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torisob [31]
3 years ago
5

The length of a 1-inch paperclip is about 1.6 x 10−5 miles. It takes about 1.5 x 1010 paperclips linked together to reach the mo

on. What is the approximate distance to the moon?
Mathematics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

Distance from the moon is 2.4 × 10⁵ miles

Step-by-step explanation:

Length of 1-inch paperclip = 1.6\times 10^{-5} miles

Number of paperclips used to reach the moon = 1.5 × 10¹⁰

Total distance from the moon = Number of paper clips used × Length of one clip

= (1.5 × 10¹⁰) × (1.6 × 10⁻5)

= (1.5 × 1.6) (10¹⁰× 10⁻⁵)

= 2.4 × 10⁽¹⁰⁻⁵⁾ miles

= 2.4 × 10⁵ miles

Distance from the moon is 2.4 × 10⁵ miles.

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Find the probability of rolling an odd sum or a sum less than 7 when a pair of dice is rolled
Nutka1998 [239]

Answer:

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

Step-by-step explanation:

For this case when a pair of dice is rolled we have the following sample pace for the outcomes:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

We see 36 possible outcomes and we want to find how many of these pairs we got rolling an odd sum or a sum less than 7

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818

7 0
3 years ago
Every week, a car dealer ship sells at least 5 minivans. Write an inequality that represents this situation. Let c represent the
kvv77 [185]

Answer:

we conclude that an inequality 'c ≥ 5' denotes this situation.

Step-by-step explanation:

Given

  • A car dealer ship sells at least 5 minivans

When we talk about 'at least', it means we are talking about '≥' in terms of representing the 'at least' in inequality symbol.

For instance,

'm≥n' means 'm' is greater than or equal to 'n'.

It means 'm' is at least equal to 'n'.

Coming back to the question,

  • Let 'c' represent the number of minivans sold each week.

As the car dealer ship sells at least 5 minivans. so the

inequality will be: c ≥ 5

Thus, we conclude that an inequality 'c ≥ 5' denotes this situation.

8 0
3 years ago
in an experiment, the temperature dropped 2 degrees each day until the temperature reaches negative 5.5 degrees. if the starting
WITCHER [35]
32.9-(-5.5)=38.4
38.4/2=19.2
around 19 days
3 0
3 years ago
Which choice is equivalent to the product below when x>0?
V125BC [204]

Answer:

D

Step-by-step explanation:

Using the rule of radicals

\sqrt{\frac{a}{b} } = \frac{\sqrt{a} }{\sqrt{b} }

\sqrt{a\\} × \sqrt{b} ⇔ \sqrt{ab}

Given

\sqrt{\frac{6}{x} } × \sqrt{\frac{x^2}{24} }

= \frac{\sqrt{6} }{\sqrt{x} } × \frac{x^2}{24}

= \frac{\sqrt{6} }{\sqrt{x} } × \frac{x}{2 \sqrt{6\\} }

Cancel \sqrt{6} on numerator/ denominator

= \frac{1}{\sqrt{x} } × \frac{x}{2\\}

= \frac{1}{\sqrt{x} } × \frac{(\sqrt{x})^2 }{2}

Cancel \sqrt{x} on numerator/ denominator, leaving

= \frac{\sqrt{x} }{2} → D

4 0
3 years ago
Plz Help
soldier1979 [14.2K]

the answer is B)

we can check by multiply them

8 0
3 years ago
Read 2 more answers
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