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Harrizon [31]
3 years ago
14

Do the side lengths of 8, 15, and 20 forms a triangle?

Mathematics
2 answers:
liq [111]3 years ago
6 0

Answer:

yes

Step-by-step explanation:

when 1st and 2nd lengths addition is more than 3rd side

for example

6+8>10

trapecia [35]3 years ago
3 0

Answer:

Yes they do

Step-by-step explanation:

https://www.mathwarehouse.com/geometry/triangles/triangle-inequality-theorem-rule-explained.php?alenght=8&blength=15&clength=20#triangle-inequality-theorem

This demostration shows how.

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There were 24 green jelly beans which made up 8%of all the jelly beans in the bag
Zanzabum
There are 300 jelly beans in the bag.


24÷6=4
4×2=8
5×2=10
5×6=30
30×10=300


Proof:
24/300 = ?/100
24×100=2400
2400÷300=8

Hope this helps! :)
6 0
3 years ago
What two numbers multiply to get 60 and add to get 19?
marusya05 [52]
4 and 15. You can guess and check or use algebra.
4 0
3 years ago
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Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case
olga nikolaevna [1]

Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> =  Mike will show up.

Given:

P(D^{c})=0.55\\P(M^{c})=0.45

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

                                                   =1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25

Thus, the probability that neither Dave nor Mike will show up is 0.25.

6 0
3 years ago
Read 2 more answers
Can you help me please
xenn [34]
(2, -3)

The Answer is A
3 0
3 years ago
Two friends leave school at the same time Sarah is heading to do North and Beth is heading to West one hour later they are 5 mil
svet-max [94.6K]
5^-4^=25-16=9 root from 9 is 3 so it’s c) 3 miles
7 0
3 years ago
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