All categories

2 years ago

The number of boots that 25 students had in their homes in Florida were recorded

Mathematics

1 answer:

wolverine [178]2 years ago

7 0

Answer:

Min = 0; Q₁ = 0; Median = 2; Q₃ = 2; Max = 8

Step-by-step explanation:

Assume that your sorted data are

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 5, 8

1. Minimum

Min = 0

2. Median

The median is the middle value in a sorted data set. There are 25 items, so the median is the 13th item.

Median = 2

3. First quartile

Q₁ is the median of the lower half of the data set.

The median of 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2 is 0.

Q₁ = 0

4. Third quartile

Q₃ is the median of the upper half of the data set.

The median of 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 5, 8 is 3.

Q₃ = 3

5. Maximum

Max = 8

You might be interested in

An automobile producing company operates car dealerships that sell its cars. In order to assess the service costumers get at the

harina [27]

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to assess the service customers receive at the dealerships.

1) The population in this survey is:

The population is the customers of car dealerships.

2) The sample in this survey is:

The random sample taken was 20 selected car dealerships, where the customers were surveyed.

3) A parameter that may be targeted by the survey is:

To study the service the customers receive at the dealerships and if it has any influence in the customers owning or not a car of the company, one parameter to study is the population proportion of customers that own a car of the company vs the proportion of customers that are satisfied with the service received.

4) A statistic that may be used to summarize the outcome of the survey is:

The statistic to use to summarize the outcome of the survey is the sample proportion.

I hope it helps!

4 0

2 years ago

Pablo built a large wooden storage box. The box was in the shape of a rectangular prism, as shown below. He covered all the side

Naily [24]

The cost of wallpaper used to cover all sides of the storage box is given by $3 per square foot

<h3>Cost of wallpaper per square foot</h3>

length = 9 ft
Width = 8 ft
Height = 5 ft

Surface area of a box = 2(lh + wh + lw)

= 2(9×5 + 8×5 + 9×8)

= 2(45 + 40 + 72)

= 2(157)

= 314 square foot

Total cost of wallpaper = $942

Cost of wallpaper per square foot = $942 ÷ 314 square foot

= $3 per square foot

Learn more about surface area:

brainly.com/question/2500827

#SPJ1

3 0

2 years ago

The projectile motion of an object can be modeled using s(t) = gt2 + v0t + s0, where g is the acceleration due to gravity, t is

WARRIOR [948]

Answer:

s(t) = -4.9t² + 39.2t

Step-by-step explanation:

Given the projectile motion of an object can be modeled using s(t) = gt2 + v0t + s0, where g is the acceleration due to gravity, t is the time in seconds since launch, s(t) is the height after t seconds, v0 is the initial velocity, and s0 is the initial height. The acceleration due to gravity is –4.9 m/s²

Given

g = -4.9m/s²

v0 = 39.2m/s

s0 = 0m (the initial height)

On substituting into the formula;

s(t) = gt² + v0t + s0

s(t) = -4.9t² + 39.2t + 0

s(t) = -4.9t² + 39.2t

This gives the equation that models the height

4 0

1 year ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

2 years ago

Please solve this...I can;t do it do it

yanalaym [24]

Answer:

Area = 867 meters

Step-by-step explanation:

You have a rectangle. This means opposite sides are equal. If you set these expressions equal to each other, you will get two equations with two unknowns, x and y. Next, solve the system of equations to find x and y. Then, use the values to find the length and width of the rectangle. Last, multiply length × width to find the area.

8 0

2 years ago