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2 years ago

PLEASE HELP ME QUICKKKK!

Mathematics

1 answer:

telo118 [61]2 years ago

5 0

Look up what is mean and calculate it.

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Find cos(ZA).<br> Choose 1 answer:

lbvjy [14]

Answer: option A. 24 / 25

Step-by-step explanation:

Cos (Z) = Adj / Hypo

Cos (Z) = 24 / 25

3 0

3 years ago

4c=-40 i don't know this and i need help<br> ill do what i can in return

melamori03 [73]

Answer:

4c=-40

Divide boyh sides by 4

c=-40/4

c=-10

8 0

2 years ago

Read 2 more answers

11-Across plus 28-Down

Art [367]

Answer:

(11,-28)

Step-by-step explanation:

8 0

3 years ago

n a call center the number of received calls in a day can be modeledby a Poisson random variable. We know that, on average, 0.5%

guapka [62]

Answer:Pois(ln(200))

Step-byy-step explanation:

Let N be the number of received calls in a day

That is

N∼Pois(λ).

0.5% = 0.5/100 = 1/200 of no calls

P(N=0)=e^−λ=1/200

-λ=e^(1/200)

λ=in(200)

Our number of calls in a day is distributed Pois(ln(200)).

7 0

3 years ago

Read 2 more answers

There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student

Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

$\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57$

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0

3 years ago