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torisob [31]
3 years ago
11

The winning car in a race beat the second car by 19/100 of a second. The third car was 4/10 of a second behind the second car. B

y how much did the first car beat the third car?
Mathematics
1 answer:
Neko [114]3 years ago
8 0
20/5 should be correct
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The h in p will start the p and the glock I believe
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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

6 0
3 years ago
At a conference for 500 people, 30% of the participants are French, 15% are Americans, 5% are Germans, and are of other national
Olenka [21]
As said before, 160 is the correct answer
Please: Rate. Thank!
7 0
3 years ago
Read 2 more answers
Please help i have no idea with this lesson
sergey [27]

Answer:

40

Step-by-step explanation:

ABD-CBD=ABD

70-30=40

3 0
3 years ago
Read 2 more answers
Determine whether the polynomial is a difference of squares and if it is, factor it.
ladessa [460]

Answer:

(y - 4)(y + 4)

Step-by-step explanation:

y² - 16 ← is a difference of squares

Since y² and 16 are both squares separated by a difference , that is minus

A difference of squares factors as

y² - 16

= y² - 4²

= (y - 4)(y + 4)

5 0
3 years ago
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