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3 years ago

Make x the subject PLEASE HELP ILL GIVE BRAINLIEST ASAP

Mathematics

1 answer:

Phoenix [80]3 years ago

3 0

Answer:

$x = \frac{(a+1)f^3}{2e}$

Step-by-step explanation:

$\frac{a + 1}{2x} = \frac{e}{f^3}\\~\\\\a+1 = \frac{2ex}{f^3}\\~\\\frac{a + 1}{2e} = \frac{x}{f^3}\\~\\x = \frac{(a+1)f^3}{2e}$

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A fast-food company is interested in knowing the probability of whether a customer viewed an advertisement for their new special

lord [1]

Answer:

45% probability that a randomly selected customer saw the advertisement on the internet or on television

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a customer saw the advertisement on the internet.

B is the probability that a customer saw the advertisement on television.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a customer saw the advertisement on the internet but not on television and $A \cap B$ is the probability that the customers saw the advertisement in both the internet and on television.

By the same logic, we have that:

$B = b + (A \cap B)$

12% saw it on both the internet and on television.

This means that $A \cap B = 0.12$

20% saw it on television

This means that $B = 0.2$

37% of customers saw the advertisement on the internet

This means that $A = 0.37$

What is the probability that a randomly selected customer saw the advertisement on the internet or on television

$A \cup B = A + B - (A \cap B) = 0.37 + 0.2 - 0.12 = 0.45$

45% probability that a randomly selected customer saw the advertisement on the internet or on television

6 0

3 years ago

For circle O, mCD=125° and m

Burka [1]

CA is a diameter of the circle, so $m\widehat{AC}=180^\circ$ , which means $m\angle AOB=m\angle AOD=m\widehat{AD}=180^\circ-m\widehat{CD}=55^\circ$ . Then $m\boxed{\angle ABO}=90^\circ-55^\circ=35^\circ$ .

This means $m\angle CBO=55^\circ-35^\circ=20^\circ$ . Also, if $m\angle AOB=55^\circ$ , then $m\angle BOC=180^\circ-55^\circ=125^\circ$ , which in turns tells us that $m\boxed{\angle BCO}=180^\circ-20^\circ-125^\circ=35^\circ$ .

3 0

3 years ago

How would you classify the decimal form of 7/25?

olga55 [171]

The answer is C to the question

8 0

3 years ago

Read 2 more answers

Help me please urgent

asambeis [7]

$Let \: x \: be \: the \: number \: of \: hours \\ Expession: - 4.5x = - 27\Leftrightarrow x = \frac{ - 27}{ - 4.5} = 6 \: hours$

7 0

3 years ago

How many solutions does the system of linear equations have?

sattari [20]

There is no solutions to these linear equations. in order to find how many solutions there are, you first need to solve for the first variable in one of the equations, then substitute the result into the other equation. since both equations are almost the same and the same format, it won't have any solutions.

hope this helped, God bless!

3 0

3 years ago