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bekas [8.4K]
3 years ago
10

Can anyone help me find the unknown side length in this right triangle.

Mathematics
2 answers:
Maru [420]3 years ago
7 0

Answer:

9sqrd= 81

12sprd= 144

144+81= 225

\sqrt{225}= 15

so c=15

ps. use Pythagoras to help you.  

Alisiya [41]3 years ago
7 0

Answer:

c=15

Step-by-step explanation:

You can use Pythagoras theorem to find the value of c

Formula to find the hypotenuse:

c=\sqrt{a^{2}+b^{2}  }

The value of a in the formula will be the height & the value for b in the formula is the base.

Now, add the values in the formula:

c=\sqrt{9^{2}+12^{2}  }

c=\sqrt{81+144 }

c=\sqrt{225}

c=15.96871942

You can simply say that:

c=15

I hope this helps :)

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Apply each of the four counting/sampling methods (with replacement and with ordering, without replacement and with ordering, wit
dusya [7]

Answer:

Step-by-step explanation:

I will illustrate this solution with a unique birthday party situation between Jane (the celebrant) and her 10 friends.

In the said party , we will assume that Jane only has 5 chocolates to share among her friends

i. Assuming that the 5 chocolates are of the same type, if she doesn't want to give any friend more than one piece of chocolate, the situation here is said to be WITHOUT REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

(\left {n} \atop {k}} \right. )=(\left {10} \atop {5}} \right. )=252

ii. Assuming that the 5 pieces of chocolate are of the same type and she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

(\left {n+k-1} \atop {k}} \right. )=(\left {14} \atop {5}} \right. )=2002

iii. Assuming that the 5 pieces of chocolate are of different types and she isn't willing to give any friend more than one piece of chocolate, the situation is said to be WITHOUT REPLACEMENT & ORDERED

n = 10 & k = 5

Total number of possible combinations becomes,

P\left {n} \atop {k}} \right=P\left {10} \atop {5}} \right. =30240

iv. Assuming the the 5 pieces of chocolate are of different types, if she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & ORDERED

n = 10 AND k = 5

Total number of combinations becomes

n^k=10^5=100,000

Sampling with replacement means that one friend can be sampled more than once i.e: A friend receives a piece of chocolate more than once

The order dictates how the sampling is applied.

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Step-by-step explanation:

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3 years ago
A practice law exam has 100 questions, each with 5 possible choices. A student took the exam and received 13 out of 100.If the s
Cloud [144]

Answer:

z=\frac{13-20}{4}=-1.75

Assuming:

H0: \mu \geq 20

H1: \mu

p_v = P(Z

Step-by-step explanation:

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Let X the random variable of interest (number of correct answers in the test), on this case we now that:

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The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

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We need to check the conditions in order to use the normal approximation.

np=100*0.2=20 \geq 10

n(1-p)=20*(1-0.2)=16 \geq 10

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

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\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2(1-0.2)}=4

So we can approximate the random variable X like this:

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The z score is given by this formula:

z=\frac{x-\mu}{\sigma}

If we replace we got:

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We want to check is the score for the student is significantly less than the expected value using random guessing.

So on this case since we have the statistic we can calculate the p value on this way:

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3 years ago
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