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2 years ago

Question 1.

Mathematics

1 answer:

tangare [24]2 years ago

7 0

Answer:

1. 8.03 x 108 2. 6.8 × 10-9 3. 135,900 4. 2.5x10^-11

Step-by-step explanation:

Step 1

To find a, take the number and move a decimal place to the right one position.

Original Number: 803,000,000New Number: 8.03000000Step 2

Now, to find b, count how many places to the right of the decimal.

New Number:8.03000000Decimal Count:12345678

There are 8 places to the right of the decimal place.

Step 3

Building upon what we know above, we can now reconstruct the number into scientific notation.

Remember, the notation is: a x 10b

a = 8.03 (Please notice any zeroes on the end have been removed)

b = 8

Now the whole thing:

8.03 x 108

Step 4

Check your work:

108 = 100,000,000 x 8.03 = 803,000,000

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Explanation:

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3 years ago

-2(5 + 6n) < 6(8 - 2n) solve for n

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Answer:

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Step-by-step explanation:

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3 years ago

In 2018, the population will grow over 700,000.The population of a city in 2010 was 450,000 and was growing at a rate of 5% per

Firlakuza [10]

Answer:

The year is 2020.

Step-by-step explanation:

Let the number of years passed since 2010 to reach population more than 7000000 be 'x'.

Given:

Initial population is, $P_0=450,000$

Growth rate is, $r=5\%=0.05$

Final population is, $P=700,000$

A population growth is an exponential growth and is modeled by the following function:

$P=P_0(1+r)^x$

Taking log on both sides, we get:

$\log(P)=\log(P_0(1+r)^x)\\\log P=\log P_0+x\log (1+r)\\x\log (1+r)=\log P-\log P_0\\x\log(1+r)=\log(\frac{P}{P_0})\\x=\frac{\log(\frac{P}{P_0})}{\log(1+r)}$

Plug in all the given values and solve for 'x'.

$x=\frac{\log(\frac{700,000}{450,000})}{\log(1+0.05)}\\x=\frac{0.192}{0.021}=9.13\approx 10$

So, for $x > 9.13$ , the population is over 700,000. Therefore, from the tenth year after 2010, the population will be over 700,000.

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3 years ago

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

3 years ago

Please help me ASAP.

Sladkaya [172]

Sending help your way :)

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3 years ago