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3 years ago

10

i really need help with my math assignment it’s normal distributions and i need this to graduate tomorrow if anyone can help it

would be deeply appreciated

1 answer:

Vika [28.1K]3 years ago

7 0

Where’s the work you need help with :)

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In circle J with mZHIK = 54 and HJ = 13 units find area of sector

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I circled the formula for the area of a sector.

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3 years ago

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luda_lava [24]

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Complete the assignment on a separate sheet of paper<br><br> Please attach pictures of your work.

Irina18 [472]

Answer:

<u>TO FIND :-</u>

Length of all missing sides.

<u>FORMULAES TO KNOW BEFORE SOLVING :-</u>

$\sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}$
$\cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}$
$\tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}$

<u>SOLUTION :-</u>

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

$=> \sin 16 = \frac{7}{x}$

$=> \frac{7}{x} = 0.27563......$

$=> x = \frac{7}{0.27563....} = 25.39568.....$ ≈ 25.3

2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

$=> \sin 29 = \frac{6}{x}$

$=> \frac{6}{x} = 0.48480......$

$=> x = \frac{6}{0.48480....} = 12.37599.....$ ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

$=> \sin 30 = \frac{x}{11}$

$=> \frac{x}{11} = 0.5$

$=> x = 0.5 \times 11 = 5.5$

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

$=> \cos 43 = \frac{x}{12}$

$=> \frac{x}{12} = 0.73135......$

$=> x = 12 \times 0.73135.... = 8.77624....$ ≈ 8.8

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

$=> \cos 55 = \frac{x}{6}$

$=> \frac{x}{6} = 0.57357......$

$=> x = 6 \times 0.57357.... = 3.44145....$ ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

$=> \cos 73 = \frac{8}{x}$

$=> \frac{8}{x} = 0.29237......$

$=> x = \frac{8}{0.29237.....} = 27.36242.....$ ≈ 27.3

7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

$=> \tan 69 = \frac{12}{x}$

$=> \frac{12}{x} = 2.60508......$

$=> x = \frac{12}{2.60508....} = 4.60636....$ ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

$=> \tan 20 = \frac{11}{x}$

$=> \frac{11}{x} = 0.36397......$

$=> x = \frac{11}{0.36397....} =30.22225....$ ≈ 30.2

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3 years ago

Match the English word or phrase to its math equivalent.

nydimaria [60]

Sum is addition, less than is subtraction, equals is is, division is of and multiplication is per

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3 years ago

In the 2009-2010 school year in country A, there were 92,000 foreign students from country B. This number I 23% more than the nu

max2010maxim [7]

Answer:

$\approx74796$

Step-by-step explanation:

GIVEN: In the $2009-10$ school year in country A, there were $92,000$ foreign students from country B. This number is $23\%$ more than the number of students from country C.

TO FIND: How many foreign students were from country C.

SOLUTION:

Total foreign students from country B $=92000$

let the foreign students from country C be $x$

As the students from country B is $23\%$ more than the country C

Students from country B

$92000=\frac{123}{100}x$

$x=\frac{9200000}{123}$

$x\approx74796$

Hence the number of students from country C are $74796$

8 0

3 years ago