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Lapatulllka [165]
3 years ago
10

If Kate has an apple, an orange, a pear, a banana, and a kiwi at home and she wants to bring three pieces of fruit to school, ho

w many combinations of fruit can she bring?
a. 5!
b. 3!
c. 125
Mathematics
2 answers:
qaws [65]3 years ago
8 0

Step-by-step explanation:

i think A is correct

hope it helps

Zepler [3.9K]3 years ago
3 0
I think it’s A sorry if i’m wrong goodluck though
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What is x^2 -7x + 12 = 0 what is using method of factoring
statuscvo [17]
x^2 -7x + 12 = 0 \\ \\ (x - 4)(x - 3) = 0 \\ \\ x = 4, 3 \\ \\

The final result is: x = 4, 3.
6 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
94 plus 20 subract 8
Marina86 [1]
The answer is 106 your welcome :-)
7 0
3 years ago
Read 2 more answers
What is a dot product.
kifflom [539]

Answer:

In mathematics, the dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. ... Algebraically, the dot product is the sum of the products of the corresponding entries of the two sequences of numbers

kamsahamnida have a great day

7 0
3 years ago
Read 2 more answers
At a Psychology final exam, the scores are normally distributed with a mean 73 points and a standard deviation of 10.6 points. T
USPshnik [31]

Answer:

The score that separates the lower 5% of the class from the rest of the class is 55.6.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

\mu = 73, \sigma = 10.6

Find the score that separates the lower 5% of the class from the rest of the class.

This score is the 5th percentile, which is X when Z has a pvalue of 0.05. So it is X when Z = -1.645.

Z = \frac{X - \mu}{\sigma}

-1.645 = \frac{X - 73}{10.6}

X - 73 = -1.645*10.6

X = 55.6

The score that separates the lower 5% of the class from the rest of the class is 55.6.

3 0
3 years ago
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