<h2><u>
PLEASE MARK BRAINLIEST!</u></h2>
Answer:
Let's see...
Step-by-step explanation:
- The first coin is a quarter.
- The second, third, fourth, and fifth coins are dimes.
- The sixth, seventh, and eighth coins are nickels.
That being said, what is the total value of the cents?
cents = ¢
<u>Values:</u>
Quarter = 25 ¢
Dime = 10 ¢
Nickel = 5 ¢
So,
1 quarter + 4 dimes + 3 nickels = total cents
25 + 10 + 10 + 10 + 10 + 5 + 5 + 5 = total cents
25 + 40 + 5 + 5 + 5 = total cents
25 + 40 + 15 = total cents
25 + 55 = total cents
80 = total cents
<h3><em>Our answer is 80¢</em></h3><h3><em /></h3>
I hope this helps!
- sincerelynini
The probability that at most 8 of them take the bus to school is 0.925, written in percentage form this is 92.5%
<h3>
How to find the probability?</h3>
We know that roughly 75% of the students take the bus, then, if we select a student at random.
- There is a probability of 0.75 that the student takes the bus.
- There is a probability of 0.25 that the student does not take the bus.
The probability that at most 8 out of 9 students take the bus, is equal to one minus the probability of the 9 taking the bus, which is:
p = (0.75)^9 = 0.075
Then we have:
P = 1 - 0.075 = 0.925
The probability that at most 8 of them take the bus to school is 0.925, written in percentage form this is 92.5%
If you want to learn more about probability, you can read:
brainly.com/question/251701
Answer:
GCF = 3
Step-by-step explanation:
Express the numbers as a product of their primes.
42 = 2 × 3 + 7
30 = 2 × 3 × 5
45 = 3 × 3 × 5
Identify the prime factors common to all 3 numbers.
common prime factor = 3 , thus
GCF = 3
Answer:
Answer:
3.14159
Step-by-step explanation:
This number is pi. Pi goes on for millions of digits on end.
Step-by-step explanation:
9514 1404 393
Answer:
Step-by-step explanation:
There are <em>an infinite number of possibilities</em>. Any vector whose dot-product with p is zero will be perpendicular to p.
Let m = 0i +1j +ak. Then we require ...
m·p = 0 = 0×1 +1×2 +a(-2) ⇒ 0 = 2 -2a ⇒ a = 1
m = 0i +1j +1k
__
Let n = 2i +0j +bk
n·p = 0 = 2×1 +0×2 +b(-2) ⇒ 2 -2b = 0 ⇒ b = 1
n = 2i +0j +1k
