The ratio of the sides of a rectangle is 2/3.

Company A: A 2-column table with 5 rows. The first column is labeled hours with entries 5, 12, 20, 29, 42. The second column is

Gnom [1K]

Answer:

86 & tends to increase

Step-by-step explanation:

Goodluck :)

3 0

2 years ago

Rewrite the problem. Multiply and reduce if needed. Use the (') as the divisor li

Alik [6]

Answer:

oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

Step-by-step explanation:

8 0

3 years ago

Rogers Company reported net income of $35,000 for the year. During the year, accounts receivable increased by $7,000, accounts p

storchak [24]

Answer:

correct answer is option C.

Step-by-step explanation:

net income of a year = $35,000

accounts receivable is increased (AR)= $7,000

accounts payable decrease(AP) = $3,000

depreciation expense = $8,000

net cash provided = net income - increase in current asset (AR) - decrease in current asset (AP) +non cash flow

net cash provided = $35,000 -$7,000-$3,000+ $8,000

= $33000

hence the correct answer is option C.

4 0

4 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

The solutions to the inequality y<-x+1 are shaded on the graph. Which point is a solution?

suter [353]

Answer:
B) (3, –2)

Explanation:
The inequality is y ≤ –x + 1
There are two ways to do this. You can try the four options by seeing where they lie on the graph, or by inputting them into the inequality and seeing if they check out. I am going to do a bit of both.

I know that the solution cannot have two positive coordinates because the first quadrant is not part of the solution, so I won't guess A or C.
I'll try (3, –2) (which is option B).
On the graph, (3, –2) is on the line, which means it is part of the solution because the line is solid and the inequality is a greater than or equal to sign.
Try it in the inequality:
y ≤ –x + 1
–2 ≤ –3 + 1
–2 ≤ –2 yes this checks out.

9 0

4 years ago