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3 years ago

Pls help!!!!!!!!!!!!!!!!

Mathematics

1 answer:

MakcuM [25]3 years ago

3 0

Answer:

6562331215

Step-by-step explanation:

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An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. The ACME Manuf

iren [92.7K]

Answer:

a) P(ACME) = 0.7

b) P(ACME/D) = 0.5976

Step-by-step explanation:

Taking into account that ACME manufacturing company makes 70% of the ELTs, if a locator is randomly selected from the general population, the probability that it was made by ACME manufacturing Company is 0.7. So:

P(ACME) = 0.7

Then, the probability P(ACME/D) that a randomly selected locator was made by ACME given that the locator is defective is calculated as:

P(ACME/D) = P(ACME∩D)/P(D)

Where the probability that a locator is defective is:

P(D) = P(ACME∩D) + P(B. BUNNY∩D) + P(W. E. COYOTE∩D)

So, the probability P(ACME∩D) that a locator was made by ACME and is defective is:

P(ACME∩D) = 0.7*0.035 = 0.0245

Because 0.035 is the rate of defects in ACME

At the same way, P(B. BUNNY∩D) and P(W. E. COYOTE∩D) are equal to:

P(B. BUNNY∩D) = 0.25*0.05 = 0.0125

P(W. E. COYOTE∩D) = 0.05*0.08 = 0.004

Finally, P(D) and P(ACME/D) is equal to:

P(D) = 0.0245 + 0.0125 + 0.004 = 0.041

P(ACME/D) = 0.0245/0.041 = 0.5976

5 0

3 years ago

This is an area in which objects will interact with a magnet

Basile [38]

Answer: magnetic field

7 0

2 years ago

F(x) = x2 – 2x + 3; f(x) = -x + 5

creativ13 [48]

Dhvbvheusicfhiugbbvuchdrhbjbjgjgnjrmenwmsmxm nvbhbh h jcndnfhr£xjvybuvcjdnf yvuduf

8 0

3 years ago

"A laboratory tested 73 chicken eggs and found that the mean amount of cholesterol was 230 milligrams with LaTeX: \sigmaσ = 17.4

Scrat [10]

Answer:

The 95% confidence interval for the true mean cholesterol content, μ, of all such eggs is between 226.01 and 233.99 milligrams.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{17.4}{\sqrt{73}} = 3.99$

The lower end of the interval is the sample mean subtracted by M. So it is 230 - 3.99 = 226.01

The upper end of the interval is the sample mean added to M. So it is 230 + 3.99 = 233.99.

The 95% confidence interval for the true mean cholesterol content, μ, of all such eggs is between 226.01 and 233.99 milligrams.

4 0

3 years ago

The number of cars sold per month can be modeled by f(x) = 0.04x^3 + 0.2x^2 + 0.5x + 60, where x represents the number of months

Karolina [17]

Answer:

0.04x^3+0.2x^2+0.5x+90 vertical shift 30 units up, increase of 30 units per month

Step-by-step explanation:

You don't need an explanation just take the answer.

6 0

3 years ago