩P'∩H) be those that had only hiking trails = 20

n(C'∩P∩H') be those that had only picnicking = 35

Find the calculation in the attached file

Final answer:

The number of parks that had at least one of the listed features was 135.

The number of parks that had all three features was 20.

The number of parks that did not have any of these features was 165.

Explanation:

To determine the number of parks that had at least one of the listed features, we can add up the numbers of parks that had only camping, only hiking trails, and only picnicking. Then we subtract the parks that had two or three of these features, as they were already counted in the previous step. Doing this calculation, we get:

  1. Parks with at least one feature: 15 + 20 + 35 + (185 - 140 - 125) + (140 - 125) + (210 - 140 - 125) = 135

To find the number of parks that had all three features, we need to subtract the parks that had only camping, only hiking trails, only picnicking, or none of these features from the total number of parks (300). Doing this calculation, we get:

  1. Parks with all three features: 300 - 15 - 20 - 35 - (185 - 140 - 125) - (140 - 125) - (210 - 140 - 125) - (300 - 135) = 20

To determine the number of parks that did not have any of these features, we subtract the parks with at least one feature from the total number of parks (300). Doing this calculation, we get:

  1. Parks with no features: 300 - 135 = 165

To calculate the number of parks that had exactly two features, we add the intersections of each pair of features and subtract the parks that had all three features. Doing this calculation, we get:

  1. Parks with exactly two features: (185 - 140 - 125) + (140 - 125) + (210 - 140 - 125) - (300 - 20) = 60

Learn more about counting here:

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define the null and alternative null hypothesis for the following. also, explain what it would mean to make a type 1 error and explain what it would mean to make a type 2 error. the newspaper in a certain city had a circulation of 15,000 per day in 2010. you believe that the newspaper circulation is more than 15,000 today

Answers

Answer:

(A)

Null hypothesis: Newspaper circulation in the city per day was = 15,000 in 2010

(B)

Alternative hypothesis: Newspaper circulation in the city today is > 15,000

(C)

Type 1 Error: This is the rejection of a true null hypothesis. It is the acceptance of the alternative hypothesis when the null hypothesis is true.

(D)

Type 2 Error: This is the non-rejection of a false null hypothesis. It is the acceptance of a null hypothesis when it is false.

Step-by-step explanation:

In statistical theory, the complete absence of any of these errors is virtually impossible.

Can someone help me with isosceles triangles???

Answers

Answer:

1) x = 75°

2) x = 180° - 2×40° = 180° - 80° = 100°

3) x = 180° - 2×73° = 180° - 146° = 34°

4) x = (180° - 122°) : 2 = 58° : 2 = 29°

5) x = 90° - (180° - 80°) : 2 = 90° - 100° : 2 = 90° - 50° = 40°

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 4 right angle0,−4 in the directions parallel to and normal to the plane that makes an angle of StartFraction pi Over 3 EndFraction π 3 with the positive​ x-axis. Show that the total force is the sum of the two component forces.

Answers

Answer:

F_p = < - √(3) , -3 >\n\nF_o = < √(3) , -1 >

Step-by-step explanation:

- A plane is oriented in a Cartesian coordinate system such that it makes an angle of ( π / 3 ) with the positive x - axis.

- A force ( F ) is directed along the y-axis as a vector < 0 , - 4 >

- We are to determine the the components of force ( F ) parallel and normal to the defined plane.

- We will denote two unit vectors: ( u_p ) parallel to plane and ( u_o ) orthogonal to the defined plane. We will define the two unit vectors in ( x - y ) plane as follows:

- The unit vector ( u_p ) parallel to the defined plane makes an angle of ( 30° ) with the positive y-axis and an angle of ( π / 3 = 60° ) with the x-axis. We will find the projection of the vector onto the x and y axes as follows:

                         u_o = < cos ( 60° ) , cos ( 30° ) >

                         u_o = < (1)/(2) ,  (√(3) )/(2) >

- Similarly, the unit vector ( u_o ) orthogonal to plane makes an angle of ( π / 3 ) with the positive x - axis and angle of ( π / 6 ) with the y-axis in negative direction. We will find the projection of the vector onto the x and y axes as follows:

                        u_p = < cos ( (\pi )/(6)  ) , - cos ( (\pi )/(3) ) >\n\nu_p = < (√(3) )/(2)  , -(1)/(2)  >\n

- To find the projection of force ( F ) along and normal to the plane we will apply the dot product formulation:

- The Force vector parallel to the plane ( F_p ) would be:

                          F_p = u_p(F . u_p)\n\nF_p = < (1)/(2) , (√(3) )/(2) > [  < 0 , - 4 > . < (1)/(2) , (√(3) )/(2) > ]\n\nF_p = < (1)/(2) , (√(3) )/(2) > [ -2√(3)  ]\n\nF_p = < -√(3)  , -3 >\n

- Similarly, to find the projection of force ( F_o ) normal to the plane we again employ the dot product formulation with normal unit vector (  u_o  ) as follows:

                         F_o = u_o ( F . u_o )\n\nF_o = < (√(3) )/(2) , - (1)/(2) > [ < 0 , - 4 > . < (√(3) )/(2) , - (1)/(2) > ] \n\nF_o = < (√(3) )/(2) , - (1)/(2) > [ 2 ] \n\nF_o = < √(3) , - 1 >

- To prove that the projected forces ( F_o ) and ( F_p ) are correct we will apply the vector summation of the two orthogonal vector which must equal to the original vector < 0 , - 4 >

                       F = F_o + F_p\n\n< 0 , - 4 > = < √(3), -1 > + < -√(3), -3 >  \n\n< 0 , - 4 > = < √(3) - √(3) , -1 - 3 > \n\n< 0 , - 4 > = < 0 , - 4 >  .. proven                    

What is 62 divided by 17 plus 29


I will mark brainliest

Answers

62/17=3.647059+29= 32.64059 but if you round just say 33!

Answer:

32.647

Step-by-step explanation: