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Lelu
1 month ago
11

what is the typical distance a moped can be driven on a single tank of gas? a statistical question/ explain your reasoning

Mathematics
2 answers:
Zina [12.3K]1 month ago
8 0

Response:

Step-by-step reasoning:

the bfjfnd hat nsth dum

AnnZ [12.3K]1 month ago
5 0

Response:

This inquiry is indeed statistical.

Step-by-step reasoning:

A statistical question allows for an answer through the collection of variable data,

This question is statistical since it can yield various answers, influenced by multiple factors.

A single integer value cannot answer this question definitively.

You might be interested in
A rectangular garden is 6 feet long and 4 feet wide. A second rectangular garden has dimensions that are double the dimensions o
Svet_ta [12734]

Response:

100 percent increase

Detailed explanation:

1st garden

Length = 6 ft

Width = 4 ft

Perimeter = 2 (l+w)

                 = 2 (6+4) = 2(10) = 20

2nd garden

The length and width of this garden are double those of the first

Length = 2 *6 = 12

Width = 2 *4 = 8

Perimeter = 2 (l+w)

                 = 2 (12+8) = 2(20) = 40

Percent change = (new - old )/old * 100 percent

The first garden represents the old garden = 20  and the second refers to the new garden = 40

Substituting values in

Percent change = (40-20)/20 = 20/20 =100 *100 percent

                            = 100 percent increase

5 0
18 days ago
Including Jose, there are eight people in his family.
lawyer [12517]

Answer:

840

Step-by-step explanation:

As the arrangement matters, we apply the permutations formula to find the solution.

Permutations formula:

The count of possible arrangements of x items chosen from a total of n items is defined by this formula:

P_{(n,x)} = \frac{n!}{(n-x)!}

For this problem:

Jose occupies the first seat.

The other four can be arranged among the remaining 7 family members. Thus

P_{(7,4)} = \frac{7!}{(7-4)!} = 840

Hence, the final answer is:

840

7 0
1 month ago
Read 2 more answers
Determine the value of base x if (211)x = (6A)16
AnnZ [12381]
This equation can be solved with two solutions, corresponding to x=7 and x=-\dfrac{15}2.
4 0
29 days ago
Tim was given a large bag of sweets and ate one third of the sweets before stopping as he was feeling sick. The next day he ate
Leona [12618]

Answer: Initially, he had 27 sweets.

Step-by-step explanation: The most logical approach is to work backwards from what remained after the third day to the start of the first day.

On the third day, he consumed one-third of his sweets and was left with 8. If we let the total sweets on day three be denoted as a, then one-third of a equals what he ate and the two-thirds left equals 8, giving us:

8/a = 2/3

By cross-multiplying, we find:

8 x 3 = 2a

Therefore, 24 = 2a

This leads to a = 12.

Let the sweets on day two be represented as b. If he consumed one-third of b and was left with 12, we have the same structure; hence:

12/b = 2/3

Cross-multiplying gives:

12 x 3 = 2b

So, 36 = 2b, leading to b = 18.

Denote the number of sweets on day one as x. If one-third of x was eaten and 18 remained, we can set up the equation:

18/x = 2/3

Again, cross-multiplying results in:

18 x 3 = 2x

Which simplifies to 54 = 2x, yielding x = 27.

Thus, Tim received 27 sweets at the start.

3 0
1 month ago
The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th
AnnZ [12381]

Respuesta:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Explicación paso a paso:

La probabilidad (P) de encontrar la partícula está dada por:

P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx  

P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx     (1)

La solución de la integral de la ecuación (1) es:

P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}  

(a) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}    

(b) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}  

(c) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}    

(d) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5

(e) La probabilidad de encontrar la partícula en el tercio central de la caja, es decir, entre x = 0 nm y 100/6 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}

Espero que te ayude.

3 0
25 days ago
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