Hmmm wait let me try to do it
Answer:
1. 5700mg
2. 1000cm
3. 0.0453m
4. 37.5 kg/L
Step-by-step explanation:
1. 1gram (g) =1000milligrams(mg)
5.70×1000=5700mg
2. 1 m= 100cm
10×100=1000cm
3. 1000mm=1m
45.3/1000=0.0453m
4. 1000g=1kg
37.5/1000=0.0375
1000ml=1L
1ml= 1/1000= 0.001L
0.0375/0.001= 37.5 kg/L
<h3>Answer: 0.47178 Step-by-step explanation:
Find the probability for each p(X=x) up to 5 using the equation: (x-1)C(r-1)*p^r * q^x-r,
where x is number of days, p = .3 (prob of rain). q=.7 (prob of not rain), and r=2 (second day of rain). also C means choose.
So p(X=1) = 0
p(X=2) = 1C1 * .3^2 * .7^0 = .09
P(X=3) = 2C1 * .3^2 * .7^1 = .126
P(X=4) = 3C1 * .3^2 * .7^2 = .1323
P(X=5) = 4C1 * .3^2 * .7^3 = .12348
Then add all of them up
0+.09+.126+.1323+.12348 = .47178</h3>
80% = 6 in
100%=?
100/x 80/6 (proportional relationship)
then cross multiply
600 = 80x
600/80= 7.5
x=7.5
Answer:
Part c: Contained within the explanation
Part b: gcd(1200,560)=80
Part a: q=-6 r=1
Step-by-step explanation:
I will start with c and work my way up:
Part c:
Proof:
We want to shoe that bL=a+c for some integer L given:
bM=a for some integer M and bK=c for some integer K.
If a=bM and c=bK,
then a+c=bM+bK.
a+c=bM+bK
a+c=b(M+K) by factoring using distributive property
Now we have what we wanted to prove since integers are closed under addition. M+K is an integer since M and K are integers.
So L=M+K in bL=a+c.
We have shown b|(a+c) given b|a and b|c.
//
Part b:
We are going to use Euclidean's Algorithm.
Start with bigger number and see how much smaller number goes into it:
1200=2(560)+80
560=80(7)
This implies the remainder before the remainder is 0 is the greatest common factor of 1200 and 560. So the greatest common factor of 1200 and 560 is 80.
Part a:
Find q and r such that:
-65=q(11)+r
We want to find q and r such that they satisfy the division algorithm.
r is suppose to be a positive integer less than 11.
So q=-6 gives:
-65=(-6)(11)+r
-65=-66+r
So r=1 since r=-65+66.
So q=-6 while r=1.
