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ki77a [65]
9 months ago
12

Could the solutions of a system of inequalities be a rectangular region? If so, give an example.

Mathematics
2 answers:
Flura [38]9 months ago
7 0

Answer: i dont know

thats the same question i have

Step-by-step explanation:

Arisa [49]9 months ago
4 0
The person above is correct
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What is the value of y? Enter your answer in the box. y = An isosceles triangle with vertices labeled A, B, and C. Side B C is t
masha68 [24]

Answer:

y = 5

Step-by-step explanation:

Given:

An isosceles triangle ABC with AB = AC.

∠A = (4y+10)\°

∠B = 75\°

∠C = (3x)\°

We know that, for an isosceles triangle, the angles opposite equal sides are also equal to each other.

Therefore, ∠B = ∠C

⇒ 75 = 3x

⇒ x=\frac{75}{3}=25

∴ ∠C = 3\times 25=75°

Now, the sum of all the interior angles of a triangle is always 180°.

\angle A + \angle B + \angle C=180\\(4y+10)+75+75=180\\4y+10+150=180\\4y+160=180\\4y=180-160\\4y=20\\y=\frac{20}{4}=5

Therefore, the value of 'y' is 5.

5 0
2 years ago
Find the 23rd term of the arithmetic sequence 17, 19, 21, 23, ...
olga55 [171]

Answer:

260 ok

Step-by-step explanation:

4 0
2 years ago
Find the product (-2)(3)
Umnica [9.8K]

Answer:

Multiply -2 with 3

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Find the latitude and longitude of the location on Earth precisely opposite of a town located at latitude 41 degrees S​, longitu
hjlf

Answer:

41° N 56° W

Step-by-step explanation:

For the latitude, change the south to north.

41° N

For the longitude, subtract from 180 and change east to west.

(180° − 124°) W

56° W

4 0
3 years ago
Vehicles entering an intersection from the east are equally likely to turn left, turn right, or proceed straight ahead. If 50 ve
ira [324]

Answer:

The probability that at least two-third of vehicles in the sample turn is 0.4207.

Step-by-step explanation:

Let <em>X</em> = number of vehicles that turn left or right.

The proportion of the vehicles that turn is, <em>p</em> = 2/3.

The nest <em>n</em> = 50 vehicles entering this intersection from the east, is observed.

Any vehicle taking a turn is independent of others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 50 and <em>p</em> = 2/3.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>X</em> if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

np=50\times \frac{2}{3}=33.333>10\\\\n(1-p)=50\times \frac{1}{3}= = 16.667>10

Thus, a Normal approximation to binomial can be applied.

So, X\sim N(np, np(1-p))

Compute the probability that at least two-third of vehicles in the sample turn as follows:

P(X\geq \frac{2}{3}\times 50)=P(X\geq 33.333)=P(X\geq 34)

                        =P(\frac{X-\mu}{\sigma}>\frac{34-33.333}{\sqrt{50\times \frac{2}{3}\times\frac {1}{3}}})

                        =P(Z>0.20)\\=1-P(Z

Thus, the probability that at least two-third of vehicles in the sample turn is 0.4207.

6 0
3 years ago
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