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Nana76 [90]
3 years ago
11

Determine the value of the missing sides for the following triangle

Mathematics
1 answer:
Hunter-Best [27]3 years ago
3 0

Answer:

90-30 = 60 degree.

sin90/14 = sin60/m

m= 14sin60/sin90

m= 12.1

sin60/12 = sin30/sin90

n =12sin30/sin90

n= 6

Step-by-step explanation:

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3) Sue bought a triangular shaped piece of wood
olga_2 [115]

Answer:

12 inches of fabric

Step-by-step explanation:

Area of a triangle = hight x base

so 6 x 2 = 12

12 inches of fabric is your answer

Hope this helps :)

6 0
2 years ago
X-2y=0 ,x²-y²=3<br>by step​
masya89 [10]

Answer:

x=2, y=1

x=-2, y = -1

Step-by-step explanation:

x - 2y = 0

x = 2y

x² - y² = 3

(2y)² - y² = 3

3y² = 3

y² = 1

y = 1 or y = - 1

x = 2 or x = -2

5 0
3 years ago
PLEASE HELP QUICKLY!!
choli [55]

Answer:

By excluding (10, 15), a better description can be given for the data set.

Step-by-step explanation:

Overall, this data seems to suggest a decreasing trend.  The value (10, 15) is an outlier; if we remove it, the trend appears to be stronger.  Thus if we exclude this value we can give a better description of the data set.

6 0
3 years ago
Read 2 more answers
A student rolls a standard number cube. What is the probability the student rolled a 4, given that the number rolled was greater
Temka [501]

Answer:1/5

Step-by-step explanation:

8 0
3 years ago
An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

P (x≤100) = P(z = P(z<0.34) = 0.63307.

6 0
3 years ago
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