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4 years ago

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Someone please help me

1 answer:

Karolina [17]4 years ago

3 0

NO = ON: reflexive property
MN= PO; MO = PN; NO = ON; SSS

i believe the answer is the last one
Reflexive property SSS

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Glenda and Fiero are at a restaurant eating meal that costs $30. They have a discount coupon that gets them 10% off the cost. If

yaroslaw [1]

The cost of meal = $$30$

The tip to the waiter = $8%$ %

Discount = $10%$ %

Cost = $30 + 0.08 * 30 - 0.1 * 30$

Cost = $30(1 + 0.08 - 0.1)$

Cost = $ $$29.4$ 0

8 0

2 years ago

It performs the following operations of the complex numbers would greatly appreciate friends...:)

Assoli18 [71]

(5i+4) + (4i-3)
5i+4i-3+4
9i - 1

(2+i) - (3+4i)
(2+i) + (-3 - 4i)
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Each part, real numbers and imaginary, are to be taken as separate pieces; add them with and subtract them by each other as you normally would. :)

3 0

3 years ago

Read 2 more answers

Create a graph of y=2x−6 y = 2x-6.

cupoosta [38]

<span>y = mx + b</span><span> m = slope = 2
b = y intercept = -6

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<span>I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.</span>

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5 0

3 years ago

Express as a percentage<br>25

motikmotik

Answer:

25000%

Step-by-step explanation:

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3 years ago

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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