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3 years ago

2. What is the external probability of rolling a 4

Mathematics

1 answer:

Nataly_w [17]3 years ago

6 0

Answer:

2. 1- Experimental probability of rolling a 4 = 40%

3. 2- Theoretical probability is 3% greater than experimental probability.

Step-by-step explanation:

Experimental probability of rolling a 4 = 100 × $\frac{8}{20}$

= 100 × 0.4

= 40%

Experimental probability of getting at least one tail = $\frac{72}{100}$

= 0.72

Theoretical probability of getting at least one tail = $\frac{3}{4}$

= 0.75

Theoretical probability is 3% greater than experimental probability.

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Work out the volume of this sphere?

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Step-by-step explanation:

$v = \frac{4}{3} \pi {( \frac{19}{2} )}^{3} \\ v = \frac{4}{3} \pi \: 857.375 = 3591.364 \\ = 3591.4 \: {cm}^{3}$

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3 years ago

Luna collected two random samples of 100 citizens each, with the results listed below. She made the inference that the least pop

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Answer:

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Step-by-step explanation:

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2 years ago

The area of the square is 36 square cm. Find its side.

bonufazy [111]

Answer:

s = 6 cm

Step-by-step explanation:

the area (A) of a square is calculated as

A = s² ( s is the side length )

given A = 36 , then

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2 years ago

Read 2 more answers

Solve by substitution. Please show work -x+9y=-5 x-5y=1

kkurt [141]

Let's solve for x.

−x+9y=−5

Step 1: Add -9y to both sides.

−x+9y+−9y=−5+−9y

−x=−9y−5

Step 2: Divide both sides by -1.

−x ÷ −1 = −9y−5 ÷ −1

x=9y+5

THE ANSWER FOR THE OTHER EQUATION: (x-5y=1)

Let's solve for x.

x−5y=1

Step 1: Add 5y to both sides.

x−5y+5y=1+5y

x=5y+1

Answer:

x=5y+1

I hope this helped I was a little confused on what your problem meant...so if this is not what you asked for just lmk so I can fix it for you :)

3 0

2 years ago

5. A yacht is cruising at 20 knots at bearing of 185° when a 15 knot wind starts to blow at a bearing of 290°. Find the directio

kicyunya [14]

Answer:

Direction of the boat is 43.05° and the speed is 21.66 knot

Step-by-step explanation:

yacht speed = 20 knots

yacht bearing = 185° = 85° below the negative horizontal x-axis

wind speed = 15 knots

wind bearing = 290° = 20° above the negative horizontal x-axis

we find the x and y components of the boat velocities

for yacht,

x component = -20 cos 85° = -1.74 knots

y component = -20 sin 85° = -19.92 knots

for the wind,

x component = -15 cos 20° = -14.09 knots

y component = 15 sin 20° = 5.13 knots

total x component Vx = -1.74 + (-14.09) = -15.83 knots

total y component Vy = -19.92 + 5.13 = -14.79 knots

Resultant speed of the boat = $\sqrt{Vx^{2} + Vy^{2} }$

==> $\sqrt{15.83^{2} + 14.79^{2} }$ = 21.66 knot

direction of boat = $tan^{-1} \frac{Vy}{Vx}$

==> $tan^{-1} \frac{14.79}{15.83}$ = 43.05°

3 0

3 years ago