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For part (a), there is a third value for x, x = 0.

For part (b), f(x) = g(x) when x= 0 and so cannot be included in the interval.

"Two days ago" or "two years ago"? if "two days ago", then Robert must be 18, if he is going to be 19 next year. if "two years ago", then Robert is 18 now and will be 19 next years.

His birthday appears to be "two days ago", which was March 23 and was born March 23, 2005. Next March 23, 2024, he will be 19 years old.

The best possible answer is 17. If the biologist and others are seated Ina table so we group them as one and the answer is 5. So 5 multiplied by the number of people= 85multiplied by 2= 170 so there are 170 different ways to rearrange them

The 99​% confidence interval estimate is 2.7 ​mi/h<σ<8.7 ​mi/h.

A)99 digits for sam's 

B)0 because Sam did not include a zero

C)4934 is the total of sam's written digits

You have 17 people. 

Seat one biologist at the head of table. That leaves 16 people. 5 of them are scientists who want to sit together. We will consider them as 1 person. So, we have:16 - 5 + 1=12 people who can be seated in:

12! ways. But, the 5 scientists can rearrange themselves in 5! ways. So, the total seating of all 17 people should be: 12!  x  5! =57,480,192,000 ways of seating 17 people.

Thanks a lot! Much appreciated.

For (c) I think you are supposed to add up ALL the digits he wrote down like this:

1+2+3+4+5+6+7+8+9+1+0(for ten)+1+1(for eleven)+1+2(for twelve).......and so on. You should get a total of 900 for the the sum of 189 digits he wrote down.

The same goes for b).

That's what I got for a) too, but it's shown as wrong

a) The solutions are x = -1 and x = 1.

b) The solution is -1 < x < 1, or in interval notation, (-1,1).

Let's denote the vertices of the rectangle as A, B, C, and D. Since ABCD is a rectangle, AD and BC are parallel and have the same length. Therefore, we can use the Pythagorean theorem to find the length of AD:

AD² = AP² + DP²

Similarly, we can use the Pythagorean theorem to find the length of BC:

BC² = BP² + CP²

Since AD and BC are equal in length, we have:

AP² + DP² = BP² + CP²

Substituting the given values, we get:

3² + DP² = 2² + 7²

9 + DP² = 4 + 49

DP² = 44

DP = √44

DP = 2√11

Therefore, the length of DP is 2√11.

a) Sam has total digits of 189

b) The possibility is 9/189; as the numbers 10, 20, 30 ...90 have the number 0. 

c) The sum of n = n(n+1)/ 2

-> 99(99 + 1) / 2 

-> 99(100) / 2 

-> 9900 / 2

-> 4950

90 km/h  x  1,000 m ==90,000 m/h

90,000 / 3,600 seconds ==25 m / second

25 m/s  x  45 s ==1,125 meters - length of the train.

(a) 99 digits in total

(b) 0 because Sam didn't write the number 0

(c) 4950

Hope this helps ^^

Can you explain your answer? I didn't quite understand how you got that answer.

AB = 6*sqrt(5)

Thank you!

I can only see 3 ways.

Say your have 2 reds and 4 blues.

1) You can have 4 blues together

2) You can have 3 blues together 

OR

3) you can have 2 lots of 2 blues together.

That is it I think.

Try proof reading you question.  I think there is a word missing.

For the first tile, there is a one third probability. For the second, it is just (8-1)/(24-1) = 7/23. The multiplication gets answer 7/69

To find the total number of possible sequences of 12 coin tosses, we can use the multiplication rule, which states that if there are k independent events with n1 possible outcomes for the first event, n2 possible outcomes for the second event, and so on, then the total number of possible outcomes is n1 * n2 * ... * nk.

For each of the 12 tosses, there are two possible outcomes, heads or tails. Therefore, the total number of possible sequences of 12 coin tosses is 2^12 = 4096.

To find the number of sequences with at least k heads, we can use the complement rule, which states that the probability of an event happening is 1 minus the probability of the event not happening. In this case, the probability of getting at least k heads is equal to 1 minus the probability of getting less than k heads.

Let's consider the case where k = 1, i.e., we want to find the number of sequences with at least one head. The number of sequences with no heads is just 1 (i.e., all tails), so the number of sequences with at least one head is 4096 - 1 = 4095.

For the case where k = 2, we want to find the number of sequences with at least two heads. The number of sequences with no heads or one head is 1 + 12 = 13, so the number of sequences with at least two heads is 4096 - 13 = 4083.

Using this approach, we can find the number of sequences with at least 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12 heads, by subtracting the number of sequences with fewer than k heads from the total number of sequences.

Alternatively, we can use the binomial distribution formula to calculate the number of sequences with at least k heads:

Number of sequences with at least k heads = sum from i=k to i=12 of (12 choose i) = (12 choose k) + (12 choose k+1) + ... + (12 choose 12)

Look at 1 + 2^(1/3) + 2^(2/3) and think arithmetic progression.