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2 years ago

If one angle of a triangle is 100 degrees and the other two angles are equal what is the measure of each unknown angle?

Mathematics

1 answer:

Shkiper50 [21]2 years ago

5 0

Each of the 2 angles would be 40

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Tatiana [17]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

1.) A tower casts a shadow that is 60 feet long when the angle

Fynjy0 [20]

Answer:

128.67 ft

Step-by-step explanation:

Let the height of the tower be x feet.

$\tan \: 65 \degree = \frac{x}{60} \\ \\ 2.1445069205 = \frac{x}{60} \\ \\ x = 2.1445069205 \times 60 \\ \\ x = 128.670415 \\ \\ x = 128.67 \: ft$

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2 years ago

4 × (28 ÷ 7)2 = <br> HURRYYYYY

umka21 [38]

The answer is 32

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3 years ago

Find two numbers that multiply to 17 and add to 18

Orlov [11]

Answer:

The numbers are 1 and 17

Step-by-step explanation:

The only factors of 17 are 1 and 17 since 17 is a prime number

1*17 =17

1+17 =18

6 0

3 years ago

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

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

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.

We apply the same formula but as a sum:

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

Then we have:

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

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

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

8 0

3 years ago