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3 years ago

Please help asap (:

2 answers:

7 0

For this case we first define the variable:
x = number of terms.
The equation that models the problem is:
f (x) = 3.4 - 0.6x
We have then that the first four terms are:
x = 1
f (1) = 3.4 - 0.6 (1) = 3.4 - 0.6 = 2.8
x = 2
f (2) = 3.4 - 0.6 (2) = 3.4 - 1.2 = 2.2
x = 3
f (3) = 3.4 - 0.6 (3) = 3.4 - 1.8 = 1.6
x = 4
f (4) = 3.4 - 0.6 (4) = 3.4 - 2.4 = 1
Answer:
The rule for the sequence is:
f (x) = 3.4 - 0.6x
option 1

Margarita [4]3 years ago

4 0

F(1)=2.8, f(2)=2.2, f(3)=1.6, f(4)=1
f(2)-f(1)=2.2-2.8=-0.6
f(3)-f(2)=1.6-2.2=-0.6
f(4)-f(3)=1-1.6=-0.6
Then the function decreases 0.6 units by each unit increases x: Options 1 or 2

With the first function:
x=1→f(1)=3.4-0.6(1)=3.4-0.6→f(1)=2.8 OK

With the second function:
x=1→f(1)=2.8-0.6(1)=2.8-0.6→f(1)=2.2 NO

Answer: First option f(x)=3.4-0.6x

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Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Ro

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Answer:

(a) P(0 ≤ Z ≤ 2.87)=0.498

(b) P(0 ≤ Z ≤ 2)=0.477

(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.

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

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.

$P(|z|$