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Studentka2010 [4]
3 years ago
15

Audience thinks 30% of Brandon jokes are funny what fraction of Brandon jokes does the audience think are funny

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
6 0

Answer:

3/10

Step-by-step explanation:

Percent is basically the number over 100.

In this case, 30% is equal to 30/100, which can be simplified to 3/10.

Thus, the answer is 3/10 of the audience.

Hope this helps!

sergejj [24]3 years ago
6 0

The percent was 30% and since we do percent's out of 100 you write 30/100 and you simplify it which  gives you 3/10

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The table below shows the numbers of students at 4 middle schools.
Naddik [55]

Answer:

(a) 300

(b)100

Step-by-step explanation:

(a) each figure is 100 student and there are 3 figures at Woodbridge so it's 3*100=300

(b) DuBois=3*100=300

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Lana71 [14]

9514 1404 393

Answer:

  • x = 8
  • BF = 45

Step-by-step explanation:

Using the given information, we have ...

  BH = 2·HF

  3x +6 = 2(2x -1)

  3x +6 = 4x -2

  8 = x

__

The length of the median is ...

  BF = BH + HF

  BF = (3x +6) +(2x -1) = 5x +5

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6 0
3 years ago
Suppose that from the past experience a professor knows that the test score of a student taking his final examination is a rando
DENIUS [597]

Answer:

n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =73,\sigma =10.5)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

We want to find the value of n that satisfy this condition:

P(71.5 < \bar X

And we can use the z score formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we have this:

P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z

And we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94

And by properties of the normal distribution we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z

If we solve for P(Z we got:

P(Z

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: z=-1.881

And using this we have this equality:

-1.881 = -0.14286 \sqrt{n}

If we solve for \sqrt{n} we got:

\sqrt{n} = \frac{-1.881}{-0.14286}=13.167

And then n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

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3 years ago
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Marysya12 [62]

Answer:

Step-by-step explanation:

8 0
3 years ago
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