The last anwser option is : y = x + 5

Algebra problem<br> Pls help

Lina20 [59]

Answer:

The answer to your question is:

Shelly = 35

Michele = 32

Step-by-step explanation:

Shelly = S

Michele = M

Equations

S = M + 3

S + M = 67

M + 3 + M = 67

2M = 67 - 3

M = 64/2

M = 32

S = 67 - 32

S = 35

4 0

3 years ago

The probability is 0.4 that a traffic fatality involves an intoxicated or alcohol-impaired driver or nonoccupant. In 7 traffic

Luden [163]

Answer:

a.

$P(X=3)=0.2903\\\\P(X \geq 3)=0.5801\\\\P(X\leq 3)0.7102$

b.

$P(2\leq x\leq 4 )=0.7451$

c. mean=2.8

d . standard deviation=1.2961

Step-by-step explanation:

We determine that the accident rates follow a binomial distribution. The rate of success p=0.4 and sample n=7:

$P(x)={n\choose x}p^x(1-p)^{n-x}$

#the probability of exactly three;

$P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X=3)={7\choose 3}0.4^3(0.6)^{4}\\\\=0.2903$

#At least(more than 2)

$P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 3)=1-P(X\leq 2)\\\\=1-{7\choose 0}0.4^0(0.6)^{7}-{7\choose 1}0.4^1(0.6)^{6}-{7\choose 2}0.4^2(0.6)^{5}\\\\=1-0.0280-0.1306-0.2613\\\\=0.5801$

#At most 3;

$P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X \leq 3)={7\choose 0}0.4^0(0.6)^{7}+{7\choose 1}0.4^1(0.6)^{6}+{7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}\\\\\\\=0.0280+0.1306+0.2613+0.2903\\\\=0.7102$

b. Between 2 and 4:

Using the binomial expression, this probability is calculated as:

$P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(2\leq x\leq 4 )={7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}+{7\choose 4}0.4^4(0.6)^{3}\\\\\\\\\=0.2613+0.2903+0.1935\\\\=0.7451$