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1 year ago

If your current examination mean is 97.2 and you receive a 99 on the next examination, what effect will this have?

Mathematics

2 answers:

Sonbull [250]1 year ago

6 0

If your current examination mean is 97.2 and you receive a 99 on the next examination, the mean will increases.

Mean: It is the sum (total) of all the values in a set of data, such as numbers or measurements, divided by the number of values on the list. To find the mean, add up all the values in the set. Then divide the sum by how many values there are.

We have the current examination mean, $x_o_l_d$ = 97.2

Now, we receive, $x =$ 99 on the next examination, the new mean will be:

$x_n_e_w = \frac{x_o_l_d + x}{N}$ ∴ $N= 2$

$x_n_e_w = \frac{97.2 + 99}{2}$

$x_n_e_w = \frac{196.2}{2}$

$x_n_e_w = 98.1$

∴ New mean, $x_n_e_w =$ 98.1

98.1 > 97.2

Hence, mean is increased on receiving a 99 on the next examination.

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mr Goodwill [35]1 year ago

3 0

If you have an average of 97.2 on your current exam and you get a 99 on your next exam, your average will increase.

Mean in mathematics is the sum (total) of all the values in a set of data, such as numbers or measurements, divided by the total number of values.

To find the average, sum all the values in the set. Then divide the total by the number of values.

We have the current examination mean, xold = 97.2

Now, we receive, x = 99 on the next examination, the new mean will be:

xnew = (xold + x)/N

xnew = (97.2+99)/2

xnew = 196.2/2

xnew = 98.1

The new average is 98.1

98.1 > 97.2

So if you score 99 on your next exam, your average will increase.

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What is the value of x?<br><br><br><br> Enter your answer in the box.

pshichka [43]

Hello!

The sum of all angle measures in a heptagon in 900°. We will subtract all of these angles from 900 to find our answer.

900-146-122-142-140-110-142=98

x=98°

I hope this helps!

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2 years ago

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Suppose that 40 deer are introduced in a protected wilderness area. The population of the herd

Scorpion4ik [409]

The horizontal asymptote of the function is the minimum number of deer in the area.

The equation of horizontal asymptote is: $\mathbf{y = 40}$
The horizontal asymptote means that, the number of deer will never be less than 40

The equation is given as:

$\mathbf{f(x) = \frac{40 + 2x}{1 + 0.05x}}$

Expand the numerator

$\mathbf{f(x) = \frac{40(1 + 0.05x)}{1 + 0.05x}}$

Cancel out the common factor

$\mathbf{f(x) = 40}$

Hence, the equation of horizontal asymptote is:

$\mathbf{y = 40}$

The horizontal asymptote means that, the number of deer will never be less than 40

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2 years ago

If f(x)=x-2 and g(x)=1/2x, find g(f(x))

Scilla [17]

Answer:

hello :

If f(x)=x-2 and g(x)=1/2x,

g(f(x)) = g ( x - 2 )

= 1/2(x-2)

g(f(x)) = (1/2)x - 1

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3 years ago

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Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}$

recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

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marin [14]

Answer:

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Step-by-step explanation:

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