All categories

3 years ago

Analyze tables use the table that shows the percents of blood types of 145 donors during a recent blood drive

Mathematics

1 answer:

Alenkinab [10]3 years ago

6 0

Answer:

a. 16

b. 80

c. Blood type AB

Step-by-step explanation:

Given:

O = 45%

A = 40%

B = 11%

AB = 4%

Total number of donors = 145

a. No. of donors that had type B blood = $\frac{11}{100}*145 = \frac{1,595}{100} = 15.95$

Type B blood donors ≈ 16

b. No. of blood donors without blood type O = $\frac{40 + 11 + 4}{100}*145 = \frac{55}{100}*145 = \frac{7,975}{100} = 79.75$

Those without blood type O ≈ 80

c. Find the number of people that had type AB:

Type AB = $\frac{4}{100}*145 = 5.8$ ≈ 6. This is less than 10 donors.

Therefore, blood type AB had less than 10 donors.

You might be interested in

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

What is the equation to the line that goes through the point

DochEvi [55]

Step-by-step explanation:

perpendicular slope is negative reciprocal. -1/-1 = 1

y - 4 = x +3

y = x + 7

5 0

3 years ago

If m< BOE = 130 degrees and m< cob = 80 degrees

hjlf

Answer:

then what?

Step-by-step explanation:

do we have to find the angle?

all I know is every right triangle is equal to 90°

6 0

3 years ago

Estimate 73% of 120 <br><br> Thanks if you help!

xxMikexx [17]

Answer:

73/100 (120)

answer:-87.6

3 0

3 years ago

Read 2 more answers

The hockey team won 8 out of their first 14 games. At the same rate, how many games should they except to win out of 84

babunello [35]

The answer is 48.

8/14=x/84

Cross mutiply, to get:

14x=672

Divide each side by 14

You get 48.

Hope this help!!! :)

3 0

3 years ago