1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Crazy boy [7]
3 years ago
10

Alice's average mark for 4 Science exams is 68%. Can Alice score enough marks in a fifth exam so that her average mark increases

to 75?%? Give reasons.
Mathematics
1 answer:
Keith_Richards [23]3 years ago
7 0

Answer:

It is not possible for Alice to score an average of 75% because what she needs to score in the fifth exam exceeds the maximum possible score.

Step-by-step explanation:

She scored an average 68% in all 4 exams.

This means; (68% + 68% + 68% + 68%)/4

Now, a fifth exam is added and we want to know if her average can now be 75%.

Thus, let the score in the 5th exam be x and we now have;

(68% + 68% + 68% + 68% + x)/5 = 75%

Multiply both sides by 5 to get;

(68% + 68% + 68% + 68% + x) = 375%

272% + x = 375%

x = 375% - 272%

x = 103%

The maximum score is 100% and therefore it is impossible for her to score 103%. Therefore it is not possible for Alice to score an average of 75%

You might be interested in
Can you help me with this?
amid [387]
The answer is 1:60 no it’s wrong
4 0
3 years ago
Condense and simplify, if possible: 2 In 3 + 3 ln x – 2 In y
elixir [45]

Answer:

Step-by-step explanation:

4 0
3 years ago
Is a measure of dispersion that indicates how much scores in a sample vary around the mean of the sample?
Anika [276]
<span>The measure of dispersion that indicates how much scores in a sample vary around the mean of the sample is the standard deviation.</span>
5 0
3 years ago
Read 2 more answers
7. The probability of passing an
Citrus2011 [14]

Answer:

Option B

Step-by-step explanation:

Total Probability = 1

Probability of passing = 0.77

Probability of failing = 1-0.77

=> 0.23

3 0
3 years ago
Please help me to prove this!​
Ymorist [56]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π              → A + B = π - C

                                              → B + C = π - A

                                              → C + A = π - B

                                              → C = π - (B +  C)

Use Sum to Product Identity:  cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Use the Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → Middle:</u>

\text{LHS:}\qquad \qquad \cos \bigg(\dfrac{A}{2}\bigg)+\cos \bigg(\dfrac{B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)

\text{Sum to Product:}\qquad 2\cos \bigg(\dfrac{\frac{A}{2}+\frac{B}{2}}{2}\bigg)\cdot \cos \bigg(\dfrac{\frac{A}{2}-\frac{B}{2}}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)

\text{Given:}\qquad \quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)

\text{Sum/Difference:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)

\text{Double Angle:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{2(A+B)}{2(2)}\bigg)\\\\\\.\qquad \qquad  \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)

\text{Factor:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]

\text{Cofunction:}\quad  =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]

\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)

\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{\pi -C}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)

LHS = Middle \checkmark

<u>Proof Middle → RHS:</u>

\text{Middle:}\qquad 4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)\\\\\\\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)

Middle = RHS \checkmark

3 0
3 years ago
Other questions:
  • What is the measure of angle c in degrees? *<br> 150<br> 30<br> 90<br> 60
    10·1 answer
  • Inside a square with side length 10, two congruent equilateral triangles are drawn such that they share one side and each has on
    15·1 answer
  • Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
    7·1 answer
  • Next three terms in (1024, -128, 16...) you can write them as fraction if needed
    15·1 answer
  • Help me pls it is on a time i got 30 minutes left
    12·2 answers
  • Does anyone know the exponential expression of this question if so why is my choice answer wrong ?
    15·1 answer
  • Give the slope and the y-intercept of the line y = – 6x– 4. Make sure the y-intercept is written as
    9·1 answer
  • Rewrite the number in standard notation.1 times 10^8 power
    10·2 answers
  • Siobhan needs a $7,000 personal loan.
    12·2 answers
  • What value of x makes the equation -5x - (-7 - 4x) = -2(3x - 4) tru
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!