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2 years ago

Suppose that an experiment is repeated four times. A certain event has probability 1/10 in a single repetition of the

Mathematics

1 answer:

frez [133]2 years ago

5 0

Answer:

2/5

Step-by-step explanation:

1/10 times 4

4/10 equals to 2/5

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The long division below shows the first term of the quotient. Which polynomial should be subtracted from the dividend first? x +

pychu [463]

Answer:

9x² - 6x

Step-by-step explanation:

6 0

2 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

For each sequence find the first 4 terms and the 10th term

jolli1 [7]

Answer:For these, n is equal to the term you want. So you substitute the number in for whichever term you want. This means:

a) 1 (1st term) + 5 = 6

2 (2nd term) + 5 = 7

3 (3rd term) + 5 = 8

4 (4th term) + 5 = 9

10 (10th term) + 5 = 15

And so on for b (not going to keep writing the term, I’m sure you get that by now.

b) 2(1) - 1 = (2x1) - 1 = 1

2(2) - 1 = (2x2) - 1 = 3

2(3) - 1 = (2x3) - 1 = 5

2(4) - 1 = (2x4) - 1 = 7

2(10) - 1 = (2x10) - 1 = 19

Hope this helps :)

4 0

3 years ago

QUICKKKKKKKK

zavuch27 [327]

See attached picture of graph

6 0

2 years ago

How do you simplify 15/40

Katyanochek1 [597]

Answer:3/8

Step-by-step explanation:

The simplest form of

15 /40 is 3 /8 .

Steps to simplifying fractions

Find the GCD (or HCF) of numerator and denominator

GCD of 15 and 40 is 5

Divide both the numerator and denominator by the GCD

15 ÷ 5

40 ÷ 5

Reduced fraction:

3 /8

Therefore, 15/40 simplified to lowest terms is 3/8.

3 0

2 years ago