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2 years ago

8

A little girl kicks a soccer ball. It goes 10 feet and comes back to her. How is this possible?

1 answer:

Elanso [62]2 years ago

8 0

She is standing in front of a wall which is 10 feet away when she kicks the ball and it hits the wall and comes back to her by the virtue of Newton's third Law of motion! She kicks the ball into the air vertically above her and the ball rises to 10 feet and then comes back to her by the virtue of gravity!

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4. How many real-number solutions does the equation have?

erma4kov [3.2K]

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4 years ago

What is the ratio of 6:21?

VMariaS [17]

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3 years ago

1. Flight 202's arrival time is normally distributed with a mean arrival time of 4:30

Bezzdna [24]

Answer:

Step-by-step explanation:

1) Let the random time variable, X = 45min; mean, ∪ = 30min; standard deviation, α = 15min

By comparing P(0 ≤ Z ≤ 30)

P(Z ≤ X - ∪/α) = P(Z ≤ 45 - 30/15) = P( Z ≤ 1)

Using Table

P(0 ≤ Z ≤ 1) = 0.3413

P(Z > 1) = (0.5 - 0.3413) = 0.1537

∴ P(Z > 45) = 0.1537

2) By compering (0 ≤ Z ≤ 15) ( that is 4:15pm)

P(Z ≤ 15 - 30/15) = P(Z ≤ -1)

Using Table

P(-1 ≤ Z ≤ 0) = 0.3413

P(Z < 1) = (0.5 - 0.3413) = 0.1587

∴ P(Z < 15) = 0.1587

3) By comparing P(0 ≤ Z ≤ 60) (that is for 5:00pm)

P(Z ≤ 60 - 30/15) = P(Z ≤ 2)

Using Table

P(0 ≤ Z ≤ 1) = 0.4772

P(Z > 1) = (0.5 - 0.4772) = 0.0228

∴ P(Z > 60) = 0.0228

6 0

3 years ago

Geometry help!! Find x and y.

MrRa [10]

The last answer is correct
x= 45 degrees
y= 10.9sqrt2

6 0

3 years ago

Read 2 more answers

Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno

Roman55 [17]

Answer:

a) P(X>np+3 $\sqrt{np(1-p)}$ =0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= $\sqrt{0.198}$ ≈ 0.445

a) P(X>np+3 $\sqrt{np(1-p)}$ =P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

$\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}$

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - $\frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}$ - $\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}$

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - $\frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)}$ - $\frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}$

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)= $\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}$ =0.349

P(Y>1)=1-0.349=0.651

3 0

3 years ago