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Alchen [17]
2 years ago
12

I need help please!!!!

Mathematics
1 answer:
Reika [66]2 years ago
5 0

Answer:

the first one

Step-by-step explanation:

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My calculator wasn’t working for this I’m not sure how to solve it but help would be coolio
ziro4ka [17]

We know that:

_nP_k = \dfrac{n!}{(n-k)!}

so in this case n = 5, k=3 and:

_5P_3=\dfrac{5!}{(5-3)!}=\dfrac{5!}{2!}=\dfrac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\dfrac{120}{2}=\boxed{60}

4 0
3 years ago
It hard I need help can y’all help me plz
KatRina [158]

Answer:

we dont have a question or picture..?

7 0
3 years ago
Subtract – 2.2 + 3x – 9 from 8x2 + 10x – 10.
Ksju [112]

Answer:

=3x−11

2(4x2+5x−5)

Step-by-step explanation:

both subtracted: −8x2−7x−1

6 0
2 years ago
Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ = 1/ 8. If someone arrives
Elis [28]

Answer:

a) P=0.535

b) P=0.204

c) P=0.286

Step-by-step explanation:

The exponential distribution is expressed as

F(x>t)=e^{-\lambda t}

In this example, λ=1/8=0.125 min⁻¹.

a) The probability of having to wait more than 5 minutes

F(x>5)=e^{-0.125*5}=e^{-0.625}=0.535

b) The probability of having to wait between 10 and 20 minutes

F(1020)=e^{-0.125*10}-e^{-0.125*20}\\\\F(10

c) The exponential distribution is memory-less, so it is independent of past events.

If you have waited 5 minutes, the probability of waiting more than 15 minutes in total is the same as the probability of waiting 15-5=10 minutes.

F(x>15|t^*=5)=F(x>15)/F(x>5)=\frac{e^{-0.125*15}}{e^{-0.125*5}}=e^{-0.125*(15-5)}\\\\ F(x>15|t^*=5)=e^{-0.125*(10)}=F(x>10)=0.286

5 0
3 years ago
Which of these is the quadratic parent function?
hjlf

Answer:

B. f(x)= x^{2}

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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