Determine which box each reason should go in to justify each step taken to solve

The temperature in degrees Celsius, c, can be converted to degrees Fahrenheit, f, using the equation f-c+32. Which

Basile [38]

It is a function because every Celsius temperature is associated with only one Fahrenheit temperature. Therefore, option B is true.

<h3>How to convert Celsius into Fahrenheit ?</h3>

We are given that the temperature in degrees Celsius C can be converted to degrees Fahrenheit F

We know that the formula

$\dfrac{C}{5} = \dfrac{F - 32}{9}$

If we are substituting c = -40

Then, we get

$\dfrac{C}{5} = \dfrac{F - 32}{9}\\\\\dfrac{-40}{5} = \dfrac{F - 32}{9}\\\\-8 \times 9 = F - 32\\\\F = -72 + 32\\\\F = -40$

For every Celsius temperature, we get only one value of Fahrenheit.

The function is one-one because for every value of Celsius we get a unique value of Fahrenheit.

Hence, it is a function because every Celsius temperature is associated with only one Fahrenheit temperature.

Therefore, option B is true.

Learn more about temperature scale here:

brainly.com/question/2919967

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7 0

2 years ago

I mark brainliest!! fill the table using the function rule, y=-10x+3

alexdok [17]

The answers are
13
3
-7
-47

8 0

3 years ago

3.- In a certain desert region the average number of persons who become seriously ill each year from eating a certain poisonous

Naily [24]

Answer:

$P(X\geq 5)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662$

$P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634$

$P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029$

$P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595$

$P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151$

And replacing we got:

$P(X \geq 5) =0.76493$

Step-by-step explanation:

Previous concepts

Let X the random variable that represent the number of people that will become sereiosly ill in two years. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this case the value for $\lambda$ would be:

$\lambda = 3.2 \frac{ills}{year} *2 years = 6.4$

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 5)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662$

$P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634$

$P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029$

$P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595$

$P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151$

And replacing we got:

$P(X \geq 5) =0.76493$

7 0

3 years ago

Read 2 more answers

A plane is flying at a constant altitude. The angle of depression from the center of the plane to the base of a control tower is

madam [21]

Answer:

13.16 miles

Step-by-step explanation:

The attached diagram is a scale representation of the geometry, but shown upside down from the figure given in the problem statement. The tangent ratio is useful for solving this problem, as it relates the legs of the triangle to the angle of interest.

The basic relation is ...

Tan = Opposite/Adjacent

Applying this to the given angles, we have ...

tan(36°) = EG/AG . . . . . [eq1]

tan(43°) = EG/DG = EG/(AG -4) . . . . . [eq2]

Multiplying both equations by their denominators, we can eliminate the fractions:

AG·tan(36°) = EG . . . . . [eq3]

(AG -4)·tan(43°) = EG . . . . . [eq4]

Equating expressions for EG, we have ...

AG·tan(36°) = (AG -4)·tan(43°)

4·tan(43°) = AG(tan(43°) -tan(36°))

AG = 4·tan(43°)/(tan(43°) -tan(36°)) . . . . . [eq5]

Substituting into [eq3], we find ...

EG = 4·tan(36°)·tan(43°)/(tan(43°) -tan(36°))

EG ≈ 4(0.72644)(0.93252)/(0.93252 -0.72654) ≈ 13.1573 . . . . miles

The plane is flying at an altitude of about 13.16 miles.

_____

<em>Additional comment</em>

That's about 69,500 feet. The operational ceiling of most passenger aircraft is below 51,000 feet. The U2 reconnaissance aircraft reportedly has a service ceiling of about 70,000 feet.