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1 year ago

A student takes several measurements. how can the student determine which measurement is the least percise?

1 answer:

7 0

The student can compare all the measurements and whichever one is the most different out of all the measurements will be the least precise

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Please can anyone help answer this question? It would be greatly appreciated!!! :)

Ray Of Light [21]

Answer:

0.667

Step-by-step explanation:

Angle in a circle = 360°

Area of total shaded parts = 60 + 60 = 120°

Area of total unshaded parts = 360-120 = 240°

Probability that a random selected point within the circle falls in the unshaded area

= 240/360

= 2/3

≈ 0.667

I apologize in advance if I made a mistake.

7 0

3 years ago

Read 2 more answers

A building is 20 feet tall and casts a 12 foot shadow. and a person who is standing right next to the building casts 1 3 foot sh

Nesterboy [21]

I need help in this one too

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3 years ago

Can you help me pls, this doesn't make any sense

Gala2k [10]

Area of the larger square = 12² = 144 cm²
Area of the smaller square = 7² = 49 cm²
Area of the section = 144-49 = 95 cm²

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3 years ago

The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2 respectively. Let A de

antoniya [11.8K]

Answer:

a) $P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b) $P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c) $P(A') = 1-P(A) =1-0.4=0.6$

d) $P(A \cup B) =0.4 +0.8-0.2 =1.0$

e) The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

Step-by-step explanation:

We have the following space provided:

$S= [a,b,c,d,e]$

With the following probabilities:

$P(a) =0.1, P(b)=0.1, P(c) =0.2, P(d)=0.4, P(e)=0.2$

And we define the following events:

A= [a,b,c], B=[c,d,e]

For this case we can find the individual probabilities for A and B like this:

$P(A) = 0.1+0.1+0.2 = 0.4$

$P(B) =0.2+0.4+0.2=0.8$

Determine:

a. P(A)

$P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b. P(B)

$P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c. P(A’)

From definition of complement we have this:

$P(A') = 1-P(A) =1-0.4=0.6$

d. P(AUB)

Using the total law of probability we got:

$P(A \cup B) =P(A) +P(B)-P(A \cap B)$

For this case $P(A \cap B) = P(c) =0.2$ , so if we replace we got:

$P(A \cup B) =0.4 +0.8-0.2 =1.0$

e. P(AnB)

The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

8 0

3 years ago

Help me out please geometry

Tju [1.3M]

Try this option (see the attachment, the answers are marked by red colour), note, the value of the left angle in not visible (task no. 3) >> solution not given.

4 0

2 years ago