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3 years ago

9

What supplies electricity in a circuit

1 answer:

nekit [7.7K]3 years ago

4 0

Answer:

B.the conductor

Step-by-step explanation:

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Find the exact value of cos(arcsin(one fourth)). For full credit, explain your reasoning.

photoshop1234 [79]

We are asked to give the exact value of <span>cos(arcsin(one fourth)). In this case, we shift first the setting to degrees since this involves angles. we determine first arc sin of one fourth equal to 14.48 degrees. then we take the cos of 14.48 degrees equal to 0.9682. Answer is 0.9682.</span>

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3 years ago

Please help!! I’ll give brainlist..!!

Xelga [282]

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the second, fourth, fifth, and sixth

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2 years ago

Read 2 more answers

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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3 years ago

Helpppppppppppppppppppp me

Lana71 [14]

Answer:

6x5=30

Step-by-step explanation:

30÷5=6

n=6

6×5=30

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3 years ago

Number 6 please!!!!!!

lisov135 [29]

Well if you subtract 7/8 from 2/5 you get 19/40 which is 47 muffins Maria can make

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3 years ago