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KonstantinChe [14]
3 years ago
8

Please answer!!!!!!

Mathematics
1 answer:
Crazy boy [7]3 years ago
4 0

Answer:

(this isnt a mathematics question, it's science, btw)

Indeed, you are correct. It is 1. I think.

Because you get one allele from each parent, to make up 2 alleles per offspring.

Hope that helps!

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If R is midpoint of QS and QS=40 find QR
Elodia [21]

Answer:

Given: R is the midpoint of QS.

QS=40

Now,

QR=40/2(R IS THE MIDPOINT)

=20

8 0
4 years ago
5 of the 25 players on the soccer team are 7th graders. Find the Percent of 7th graders
kati45 [8]
Answer:
20%

Step-by-step explanation:
5 divided by 25 multiplied by 100 will give you 20%.
3 0
3 years ago
Target sells 24 bottles of water for $3 and 36 bottles of water for $4 which is the better buy and by how much?
Tanya [424]
D I think don’t fault me if I’m wrong I just wanted to help
3 0
3 years ago
The system x−6y=4, 3x−18y=4 has no solution.
stealth61 [152]
Ok fist you wanna change your question into a equation so it would turn into x-6 y=4 and 4.3x - 18 y = 4 then multiple both sides and that pups give you 43x - 180 y = 40 and eliminate one variable and I don’t wanna keep explaining but the the answer is



ANSWER: (x , y ) = (- 80 over 13, and - 22 over 13
6 0
3 years ago
Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte
sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so n = 11

Each player with a 5% probability of injury, so p = 0.05

Then

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688

P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293

P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so n = 2

Each team has a 0.9848 probability of showing up to play, so p = 0.9848.

This probability is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299

2.99% probability that the cost will be paid by only one team

7 0
4 years ago
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