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3 years ago

30 Points Use Scientific Notation Put Greater Than Less Than Or Equal To In Each Box!

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

3 0

Answer:

5. >

6. >

7. <

8. <

9. =

10. >

11. >

12. >

Step-by-step explanation:

5. 9.33 x 10³ = 9330

9330 > 933

6. 2.88 x 10⁻⁵ = 0.0000288

2.9 x 10⁻⁶ = 0.0000029

0.0000288 > 0.0000029

7. 0.33 x 10⁻⁷ = 0.000000033

3.7 x 10⁻⁶ = 0.0000037

0.000000033 < 0.0000037

8. 4.9 x 10⁴ = 49000

4.79 x 10⁵ = 479000

49000 < 479000

9. 3.0 x 10⁶ = 3000000

30 x 10⁵ = 3000000

3000000 = 3000000

10. 2.2 x 10⁻³ = 0.0022

1.7 x 10⁻³ = 0.0017

0.0022 > 0.0017

11. 2.1 x 10⁷ = 21000000

2 million = 2000000

21000000 > 2000000

12. 5.2 x 10¹¹ = 520000000000

8.7 x 10¹⁰ = 87000000000

520000000000 > 87000000000

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How to solve for y??????

tia_tia [17]

What he said above yeah y

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3 years ago

At 5:00 a.M., a parking lot had 16 cars already parked in it. Throughout the course of the day, 48 more cars entered and parked

Alekssandra [29.7K]

Answer:

The situation is that; the parking lot has 64 cars at the end of the day

Step-by-step explanation:

Given

Initial Number of cars = 16

Additional Cars = 48

Required

Determine the situation for the day

From the question, we understand that no car left the park;

This implies that

$Number\ of\ cars = Initial\ Cars + Additional\ Cars$

$Number\ of\ cars = 16 + 48$

$Number\ of\ cars = 64$

So, the situation is that; the parking lot has 64 cars at the end of the day

5 0

3 years ago

Alison is learning how to walk. 50% of the time that she takes a step she falls down. Her older brother wants to design a simula

MissTica

Answer with explanation:

Alison is learning how to walk.50% of times she takes a step she falls down.

Probability of falling down in next step $=\frac{50}{100}=\frac{1}{2}$

Probability of not falling down in next step $=\frac{50}{100}=\frac{1}{2}$

From ,the four option given,we have to find the best simulation which her older brother wants to design a simulation to find the probability that she will fall down 5 out of the next 7 times she tries to take step.

While Simulating the Probability of falling down should be $=\frac{5}{7}$

And , probability of not falling down $1-\frac{5}{7}=\frac{2}{7}$

→→⇒F=Falling, NF=Non Falling

A) Roll a die letting 1 represent Alison falling and 2-6 represent Alison not falling. Roll the die five times.

$P(F)=\frac{1}{6},P(NF)=\frac{5}{6}$

Incorrect option,probabilities of falling and non falling does not match with Actual probability of falling and non falling.

B) Roll a die letting 1 represent Alison falling and 2-6 represent Alison not falling. Roll the die seven times.

$P(F)=\frac{1}{6},P(NF)=\frac{5}{6}$

Incorrect option,probabilities of falling and non falling does not match with Actual probability of falling and non falling.

C) Let Heads represent Alison falling down and Tails representing Alison not falling down. Flip the coin five times.

$P(F)=\frac{1}{2},P(NF)=\frac{1}{2}$

Simulation is done only five times , that is coin has been flipped only five times.it must have been flipped seven times.Incorrect option.

D) Let Heads represent Alison falling down and Tails representing Alison not falling down. Flip the coin seven times.

$P(F)=\frac{1}{2},P(NF)=\frac{1}{2}$

Simulation is done seven times , that is coin has been flipped seven times.So,she should fall 5 times that is head should appear 5 out of seven times, and tail should appear 2 out of seven times.correct option.

Option D

4 0

4 years ago

Read 2 more answers

Simplify Fully, then evaluate: x =2 a) (3x^2)(x^3)(x)

AfilCa [17]

Answer:

3(2)^2(2^3)(2)

(12)(8)(2)

(96)(2)= 192

8 0

3 years ago

An urn contains three white balls and two red balls. The balls are drawn from the urn, oneat a time without replacement, until a

Tpy6a [65]

Answer:

p ( X = 1 ) = 0.6 , p ( X = 2 ) = 0.3 , p ( X = 3 ) = 0.1

Verified

E ( X ) = 1.5

Step-by-step explanation:

Solution:-

- An urn contains the following colored balls:

Color Number of balls

White 3

Red 2

- A ball is drawn from urn without replacement until a white ball is drawn for the first time.

- We will construct cases to determine the distribution of the random-variable X: The number of trials it takes to get the first white ball.

- We have three following case:

1) White ball is drawn on the first attempt ( X = 1 ). The probability of drawing a white ball in the first trial would be:

p ( X = 1 ) = ( Number of white balls ) / ( Total number of ball )

p ( X = 1 ) = ( 3 ) / ( 5 )

2) A red ball is drawn on the first draw and a white ball is drawn on the second trial ( X = 2 ). The probability of drawing a red ball first would be:

p ( Red on first trial ) = ( Number of red balls ) / ( Total number of balls )

p ( Red on first trial ) = ( 2 ) / ( 5 )

- Then draw a white ball from a total of 4 balls left in the urn ( remember without replacement ).

p ( White on second trial ) = ( Number of white balls ) / ( number of balls left )

p ( White on second trial ) = ( 3 ) / ( 4 )

- Then to draw red on first trial and white ball on second trial we can express:

p ( X = 2 ) = p ( Red on first trial ) * p ( White on second trial )

p ( X = 2 ) = ( 2 / 5 ) * ( 3 / 4 )

p ( X = 2 ) = ( 3 / 10 )

3) A red ball is drawn on the first draw and second draw and then a white ball is drawn on the third trial ( X = 3 ). The probability of drawing a red ball first would be ( 2 / 5 ). Then we are left with 4 balls in the urn, we again draw a red ball:

p ( Red on second trial ) = ( Number of red balls ) / ( number of balls left )

p ( Red on second trial ) = ( 1 ) / ( 4 )

- Then draw a white ball from a total of 3 balls left in the urn ( remember without replacement ).

p ( White on 3rd trial ) = ( Number of white balls ) / ( number of balls left )

p ( White on 3rd trial ) = ( 3 ) / ( 3 ) = 1

- Then to draw red on first two trials and white ball on third trial we can express:

p ( X = 3 ) = p ( Red on 1st trial )*p ( Red on 2nd trial )*p ( White on 3rd trial )

p ( X = 3 ) = ( 2 / 5 ) * ( 1 / 4 ) * 1

p ( X = 3 ) = ( 1 / 10 )

- The probability distribution of X is as follows:

X 1 2 3

p ( X ) 0.6 0.3 0.1

- To verify the above the distribution. We will sum all the probabilities for all outcomes ( X = 1 , 2 , 3 ) must be equal to 1.

∑ p ( Xi ) = 0.6 + 0.3 + 0.1

= 1 ( proven it is indeed a pmf )

- The expected value E ( X ) of the distribution i.e the expected number of trials until we draw a white ball for the first time:

E ( X ) = ∑ [ p ( Xi ) * Xi ]

E ( X ) = ( 1 ) * ( 0.6 ) + ( 2 ) * ( 0.3 ) + ( 3 ) * ( 0.1 )

E ( X ) = 0.6 + 0.6 + 0.3

E ( X ) = 1.5 trials until first white ball is drawn.

8 0

4 years ago