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Novosadov [1.4K]
2 years ago
5

I..Estimate the product by the general rule

Mathematics
1 answer:
IgorLugansk [536]2 years ago
6 0

Answer:

a.334,880

b.1806

c.78

ll,a.140

b.136.5

c.27.73914

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An airplane takes off and then climbs 2500 feet. After 20 minutes, the airplane drops 150 feet.
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Answer:

2350 feet

Step-by-step explanation:

2500-150=2350

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What plus 10 equals 26
dexar [7]
X+ 10 = 26

-10 = -10

X = 16
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4,3

Step-by-step explanation:

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A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr
ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so \pi = 0.30

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so n = 10

We want to find P(X = 1). So:

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is P(X > 1)

We have that:

P(X \leq 1) + P(X > 1) = 1

P(X > 1) = 1 - P(X \leq 1)

We also have that:

P(X \leq 1) = P(X = 0) + P(X = 1)

P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282

So

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493

Finally

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507

There is a 85.07% probability that more than 1 has the marker.

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