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3 years ago

PLEASEE HELP WILL GIVE BRAINLEST 20 POINTS

Mathematics

1 answer:

Ket [755]3 years ago

5 0

<h3>Answer:</h3>

For each birdhouse built, the total cost increases by $3.50.

<h3>Step-by-step explanation:</h3>

Let's try some of the answers and see if they work.

The "initial cost" will be the cost when 0 birdhouses are built.

... f(0) = 3.5·0 +24 = 24 . . . . . clearly, not $3.50

Then increasing the number of birdhouses by 1, the cost is ...

... f(1) = 3.5·1 +24 = 27.5 . . . . . the cost has gone up (not down) by $3.50

Clearly the total cost to build 1 birdhouse is not $3.50. (It is $27.50.)

___

Our exploration of the formula leads us to conclude ...

... For each birdhouse built, the total cost increases by $3.50.

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3 years ago

All bags entering a research facility are screened. The screening process is not perfect so that 77% of the bags that contain fo

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Answer:

(1) 0.5933

(2) 0.8955

Step-by-step explanation:

We are given that all bags entering a research facility are screened.

Let Probability that bags entering the building contains forbidden material,

P(F) = 0.69

Probability that bags entering the building does not contains forbidden material, P(NF) = 1 - 0.69 = 0.31

Let event A = alarm gets triggered

Probability that alarm gets trigger given the bags contain forbidden material, P(A/F) = 0.77

Probability that alarm gets trigger given the bags does not contain forbidden material, P(A/NF) = 0.20

(1) Probability that a bag triggers the alarm, P(A) ;

P(A) = P(F) * P(A/F) + P(NF) * P(A/NF)

= (0.69 * 0.77) + (0.31 * 0.20) = 0.5313 + 0.062

= 0.5933

Therefore, probability that a bag triggers the alarm is 0.5933 .

(2) Probability that a bag that triggers the alarm will actually contain forbidden material is given by P(F/A) ;

Using Bayes' Theorem;

P(F/A) = $\frac{P(F) * P(A/F)}{P(F) * P(A/F) + P(NF) *P(A/NF)}$ = $\frac{0.69*0.77}{0.69*0.77+0.31*0.20}$ = $\frac{0.5313}{0.5933}$

= 0.8955

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