Of the 90% students that mr. Landon teaches 30% of them received an A on the last math test how many students received an A

Suppose there is a pile of quarters, dimes, and pennies with a total value of $1.08

WINSTONCH [101]

Answer:

a)

if 1 quarter = $ 0.25

1 dime = $ 0.10

1 penny = $ 0.01

so to make the total of $1.08 and its is also required to calculate the number of each coins present without being able to make change for a dollar

therefore we say;

1 Quarter + 8 dimes + 3 penny

: ( 1 × 0.25 ) + ( 8 × 0.10 ) + ( 3 × 0.01 )

: 0.25 + 0.80 + 0.03 = $ 1.08

b)

Now if you have a 4 Quarters, you have change for $1.

If you have 5 dimes, you have change for 2 Quarters.

If you have nickel; one of those can combine with 2 dimes to have a change for a Quarter.

If you have 5 pennies, you have enough change for 1 nickel

Therefore

(4-1)×0.25 + (5-1)×0.1 + (0×0.05) + (5-1)×0.01 = x

(3 × 0.25) + ( 4 × 0.1) + 0 + ( 4 × 0.01) = x

x = 0.75 + 0.4 + 0.04

x = $ 1.19

PROVED

6 0

3 years ago

I'm stuck, plsss help ASAPP!!! :))

Bogdan [553]

Answer:

15 units

Step-by-step explanation:

$d \: = \sqrt{(x2 - x1) ^{2} + (y2 - y1) ^{2} }$

d = 14.866

d = 15

7 0

3 years ago

Can someone help? <br><br>I'll give brainliest to the right answer!

Serggg [28]

Answer:

c is the answer

Step-by-step explanation:

i ust know

7 0

3 years ago

If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi

Dmitry [639]

Answer:

The diagram for the question is missing, but I found an appropriate diagram fo the question:

Proof:

since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle

∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

∠DOP = 22.5°

∠PDO = 67.5°

∠ADO = 22.5°

∠AOD = 67.5°

Step-by-step explanation:

Given:

AB = CD = 297 mm

AD = BC = 210 mm

BCPO is a square

∴ BC = OP = CP = OB = 210mm

Solving for OC

OCB is a right anlgled triangle

using Pythagoras theorem

(Hypotenuse)² = Sum of square of the other two sides

(OC)² = (OB)² + (BC)²

(OC)² = 210² + 210²

(OC)² = 44100 + 44100

OC = √(88200

OC = 296.98 = 297

OC = 297mm

An isosceless tringle is a triangle that has two equal sides

Therefore for △OCD

CD = OC = 297mm; Hence, △OCD is an isosceless triangle.

The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles

Since BC = OB = 210mm

∠BCO = ∠BOC

since sum of angles in a triangle = 180°

∠BCO + ∠BOC + 90 = 180

(∠BCO + ∠BOC) = 180 - 90

(∠BCO + ∠BOC) = 90°

since ∠BCO = ∠BOC

∴ ∠BCO = ∠BOC = 90/2 = 45

∴ ∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

For ΔOPD

$Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}$

Note that DP = 297 - 210 = 87mm

∠PDO + ∠DOP + 90 = 180

∠PDO + 22.5 + 90 = 180

∠PDO = 180 - 90 - 22.5

∠PDO = 67.5°

∠ADO = 22.5° (alternate to ∠DOP)

∠AOD = 67.5° (Alternate to ∠PDO)

3 0

3 years ago

A fair coin is flipped twelve times. What is the probability of the coin landing tails up exactly nine times?

seraphim [82]

Answer:

$P\left(E\right)=\frac{55}{1024}$

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

2¹² = 4096

We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

$nCr=\frac{n!}{r!\left(n-r\right)!}$

Plug in n = 12 and r = 9

$=\frac{12!}{9!\left(12-9\right)!}$

$=\frac{12!}{9!\cdot \:3!}$

$=\frac{12\cdot \:11\cdot \:10}{3!}$ ∵ $\frac{12!}{9!}=12\cdot \:11\cdot \:10$

$=\frac{1320}{6}$ ∵ $3!\:=\:3\times 2\times 1=6$

$=220$

Thus, the probability will be:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$=\frac{220}{4096}$

$=\frac{55}{1024}$

Thus, the probability of the coin landing tails up exactly nine times will be:

$P\left(E\right)=\frac{55}{1024}$

4 0

3 years ago