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alexandr1967 [171]
3 years ago
6

If P(not event B) = 0.95, what is P(event B)?

Mathematics
2 answers:
zysi [14]3 years ago
6 0

Answer:

0.05

Step-by-step explanation:

To find P(event b) you would do 1- P(not event B) which in this problem would be seen as 1-(0.95)= 0.05

Talja [164]3 years ago
5 0
<h3>Answer is   0.05</h3>

The two events of B happening, or B not happening, are complementary events. One or the other must happen. There is no overlap so both events cannot happen simultaneously. This means the two probabilities add to 1, which represents 100%

P(event B) + P(not event B) = 1

P(event B) = 1 - P(not event B)

P(event B) = 1 - 0.95

P(event B) = 0.05

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Diano4ka-milaya [45]

59,468

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4 years ago
the shop keeper at an inn receives 175 letters a week (7 days). How many letters does he receive in a day
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Step-by-step explanation:

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3 years ago
Read 2 more answers
Can someone explain this to me please
IrinaVladis [17]

Answer:

c. 36·x

Step-by-step explanation:

Part A

The details of the circle are;

The area of the circle, A = 12·π cm²

The diameter of the circle, d = \overline {AB}

Given that \overline {AB} is the diameter of the circle, we have;

The length of the arc AB = Half the the length of the circumference of the circle

Therefore, we have;

A = 12·π = π·d²/4 = π·\overline {AB}²/4

Therefore;

12 = \overline {AB}²/4

4 × 12 = \overline {AB}²

\overline {AB}² = 48

\overline {AB} = √48 = 4·√3

\overline {AB} = 4·√3

The circumference of the circle, C = π·d = π·\overline {AB}

Arc AB = Half the the length of the circumference of the circle = C/2

Arc AB = C/2 = π·\overline {AB}/2

\overline {AB} = 4·√3

∴ C/2 = π·4·√3/2 = 2·√3·π

The length of arc AB = 2·√3·π cm

Part B

The given parameters are;

The length of \overline {OF} = The length of \overline {FB}

Angle D = angle B

The radius of the circle = 6·x

The measure of arc EF = 60°

The required information = The perimeter of triangle DOB

We have;

Given that the base angles of the triangles DOB are equal, we have that ΔDOB is an isosceles triangle, therefore;

The length of \overline {OD} = The length of \overline {OB}

The length of \overline {OB} = \overline {OF} + \overline {FB} = \overline {OF} + \overline {OF} = 2 × \overline {OF}

∴ The length of \overline {OD} = 2 × \overline {OF} = The length of \overline {OB}

Given that arc EF = 60°, and the point 'O' is the center of the circle, we have;

∠EOF = The measure of arc EF = 60° = ∠DOB

Therefore, in ΔDOB, we have;

∠D + ∠B = 180° - ∠DOB = 180° - 60° = 120°

∵ ∠D = ∠B, we have;

∠D + ∠B = ∠D + ∠D = 2 × ∠D = 120°

∠D = ∠B = 120°/2 = 60°

All three interior angles of ΔDOB = 60°

∴ ΔDOB is an equilateral triangle and all sides of ΔDOB are equal

Therefore;

The length of \overline {OD} = The length of \overline {OB} = The length of \overline {DB}  = 2 × \overline {OF}

The perimeter of ΔDOB = The length of \overline {OD} + The length of \overline {OB} + The length of \overline {DB} = 2 × \overline {OF} + 2 × \overline {OF} + 2 × \overline {OF} = 6 × \overline {OF}

∴ The perimeter of ΔDOB = 6 × \overline {OF}

The radius of the circle = \overline {OF} = 6·x

∴ The perimeter of ΔDOB = 6 × 6·x = 36·x

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3 years ago
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scZoUnD [109]

Answer:

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