2 years ago

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:

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

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3 years ago

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KatRina [158]

Answer:

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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

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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.

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

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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

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