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3 years ago

How do you solve numbers 27 and 29 in the photo?

Mathematics

1 answer:

Furkat [3]3 years ago

5 0

This is what I got - csc pi/2 - 1 ( on 29 ) and cot pi/4+1 ( on 27 )

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$.87 per day is equal to how many dollars per year

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3 years ago

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Roberta's boat started farther from the dock, but John's boat moves more quickly.

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3 years ago

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Help me please thank you!!

Marta_Voda [28]

Answer:

6+1.25t

Step-by-step explanation:

W(in the 1st month)=6+1.25

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3 years ago

What is equivalent to this formula: y=3(x+5)(x-2)

nadezda [96]

Answer:

y = 3*(x^2) + 9x -30

Step-by-step explanation:

y=3(x+5)(x-2)

y = (3*x + 3*5) * (x - 2)

y = (3*x + 15) * (x - 2)

y = 3*x*x + 3*x*(-2) + 15*x + 15*(-2)

y = 3*(x^2) - 6x + 15x - 30

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3 years ago

The time it takes to completely tune an engine of an automobile follows an exponential distribution with a mean of 40 minutes.

Mazyrski [523]

Answer: a. P (x<30) = 0.5276

b. P (30<x<35) = 0.0555

Step-by-step explanation: An exponential distribution is a distribution of time in which events happens at a constant average rate.

The rate is calculated as

$\lambda=\frac{1}{\mu}$

with μ as the mean.

The probability density distribution for this type of distribution is

$f(x)=\lambda e^{-\lambda x}$

And probability is calculated as

$P(X$

For tuning of an engine, the rate is

$\lambda=\frac{1}{40}$

λ = 0.025

a. Probability of less than 30 minutes:

$P(X$

P (X < 30) = 0.5276

Probability of tuning in 30 minutes or less is 52.76%.

b. Probability of between 30 and 35 can be described as

$P(30$

$P(X$

P (X < 35) = 0.5831

P (X < 30) = 0.5276

Then:

$P(30$

P (30 < X < 35) = 0.0555

Probability of tuning an engine between 30 and 35 minutes is 5.55%.

5 0

3 years ago