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3 years ago

An experiment repeated 5 times is more reliable than an experimment repeated 20 times

Mathematics

1 answer:

shutvik [7]3 years ago

8 0

Answer:

The given statement is false

Step-by-step explanation:

Given that an experiment repeated 5 times is more reliable than an experiment repeated 20 times.

The statement is false.

Any result found by way of experiment will be more reliable if more number of trials are done.

For example, consider toss of a coin. If we try for 2 times, we may get prob for head as 0 sometimes 1/2 or sometimes 1. This is because 2 is very short number of trials to decide. But when you increase to 10, you may get more reliable. SO if number of repititions increase, the more reliable the results would be.

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How do you solve for (w-2)²-63=0 where w is a real number

Murrr4er [49]

(w-2)^2 = 63

FOIL left hand side:
(w-2)^2 = w^2 - 4w + 4

So,
w^2 - 4w + 4 = 63
w^2 - 4w - 59 = 0

Then use quadratic equation to get values of w = 2 + 3*root(7) and w = 2 - 3*root(7).
±
3
√
7

4 0

3 years ago

I need a-e to be solved I’m confused still

monitta

A) 36 means that there are 36 students taking Italian and music AT SAME TIME

b) 59 means that there are 59 students taking only music

c) 48 means that there are 48 students taking neither Italian nor music

d) Italian OR Music = 97 + 36 +59 = 192

e) Do not take Italian: 48 + 59 = 107

7 0

3 years ago

In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an

Airida [17]

This is a little long, but it gets you there.

ΔEBH ≅ ΔEBC . . . . HA theorem
EH ≅ EC . . . . . . . . . CPCTC
∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
ΔDAC ≅ ΔDAG . . . HA theorem
DC ≅ DG . . . . . . . . . CPCTC
∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
(∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)

5 0

3 years ago

12<br> Fill in the blank: What's the length of the missing<br> side?<br> Ilype your answer

Kisachek [45]

Answer:

Step-by-step explanation:

Using Pythagoras Theorem:

$a^2+b^2=c^2$

Substitute values of a and b:

$5^2+12^2=25+144=169$

$c^2=169\\c=13$

The length of the missing side is 13.

8 0

3 years ago

If the area of a park is exactly halfway between 2.4 and 2.5 acres what is the area of the park

Paul [167]

Get the distance between two acres.

2.5 - 2.4 = 0.1

Divide the diffence by 2
0.1 / 2 = 0.05 - Halfway between them.

Therefore, 0.05 must be taken from 2.5 acres. it must be added to 2.4 acres to get the exact area of the park.

2.5 - 0.05 = 2.45
2.4 + 0.04 = 2.45 acres

OR

Simply get the average of the two to get the area since it is exactly halfway between the two acres of land.

2.5 + 2.4 = 2.9 / 2 = 2.45 acres

8 0

3 years ago

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