1=5 2=25 3=325 4=4325 6=?

For each sequence shown find the growth factor or rate of change. 5, 15, 25, 35, 45

olga nikolaevna [1]

The numbers are increasing by positive 10 each time. So the rate of change is 10.

6 0

3 years ago

Sales at LL Boutique decreased 10% this month compared to last month. If sales this month were $103,581, what were the sales (in

Art [367]

Answer:

Step-by-step explanation:

(103,581) • (%110)

= (103,581) • (1.10)

= $113,939.1

7 0

3 years ago

Read 2 more answers

Write an equation. Mark buys a new iPad for $670. He makes a downpayment of $100 and pays the rest in 12 equal payments, p. What

sweet-ann [11.9K]

Step-by-step explanation: 670-100=570. 570/12

4 0

3 years ago

Read 2 more answers

Use the drawing tools to form the correct answer on the provided graph.

4vir4ik [10]

Answer: (0,1), (1,0)

Step-by-step explanation:

plug in a number for y.

ex) x=0 y=1

y=-x + 1

1= -0 + 1

1=0 + 1

5 0

2 years ago

A salesperson contacts eight potential customers per day. From past experience, we know that the probability of a potential cust

AlexFokin [52]

Answer:

(a) The probability the salesperson will make exactly two sales in a day is 0.1488.

(b) The probability the salesperson will make at least two sales in a day is 0.1869.

(c) The percentage of days the salesperson does not makes a sale is 43.05%.

(d) The expected number of sales per day is 0.80.

Step-by-step explanation:

Let X = number of sales made by the salesperson.

The probability that a potential customer makes a purchase is 0.10.

The salesperson contacts n = 8 potential customers per day.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={8\choose x}0.10^{x}(1-0.10)^{8-x};\ x=0,1,2,3...$

(a)

Compute the probability the salesperson will make exactly two sales in a day as follows:

$P(X=2)={8\choose 2}0.10^{2}(1-0.10)^{8-2}\\=28\times 0.01\times 0.5314\\=0.1488$

Thus, the probability the salesperson will make exactly two sales in a day is 0.1488.

(b)

Compute the probability the salesperson will make at least two sales in a day as follows:

P (X ≥ 2) = 1 - P (X < 2)

= 1 - P (X = 0) - P (X = 1)

$=1-{8\choose 0}0.10^{0}(1-0.10)^{8-0}-{8\choose 1}0.10^{1}(1-0.10)^{8-1}\\=1-0.4305-0.3826\\=0.1869$

Thus, the probability the salesperson will make at least two sales in a day is 0.1869.

(c)

Compute the probability that a salesperson does not makes a sale is:

$P(X=0)={8\choose 0}0.10^{0}(1-0.10)^{8-0}\\=8\times 1\times 0.4305\\=0.4305$

The percentage of days the salesperson does not makes a sale is,

0.4305 × 100 = 43.05%

Thus, the percentage of days the salesperson does not makes a sale is 43.05%.

(d)

Compute the expected number of sales per day as follows:

$E(X)=np=8\times 0.10=0.80$

Thus, the expected number of sales per day is 0.80.

7 0

3 years ago