Verify [(cscx+secx)/(cotx+tanx)]=[cosx+sinx]. Show work and explain identities used.

An automobile manufacturer is concerned about a possible recall of its best selling four-door sedan. If there was a recall there

goldfiish [28.3K]

Complete question is;

An automobile manufacturer is concerned about a possible recall of its best - selling four-door sedan. If there were a recall, there is a probability of 0.25 of a defect in the brake system, 0.18 of a defect in the transmission, 0.17 of a defect in the fuel system, and 0.40 of a defect in some other area.

(a) What is the probability that the defect is the brakes or the fueling system if the probability of defects in both systems simultaneously is 0.15?

(b) What is the probability that there are no defects in either the brakes or the fueling system?

Answer:

A) 0.27

B) 0.73

Step-by-step explanation:

Let B denote that there is a defect in the break system

Let T demote that there is a defect in transmission

Let F denote that there is a defect in the fuel system.

Let P denote that there is a defect in some other area.

Now, we are given;

P(B) = 0.25

P(T) = 0.18

P(F) = 0.17

P(O) = 0.40

A) We are given that probability of defects in both brakes and the fueling system simultaneously is 0.15.

Thus, it means; P(B ⋂ F) = 0.15

Now, we want to find the probability that the defect is the brakes or the fueling system. This is expressed as;

P(B ⋃ F) = P(B) + P(F) - P(B ⋂ F)

Plugging in the relevant values to give;

P(B ⋃ F) = 0.25 + 0.17 - 0.15

P(B ⋃ F) = 0.27

B) We want to find the probability that there are no defects in either the brakes or the fueling system.

This is expressed as;

P(B' ⋃ F') = 1 - P(B ⋃ F)

Plugging in relevant value;

P(B' ⋃ F') = 1 - 0.27

P(B' ⋃ F') = 0.73

7 0

2 years ago

A movie theater sold 40 juice boxes. The cashier noticed that 75% of the boxes were apple juice.

lozanna [386]

30 of them would be apple juice boxes.

40 x 75% = 30

3 0

3 years ago

What pattern can be observed from the data?

marin [14]

I can’t see the diagram

5 0

2 years ago

Read 2 more answers

Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believe

Stolb23 [73]

Answer:

Step-by-step explanation:

The model $N (t)$ , the number of planets found up to time $t$ , as a Poisson process. So, the $N (t)$ has distribution of Poison distribution with parameter $(\lambda t)$

a)

The mean for a month is, $\lambda = \frac{1}{3}$ per month

$E[N(t)]= \lambda t\\\\=\frac{1}{3}(24)\\\\=8$

(Here. t = 24)

For Poisson process mean and variance are same,

$Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8$

(Poisson distribution mean and variance equal)

The standard deviation of the number of planets is,

$\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828$

b)

For the Poisson process the intervals between events(finding a new planet) have independent exponential distribution with parameter $\lambda$ . The sum of $K$ of these independent exponential has distribution Gamma $(K, \lambda)$ .

From the given information, $k = 6$ and $\lambda =\frac{1}{3}$

Calculate the expected value.

$E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}\\\\=\frac{6}{\frac{1}{3}}\\\\=18$

(Here, $\alpha =k$ and $\beta=\lambda$ )

C)

Calculate the probability that she will become eligible for the prize within one year.

Here, 1 year is equal to 12 months.

P(X ≤ 12) = (1/Г (k)λ^k)(x)^(k-1).(e)^(-x/λ)

$=\frac{1}{Г (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%$

Hence, the required probability is 0.2149 or 21.49%

5 0

2 years ago

Two lines intersect to form four angles. The measure of one angle is 27 degrees. Draw a diagram to show the measures of the othe

Snowcat [4.5K]

Since there is a 180 degrees in a straight line and we know that the intersection of one angle is 27 degrees we can can say 180 -27 = 153 degrees. So the four angles would be 27, 27 153, 153

3 0

3 years ago