All categories

2 years ago

Is six pen cost $2.10 what is the unit price of the pen?

Mathematics

2 answers:

Studentka2010 [4]2 years ago

6 0

Answer:

3.20

Step-by-step explanation:

Lemur [1.5K]2 years ago

3 0

Answer:

0.35

Step-by-step explanation:

i just divided 2.10/6

You might be interested in

Scupper Molly invested $1,800 semiannually for 23 years at 8% interest compounded semiannually. What is the value of this annuit

Shtirlitz [24]

Answer:

the amount of a $3,000.00 annuity due at 12 percent compounded semiannually for 3 years. A. $334,321.00 B. $43,068.93 C. $22,180.50 D. $41,814.50

3 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

Suppose you choose a marble from a bag containing 3 red marbles, 5 white marbles, and 4 blue marbles. You return the first marbl

Veseljchak [2.6K]

Answer:

P(red and blue) = 0.1667

Step-by-step explanation:

Let A represent choosing a red marble, there are 3 red marbles

Let B represent choosing a blue marble, there are 4 blue marbles

There are 12 total marbles

Then...

P(A) = 3/12

P(B) = 4/12

choosing the first one, then replacing means the first choice has no effect on the probabilities of the second choice, so the situation is independent.

When calculating the probability of independent events, you multiply the probabilities together. There are 2 scenarios where we can get a red and blue marble..

Choosing a red, then a blue marble, the probability is

(3/12)( 4/12) = 12/144 = 1/12

Choosing a blue, then a red marble, the probability is

(4/12)(3/12) = 12/144 = 1/12

So we have a

(1/12) + (1/12) = 2/12 = 1/6 chance for that to happen, to

P(A and B) = 1/6 = 0.1667

6 0

2 years ago

in a school there are 650 girls. It is 26% of the whole students, how many boys are there in the school?

worty [1.4K]

Answer:

Step-by-step explanation:

Frt7v6c87buhinjomp,l.;

8 0

2 years ago

What is 6/87 of 45638???????

lapo4ka [179]

Answer:

3,147.44828

Step-by-step explanation:

$\frac{6}{87} \: of \: 45638 \\ \\ = \frac{6}{87} \times 45638 \\ \\ = 6 \times 524.574713 \\ \\ = 3,147.44828$

5 0

2 years ago