Graph the solution set of the inequality 1/2y

vredina [299]

answer: 10/4 reduced is 2.5/1 or 2.5.

work: what we are doing is trying to divide ten into groups of four. how many times does 10 fit into four? well, look at the multiples of four, 4, 8, 12... we can see that 8 fits into ten, but 12 doesn't. that means two whole fours can fit into ten (since 4 x 2 = 8) so we still have a 2 left over from the ten. since 2 is one half of four, we can fit in another .5. so, ten can be fit into 2.5 groups. hope this helps, have a great day!

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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.

$P(|z|$