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Annette [7]
3 years ago
9

mr. cuddy draws a triangle with a perimeter of 36cm. Principal Aranda says that the longest side measures 18 cm, How do you know

that the principal Aranda is incorrect? Explain
Mathematics
1 answer:
hichkok12 [17]3 years ago
3 0

Answer:

See the procedure

Step-by-step explanation:

we know that

<u>The Triangle Inequality Theorem</u>, states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

Let

a,b,c the lengths side of triangle

c is the greater side

The perimeter is equal to

P=a+b+c

P=36 cm

If c=18 cm

then

a+b=18

Applying the Triangle Inequality Theorem

a+b > c

18 > 18  ----> is not true

therefore

Principal Aranda is incorrect

The larger side cannot measure 18 cm

The largest side must be less than 18 cm

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A new truck costs $32,000 if the truck is worth 24,500 after three years write an explicit formula for the value of the car afte
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Answer:

Step-by-step explanation:

depreciation for three years = 32000 - 24500 = $ 7500

depreciation per year = 7500/3 = $ 2500

depreciation % = (2500/32000)*100 = 7.81%

Value of truck after n years = 32000 * (7.81%)*n

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An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de
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Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

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This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

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P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}

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P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have

P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453

The sum of the probabilities is

P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981

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