Betty had a score of an 80 on her first quiz and a score of an 88 on her second quiz. what is the percent?

Mathematics

1 answer:

lana [24]3 years ago

4 0

80 + 88 = 168 then 168 divided by 2 (cause there were two quizzes) = 84%
Answer = 84%

You might be interested in

Can anyone help me with 4 &’ 7 ?

alukav5142 [94]

Mm i think 7 you just have to divide like the number by 2 like 40 divided by 2 is 20 then divide the 20 and you get 10 idk

8 0

3 years ago

PLS HELP ASAP PLS PLS PLS

Sedbober [7]

Answer:

$15

is the answer i think

7 0

3 years ago

Please help? Graph <img src="https://tex.z-dn.net/?f=-3x%5E2%2B12y%5E2%3D84" id="TexFormula1" title="-3x^2+12y^2=84" alt="-3x^2+

mariarad [96]

$-3x^2+12y^2=84$
$12y^2=3x^2+84$
$y^2=\dfrac{x^2}4+7$
$y=\pm\sqrt{\dfrac{x^2}4+7}$

For either square root to exist, you require that $\dfrac{x^2}4+7\ge0$ . This is true for all $x$ , since $\dfrac{x^2}4$ is always non-negative. This means the domain of $y$ as a function of $x$ is all real numbers, or $x\in\mathbb R$ or $(-\infty,infty)$ .

Now, because $\dfrac{x^2}4+7$ is non-negative, and the smallest value it can take on is 7, it follows that the minimum value for the positive square root must be $\sqrt7$ , while the maximum value of the negative root must be $-\sqrt7$ . This means the range is $y\in\mathbb R\setminus(-\sqrt7,\sqrt7)$ , or $|y|\ge\sqrt7$ , or $(-\infty,-\sqrt7]\cup[\sqrt7,\infty)$ .