Convert 10 hours into minutes

List all the 3-digit numbers that fit these clues .the hundreds digit is greater than7 .the tens is 1 more than the hundreds dig

djverab [1.8K]

In this question, there are 3 conditions that need to be met
1. The number is 3-digit
2. Hundreds digit >7 (number that will fulfill it would be 8 or 9)
3. The tens is 1 more than hundred( since the hundred possibilities is 8 or 9, then 9 in hundreds can't be used since no number higher than 9)
4. The one digit <2( that mean 0, 1)

Using the list above, you can make your possible number:
890
891

3 0

3 years ago

What is an equation of the line that passes through the points (0, 3) and (5, 3)?

Natalka [10]

Answer:

y=3

Step-by-step explanation:

5 0

3 years ago

Pennetta is a wedding planner. She charges a $50 fee plus $100 per guest to plan a wedding for g, the number of guest. Write an

malfutka [58]

Answer:

C = 100g + 50

Step-by-step explanation:

5 0

3 years ago

If a number is divisible by 3 and 5 then it is divisible by 15

koban [17]

Answer:

Yes!

Step-by-step explanation:

Besides 15 and one, 3 and 5 are the greatest common factors of 15. Those numbers are therefore divisible by any product of 15.

Example:

45

45/3 = 15

45/5 = 9

45/15 = 3

6 0

3 years ago

Read 2 more answers

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago