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Digiron [165]
3 years ago
5

A University has 30,600 students think school counselor estimates that 2% of the students break three on my languages in a rando

m sample of 240 students, 20 speak three or more languages. Determine whether the counselors estimate is likely to be accurate.
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
4 0

Answer:

Counselor's estimate isn't correct.

Step-by-step explanation:

Total number of students in university = 30,600

2% break 3 of the languages

Therefore, 2/100 × 30600 = 612 students break 3 of the languages in the University according to the counselor.

In a random sample of 240 students, 20 break 3 of the languages.

% = 20/240 × 100 = 8.33%

SinceTotal number of students = 30,600

Therefore, 30,600/240 = 127.5

Which means 127.5 × 20 = 2550.

Which means 2550 students break 3 of the languages as against the 612 students stated by the counselor.

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Here is it


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If (x) = x+8 and g(x) = -4x-3, find (f+ g)(x).
Ede4ka [16]

Answer:

Answer=B

(f+g)(x)=-3x-5

Step-by-step explanation:

f(x)=x+8.

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(f+g)(x) means f(x)+g(x),So

(x+8)+(-4x-3)

(add all same terms together)

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so,

(f+g)(x)=-3x+5

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3 years ago
Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a 5 2 per square yard. How
HACTEHA [7]

In this question, the Poisson distribution is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Parameter of 5.2 per square yard:

This means that \mu = 5.2r, in which r is the radius.

How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?

We want:

P(X \geq 1) = 1 - P(X = 0) = 0.99

Thus:

P(X = 0) = 1 - 0.99 = 0.01

We have that:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-5.2r}*(5.2r)^{0}}{(0)!} = e^{-5.2r}

Then

e^{-5.2r} = 0.01

\ln{e^{-5.2r}} = \ln{0.01}

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Another example of a Poisson distribution is found at brainly.com/question/24098004

3 0
3 years ago
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Fantom [35]

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4 0
3 years ago
ANSWER FOR 30 POINTS!
Darya [45]

Answer:

48

Step-by-step explanation:

27+36-15=63-15=48

5 0
3 years ago
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