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3 years ago

The number in scientific notation is written as a decimal. 5.03x10^5

Mathematics

2 answers:

dusya [7]3 years ago

8 0

I’m pretty sure it’s 503,000

DiKsa [7]3 years ago

4 0

Answer:

50,300,000,000

Step-by-step explanation:

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12. Write the first five terms of two different sequences that have 10 as the

kodGreya [7K]

Answer:

a) The arithmetic sequence with common difference 2 that has 8 as the first term.

b) The arithmetic sequence of common difference -5 and first term 15.

Step-by-step explanation:

Let's use for example the arithmetic sequence with common difference 2 that has 8 as the first term. Then the first two terms of this sequence are:

8, and (8+2) = 10 Therefore the second term is 10.

Another arithmetic sequence of common difference -5 and first term 15. The firs two terms of this sequence are:

15, and (15 - 5) = 10. Therefore again a 10 as second term.

6 0

3 years ago

Geometry math question please help

Umnica [9.8K]

You doing virtual school too

3 0

3 years ago

0.3% of what number is $0.66?

UNO [17]

$0.66 / 0.3% = 2.2
Check:
2.2 * 0.3 = 0.66

Hope this helps!

8 0

3 years ago

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Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago

20 points BRAINLIEST

kolezko [41]

Answer:

1.6

Step-by-step explanation:

4 0

3 years ago

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