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s2008m [1.1K]
3 years ago
10

In ΔGHI, h = 760 inches, ∠G=43° and ∠H=26°. Find the length of i, to the nearest inch.

Mathematics
1 answer:
Gwar [14]3 years ago
5 0

Answer:

1619

Step-by-step explanation:

I = 180 - 43 - 26 = 111

using sine rule

\frac{i}{sin 111} = \frac{760}{sin 26}

i = 1618.5 = 1619

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X-6/2=2x/7 solve the equation​
Triss [41]

Answer:

x-6/2=2x/7

7x-42=4x

7x-4x=42

3x= 42

X = 42/3

3 0
3 years ago
X^2+10x+19=0 complete the square PLEASE EXPLAIN
wlad13 [49]

Answer:

Step-by-step explanation

In the form ax^2+bx+c

a x c should equal two factors of b

However, in this equation

1 x 19 has no factors that = 10

so you must use the Quadratic Formula which is,

x = \frac{-b+-\sqrt{b^2-4ac} }{2a}

so the answer should be

x = \frac{-10+-\sqrt{10^2-4(1)(19)} }{2(1)}

x = \frac{-10+-\sqrt{24} }{2}

x = \frac{-10+2\sqrt{6} }{2}      and    x = \frac{-10-2\sqrt{6} }{2}

8 0
3 years ago
Which relation is a direct variation that contains the ordered pair (2, 7)?
Tasya [4]
Lets write equation of a function:

y = kx + n

Direct variation in simple is equation of a line which has n=0 or in other words which y to x ratio is k.

First option gets 7=7 but it isnt direct variation because n is not equal to 0

third option is indeed correct. once we implement coordinates (2,7) we get 7=7

Answer is 
y = 7/2x
8 0
3 years ago
Write the equation of the line in fully simplified slope-intercept form.
Alona [7]
Y=3/1x+-1 hope this is correct
5 0
3 years ago
A recent study found that the average length of caterpillars was 2.8 centimeters with a
pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

\mu = 2.8, \sigma = 0.7.

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{4 - 2.8}{0.7}

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0
2 years ago
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