Simplify the expression to a + bi form

A clothing vendor estimates that 78 out of every 100 of its online customers do not live within 50 miles of one of its physical

Korolek [52]

Answer:

78% probability that a randomly selected online customer does not live within 50 miles of a physical store.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

Total outcomes:

100 customers

Desired outcomes:

A clothing vendor estimates that 78 out of every 100 of its online customers do not live within 50 miles of one of its physical stores. So the number of desired outcomes is 78 customers.

Using this estimate, what is the probability that a randomly selected online customer does not live within 50 miles of a physical store?

$p = \frac{78}{100} = 0.78$

78% probability that a randomly selected online customer does not live within 50 miles of a physical store.

4 0

3 years ago

Alex accidentally forgot to stock up on toilet paper before the stay-at-home order. Now he has to buy toilet paper on the black

gladu [14]

<h2>Alex accidentally forgot to stock up on toilet paper before the stay-at-home order. Now he has to buy toilet paper on the black market. Though the price of toilet paper on the black market has mostly stabilized, it still varies from day to day. The daily price of a generic brand 12-pack, X, and the daily price of a generic brand 6-pack, Y, (in rubles) jointly follow a bivariate normal distribution with: </h2><h2>μx = 2,470, σx = 30, μy = 1,250, σ = 25, p = 0.60. </h2><h2>(a) What is the probability that 2 (two) 6-packs cost more than 1 (one) 12-pack? (b) To ensure that he will not be without toilet paper ever again, Alex buys 7 (seven) 12-packs and 18 (eighteen) 6-packs. What is the probability that he paid more than 40,000 rubles? </h2><h2>(c) Suppose that today's price of a 12-pack is 2,460 rubles. What is the probability that a 6-pack costs less than 1,234 rubles today? [1 US dollar is approximately 75 rubles ]</h2>

5 0

3 years ago

The ages of students enrolled in two math classes at the local community college, Class A and Class B, are listed in order below

nlexa [21]

Answer:

The true statement about Class B is that Class B has a smaller median and the same inter quartile range.

Step-by-step explanation:

We are given the ages of students enrolled in two math classes at the local community college, Class A and Class B, below;

Class A: 20, 20, 20, 21, 22, 23, 23, 25, 27, 29, 30, 31, 34, 35, 36, 39, 40

Class B: 16, 17, 18, 18, 20, 22, 22, 24, 26, 26, 28, 29, 30, 34, 37, 40, 42

1) <u>Firstly, we will calculate Median for Class A;</u>

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 27

Hence, the median of class A is 27.

2) <u>Now, we will calculate Median for Class B;</u>

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 26

Hence, the median of class B is 26.

3) Now, we will calculate the Inter quartile range for Class A;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $21+ 0.5[22- 21]$

= 21.5

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $34+ 0.5[35- 34]$

= 34.5

Therefore, Inter quartile range for Class A = 34.5 - 21.5 = 13.

4) Now, we will calculate the Inter quartile range for Class B;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $18+ 0.5[20- 18]$

= 19

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $30+ 0.5[34- 30]$

= 32

Therefore, Inter quartile range for Class B = 32 - 19 = 13.

Hence, the true statement about Class B is that Class B has a smaller median and the same inter quartile range.

4 0

4 years ago

Which expression is negative choose 1 answer: A a+1 B a-(-b)

kiruha [24]

Answer:

B. a- (-b)

Step-by-step explanation:

6 0

3 years ago

Select the correct answer. Choose the system of inequalities that best matches the graph below. у 2 4 3 X Х -6 3 -2 -1 ON 2 -1 -

AnnZ [28]

Answer:

There is not graph

Step-by-step explanation:

8 0

3 years ago