HELP ME PLEASE!!! ASAP THANK YOU SO MUCH

butalik [34]

Answer: 11 (i think )

hope this helps

7 0

3 years ago

Product A is a 12-ounce bottle of generic mouthwash that sells for $1.15. Product B is a 24-ounce bottle of mouthwash that costs

adoni [48]

12 oz = 1.15
24 oz = 2.79
Let's see if 2x12oz is equal, less, or more than the 24oz.
1.15 x 2 = 2.30
2.30<2.79
The 12 oz is a better buy.

3 0

3 years ago

Consider the diagram.

adelina 88 [10]

Answer:

skew.

Step-by-step explanation:

Lines e and c are skew lines.

<u>Reason:</u>

They do not intersect with each other.

They are in different planes.

They are nor parallel to one another....

4 0

3 years ago

Read 2 more answers

A car rental company rents cars for x dollars per day plus y dollars for each mile driven. Carlos rented a car for 4 days, drove

topjm [15]

Answer:

x = $20/day, y = $0.35/mile

Step-by-step explanation:

x = $/day, y = $/mile

4x + 160y = 136

1x + 240y = 104

You can solve by elimination or substitution.

4 0

3 years ago

A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after

Harrizon [31]

Answer:

(a) The probability that a randomly selected parcel arrived late is 0.026.

(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

Step-by-step explanation:

Consider the tree diagram below.

(a)

The law of total probability sates that: $P(A)=P(A|B)P(B)+P(A|B')P(B')$

Use the law of total probability to determine the probability of a parcel being late.

$P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026$

Thus, the probability that a randomly selected parcel arrived late is 0.026.

(b)

The conditional probability of an event A provided that another event B has already occurred is:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

$P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

4 0

3 years ago