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3 years ago

8

Suppose 60% of a graduating class in a certain high school goes on to college. if 240 stuare going on to college, how many stude

nts are in the graduating class?

1 answer:

Marina86 [1]3 years ago

6 0

400 students

hope this helps

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The table shows the monthly low temperature for several cities in December. The monthly low temperature in Toledo in December wa

alex41 [277]

Of these cities, the place that has the lowest temperature than Toledo is Youngstown.

<h3>What is the temperature of Youngstown?</h3>

The town is known to have a temperature that is less than 20. This shows that the temperature has fallen below all other temperatures in the table.

Hence it can be concluded that the lowest table here is the temperature of Youngstown at -20 degrees.

Read more on temperature here:

brainly.com/question/25677592

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2 years ago

What is the equation of the line that passes through the point (5, 0) and has a<br> slope of -1?

vazorg [7]

Answer:

y-0= -1(x-5)

y= -x+5

x+y-5=0

4 0

3 years ago

Read 2 more answers

What is the factor of 12

balu736 [363]

The factor of 12 is 1, 2 , 3 , 4 , 6 , 12

Hope it helps.

7 0

3 years ago

As notas de dois alunos X e Y estão representadas no quadro abaixo. Por meio da variância, qual deles apresentou ser o mais regu

Reil [10]

Answer:

Aluno X

Step-by-step explanation:

A nota média para os alunos X e Y são, respectivamente:

$X_m=\frac{5+3+5+7}{4}=5 \\Y_m=\frac{4+9+4+3}{4}=5$

A variância é definida pela seguinte expressão:

$V=\frac{\sum(X_i-X_m)^2}{n-1}$

Onde n é o numero de termos, neste caso, n = 4.

As variâncias para cada aluno são:

$V_x=\frac{(5-5)^2+(3-5)^2+(5-5)^2+(7-5)^2}{4-1} \\V_x= 2.7\\V_y=\frac{(4-5)^2+(9-5)^2+(4-5)^2+(3-5)^2}{4-1} \\V_y= 7.3$

O aluno X apresentou uma variância menor do que o aluno Y em suas notas e, portanto, apresentou ser o mais regular.

6 0

3 years ago

A rocket is fired upward from some initial distance above the ground. Its height in feet, h,above the ground, t seconds after

kirill115 [55]

Check the picture below.

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h(t) = &{{ -16}}t^2&{{ +96}}t&{{ +2560}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so.. the maximum point is at $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}~~~ft$

and it happened after $\bf -\cfrac{{{ b}}}{2{{ a}}}~~~seconds$

--------------------------------------------------------------------------

now, when the rocket reaches the ground, as you saw on the picture, y = 0, thus

$\bf h(t)=-16t^2+96t+2560\implies 0=-16t^2+96t+2560
\\\\\\
0=-16(t^2-6t-160)\implies 0=t^2-6t-160
\\\\\\
0=(t-16)(t+10)\implies t=
\begin{cases}
\boxed{16}\\
-10
\end{cases}$

"t" is an amount in seconds, for this specific phenomena, it cannot be negative, so -10 is not a feasible value.

8 0

3 years ago

Read 2 more answers