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Anton [14]
3 years ago
6

Someone please help!! find the value of y in this triangle

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
4 0

Answer:

30

Step-by-step explanation:

Triangle containing 120° and y° is an isosceles triangle.

Therefore,

y° + 120° + y° = 180°

2y° = 180° - 120°

2y° = 60°

y° = 60°/2

y ° = 30°

y = 30

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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
The explicit rule for a sequence is an=14-9n <br> What is the recursive rule for the sequence?
natulia [17]
You have to add it then divide it 
6 0
3 years ago
Read 2 more answers
You purchase one microsoft july $72 put contract for a premium of $1. 32. what is your maximum possible profit?
goldenfox [79]

The maximum possible profit = $7068

For given question,

One Microsoft July $72 put contract for a premium of $1.32

The payoff arise from put option is max (K - S, 0) - P

Now it would be maximum at S = 0

And, the maximum payoff is

K - 0 - P

= K - P

= 72 - 1.32

= $70.68

We assume that for each and every contract the number of shares is 100

So, the maximum profit gained from this strategy is

= $70.68 × 100 shares

= $7068

The maximum profit that will be gained from this strategy is $7068

Therefore, the maximum possible profit = $7068

Learn more about the profit here:

brainly.com/question/20165321

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4 0
2 years ago
What is 0.0008235 in scientific notation
larisa [96]
I think this is the Answer

8.235*10^-4

7 0
4 years ago
3.3 = ? as a friction
Allushta [10]

Answer:

33/10

Step-by-step explanation:

I used a calculator sorry if I am wrong

6 0
3 years ago
Read 2 more answers
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