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4 years ago

The radii of the planets in the solar system are given. Arrange the planets in increasing order by the lengths of their radii.

Mathematics

1 answer:

kakasveta [241]4 years ago

4 0

Answer:

answer below. :)

Step-by-step explanation:

mercury

neptune

uranus

mars

saturn

venus

earth

jupiter

hope this helps plz give brainliest :(

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PLEASE HELP!!!! The question is in the picture.

oee [108]

2 and 3 are both 58 as well. For 4 do 180-58 because a straight line is 180 degrees.

8 0

4 years ago

Vector performed an experiment at a local farm to determine if the addition of bananas to his biomass cow manure recipes helps p

evablogger [386]

Answer:

About 75% of the time, the mixture of cow manure and mashed bananas IS LIKELY to produce MORE biogas than pure cow manure on THIS local farm.

Iready 100%

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3 years ago

What is the solution 9s =252

raketka [301]

All you need to do is find s, and get it by itself.

So, divide 9 by 252.

252/9= 28

s= 28

I hope this helps!
~kaikers

6 0

3 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago

A straw is placed inside a rectangular box that is 9 inches by 5 inches by 4 inches, as shown. If the straw fits exactly into th

IrinaK [193]

Answer:

9÷5by4 =965

Step-by-step explanation:

storm down Williams uojdg Ellen Soleil

4 0

3 years ago

Read 2 more answers