Answer and Step-by-step explanation: With the constant velocity motion formula, we can determine constant velocity of the object in motion whose data we collected:

x = x₀ + vt

Velocity can be calculated as:

$v=\frac{x-x_{0}}{t}$

$v=\frac{70-40}{10}$

v = 3 m/s

The beginning of the data collect, object is 40m away, then x₀ = 40.

So, equation modeling the object's path is x = 40 + 3t.

5 0

2 years ago

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What

galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

$P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

$P(D_i)=\frac{1}{6}$ for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

$P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}$

$P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}$

$P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}$

$P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}$

$P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}$

$P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0$

Part A:

The probability that all of the balls selected are white:

$P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)$

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find $P(D_3|A)$

The data required is calculated above:

$P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

7 0

2 years ago

Graph the first six terms of a sequence where a1 = −10 and d = 3.

Neko [114]

notice the first term is -10, and the common difference is 3, so we add 3 to get the next term.

$\bf \begin{array}{cccll}\stackrel{x}{n}&term&\stackrel{y}{value}\\\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\1&a_1&-10\\\\2&a_2&\stackrel{-10+3}{-7}\\\\3&a_3&\stackrel{-7+3}{-4}\\\\4&a_4&\stackrel{-4+3}{-1}\\\\5&a_5&\stackrel{-1+3}{2}\\\\6&a_6&\stackrel{2+3}{5}\end{array}$

check the picture below.

8 0

2 years ago

What's y=-4+15 in standard form

kotykmax [81]

Answer:

y=1.1×10¹

Step-by-step explanation:

y=-4+15

y=11

y=1.1×10¹

5 0

2 years ago