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Alexeev081 [22]
3 years ago
10

Kaylin buys a greeting card for $3.79. She then buys 4 postcards that all cost the same amount. The total cost is $5.11. How muc

h is each postcard?
Mathematics
2 answers:
Maksim231197 [3]3 years ago
6 0

Answer:

$0.33

Step-by-step explanation:

5.11-3.79=1.32 1.32/4

Hope this helps :)

Scrat [10]3 years ago
4 0

Answer:

each postcard cost .33 or 33 cent

Step-by-step explanation:

5.11(total cost)-3.79(greeting card)= 1.32

1.32(cost of all postcards)/4= .33

You might be interested in
2.
Fantom [35]

Answer:

the mode remains the same

Step-by-step explanation:

42 is outlier.

mod is the most repetitive number of a number string, and outlier does not change it.

the mode remains the same.

7 0
3 years ago
Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving th
DedPeter [7]

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that A \cap B = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

A + (A \cap B) = 0.40

A + 0.17 = 0.40

A = 0.23

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

B + (A \cap B) = 0.35

B + 0.17 = 0.35

B = 0.18

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

5 0
3 years ago
(1,3) (-1,0) and (5,0) (3,-3) parallel or perpendicular
nalin [4]
Parallel bc the slopes the same!
4 0
3 years ago
What number equals 2 4/5?<br> O 0.28<br> O 0.357<br> O 2.45<br> O 2.8
Travka [436]

I think its 2.8.

Reason why:

4/5 = 0.8

Then the 2 would a whole number...

so 2.8.

5 0
3 years ago
Factor the expression.
valentinak56 [21]
It is (d+5)(d+5) because 5•5 is 25 and 5+5 is 10
4 0
3 years ago
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