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katen-ka-za [31]
3 years ago
10

Are two triangles ABC and A’B’C’ congruent ich hc = h‘c and sc = s‘c and alpha= alpha

Mathematics
1 answer:
vredina [299]3 years ago
4 0

Step-by-step explanation:

are two triangles ABC and A’B’C’ congruent ich hc = h‘c and sc = s‘c and alpha= alpha

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Two boats left a base camp at bearing of S60°E and S50°W at speeds of 50km/h and 70km/h respectively. Find the distance between
ivolga24 [154]

Answer:

  • 197.93 km

Step-by-step explanation:

Let the base camp is point A and boats' locations after two hours are points B and C.

By connecting the three points together we get a triangle ABC with sides:

  • AB = 50*2 = 100 km
  • AC = 70*2 = 140 km

The angle between AB and AC is:

  • 60 + 50 = 110 degrees (opposite directions from south)

We are looking for the distance BC, which can be found by using the law of cosines:

  • BC² = AB² + AC² - 2AB*AC*cos ∠BAC
  • BC² = 100² + 140² - 2*100*140*cos 110°
  • BC²  = 39176.56 (rounded)
  • BC = √39176.56 = 197.93 km (rounded)

The distance between the boats is 197.93 km.

7 0
2 years ago
Given triangle 'ABC' with a = 4, b = 11, and measurement of A = 41 Degrees , find the number of distinct solutions.
n200080 [17]
It should be A, since b=11> acosA
3 0
3 years ago
The television show CSI: Shoboygan has been successful for many years. That show recently had a share of 18, meaning that among
jeyben [28]

Answer:

(a) The value of P (None) is 0.062.

(b) The value of P(at least one) is 0.938.

(c) The value of P(at most one) is 0.253.

(d) The event is not unusual.

Step-by-step explanation:

Let <em>X</em> = number of households watching the show.

The probability of the random variable <em>x</em> is, P (X) = <em>p</em> = 0.18.

The sample selected for the survey is of size, <em>n</em> = 14

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 14 and <em>p</em> = 0.18.

The probability of a Binomial distribution is computed using the formula:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,...

(a)

Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:

P(X=0)={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}=1\times1\times0.06214=0.062

Thus, the value of P (None) is 0.062.

(b)

Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-0.062\\=0.938

Thus, the value of P(at least one) is 0.938.

(c)

Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

             ={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}+{14\choose 1}(0.18)^{1}(1-0.18)^{14-1}\\=0.062+0.191\\=0.253

Thus, the value of P(at most one) is 0.253.

(d)

An event that has a very low probability of occurrence is known as an unusual event.

The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.

This probability value is not low.

Hence, the event is not unusual.

5 0
3 years ago
A school baseball team earned $3416.90 from selling 5716 tickets to their game. If grandstand tickets sold for 65 cents each and
expeople1 [14]
If all were grandstand tickets, revenue would be 0.65*5716 = 3715.40. It was actually 298.50 less than that. Each bleacher ticket sold drops the revenue by .25, so there were 298.50/.25 = 1194 bleacher tickets sold.
3 0
3 years ago
Need this ASAP please help
Amanda [17]

Answer:

Step-by-step explanation:

b

7 0
3 years ago
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