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3 years ago

A dozen fidget spinners cost 9.60 what is the unit rate of the spinners

1 answer:

8 0

80 cents. To find the unit rate you need to know how much 1 fidget spinner costs, when all the fidget spinners cost the same amount. To this you have to divide 9.6 (total cost) by 12 (total # of fidget spinners). You get .8 (the cost of 1 fidget spinner).

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What is the value of X in the triangle below to the nearest degree? The triangle is not drawn to scale. I selected an answer but

11Alexandr11 [23.1K]

Answer:

x = 25

Step-by-step explanation:

To find the value of x we use the formula:

tan x° = opposite side /adjacent side

Opposite Side = 8

Adjacent Side = 17

tan x° = 8/17

tan x° = 0.4705882352941

$Tan^{-1}$ of $0.4705882352941$

x° = 25

4 0

3 years ago

Choice the correct answer

PSYCHO15rus [73]

Answer:

wheres A

Step-by-step explanation:

8 0

2 years ago

he closing price of a share of stock in Company XYZ is $25.69 on Thursday. If the change from the closing price on Wednesday is

mamaluj [8]

Answer:

Option C - $26.44

Step-by-step explanation:

To find the closing proce on Wednesday

Company XYZ is $ 25.69 on Thursday

Company XYZ is - $0.75 on Wednesday

This means the calculation should be as following,

25.69 + 0.75 = 26.44$ the closing price for Wednesday

6 0

3 years ago

Write the statement as an absolute value equation or inequality n is no more than 9 units from 7

Ipatiy [6.2K]

Its asking to find an answer between 9 and 7 so the answer would be 8

8 0

3 years ago

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = The email is valid.

The information provided to us are:

The probability of the word free occurring in a spam message is, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0

3 years ago