All categories

2 years ago

Help brainlist to first awnser

Mathematics

2 answers:

pishuonlain [190]2 years ago

8 0

Answer:

the answer is A

since the ball had the most energey happening to it right then

Hope This Helps!!!

Digiron [165]2 years ago

8 0

Answer:

the answer is " right before you catch it after it falls back down"

You might be interested in

The rate at which Pete’s Bakery can make pies is shown in the graph.

Mashutka [201]

I think the answer would be 17 pies

5 0

3 years ago

Read 2 more answers

What is the next number in the series 4 12 6 9 8 6

NNADVOKAT [17]

4,6,8 are being counted by twos so the next one is 10
4•3= 12, 6+3=9, 8-3=6, the next operation I assume would be division so 10/3= 3.34

8 0

3 years ago

Read 2 more answers

What is 1/3 plus 1/6?

Leya [2.2K]

The decimal is 0.0555555

4 0

3 years ago

Read 2 more answers

ILL GIVE BRAINLIESIT HELP

krok68 [10]

D is your answer

Have a good day :)

5 0

2 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

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

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

3 years ago