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baherus [9]
3 years ago
7

Roberto will roll a number cube and flip a coin. The faces of the number cube are labeled 1 through 6. The coin can land on head

s or tails. What is the probability that the number cube lands on an even number and the coin lands on tails?
Mathematics
1 answer:
Lyrx [107]3 years ago
6 0

Answer: There are 3 even numbers on the cube. There are 2 sides to a coin. there is a 1/2 chance of the cube landing on an even number. There is a 1/2 chance of the coin landing on tails.

There is a 1/4 chance of these events occurring together.

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I need help I don't understand this.
Ludmilka [50]

Answer:

gmvlɛpuyhgn

gteeyolpp

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6 0
3 years ago
A spherical balloon is being deflated at a rate of 5 cubic centimeters per second. At what rate is the radius of the balloon cha
Dvinal [7]

Answer:

0.0048cm/s

Step-by-step explanation:

Volume of the spherical balloon is expressed as;

V = 4/3 \pi r^3\\

dV/dt = dV/dr * dr/dt

Given

dV/dt = 5cm³/s

dV/dr = 4πr²

Since V = 972picm³

972π = 4/3πr³

972 = 4/3r³

4r³ = 972 * 3

r³ = (972 *3)/4

r³ = 729

r = ∛729

r = 9cm

dV/dr = 4π(9)²

dV/dr = 324π

dV/dt = dV/dr * dr/dt

5 = 324πdr/dt

dr/dt = 5/324π

dr/dt = 5/324(3.14)

dr/dt = 5/1017.36

dr/dt = 0.0048cm/s

5 0
3 years ago
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Ro
puteri [66]

Answer:

(a) P(0 ≤ Z ≤ 2.87)=0.498

(b) P(0 ≤ Z ≤ 2)=0.477

(c) P(−2.20 ≤ Z ≤ 0)=0.486

(d) P(−2.20 ≤ Z ≤ 2.20)=0.972

(e) P(Z ≤ 1.01)=0.844

(f) P(−1.95 ≤ Z)=0.974

(g) P(−1.20 ≤ Z ≤ 2.00)=0.862

(h) P(1.01 ≤ Z ≤ 2.50)=0.150

(i) P(1.20 ≤ Z)=0.115

(j) P(|Z| ≤ 2.50)=0.988

Step-by-step explanation:

(a) P(0 ≤ Z ≤ 2.87)

In this case, this is equal to the difference between P(z<2.87) and P(z<0). The last term is substracting because is the area under the curve that is included in P(z<2.87) but does not correspond because the other condition is that z>0.

P(0 \leq z \leq 2.87)= P(z

(b) P(0 ≤ Z ≤ 2)

This is the same case as point a.

P(0 \leq z \leq 2)= P(z

(c) P(−2.20 ≤ Z ≤ 0)

This is the same case as point a.

P(-2.2 \leq z \leq 0)= P(z

(d) P(−2.20 ≤ Z ≤ 2.20)

This is the same case as point a.

P(-2.2 \leq z \leq 2.2)= P(z

(e) P(Z ≤ 1.01)

This can be calculated simply as the area under the curve for z from -infinity to z=1.01.

P(z\leq1.01)=0.844

(f) P(−1.95 ≤ Z)

This is best expressed as P(z≥-1.95), and is calculated as the area under the curve that goes from z=-1.95 to infininity.

It also can be calculated, thanks to the symmetry in z=0 of the standard normal distribution, as P(z≥-1.95)=P(z≤1.95).

P(z\geq -1.95)=0.974

(g) P(−1.20 ≤ Z ≤ 2.00)

This is the same case as point a.

P(-1.20 \leq z \leq 2.00)= P(z

(h) P(1.01 ≤ Z ≤ 2.50)

This is the same case as point a.

P(1.01 \leq z \leq 2.50)= P(z

(i) P(1.20 ≤ Z)

This is the same case as point f.

P(z\geq 1.20)=0.115

(j) P(|Z| ≤ 2.50)

In this case, the z is expressed in absolute value. If z is positive, it has to be under 2.5. If z is negative, it means it has to be over -2.5. So this probability is translated to P|Z| < 2.50)=P(-2.5<z<2.5) and then solved from there like in point a.

P(|z|

7 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!!!!
erastova [34]

Answer:

B = (3, 3)

Step-by-step explanation:

Since the line segment is fifteen units long, from the looks of it, chopping ⅓ of A's side, led to the ordered pair of (3, 3).

I hope this helps you out alot, and as always, I am joyous to assist anyone at any time.

7 0
3 years ago
Can someone help me with this
Alexxandr [17]

Answer:

B

Step-by-step explanation:

the perimeter (P) of a rectangle is calculated as

P = 2l + 2w ( l is the length and w the width ) , then

P = 2(3\sqrt{15} ) + 2(\sqrt{15} ) ← distribute both parenthesis )

   = 6\sqrt{15} + 2\sqrt{15}

   = 8\sqrt{15}

5 0
2 years ago
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