If Anna's test grades are 66, 90, and 99, what is her test average?

Mathematics

2 answers:

marysya [2.9K]3 years ago

4 0

Answer:

90 because it is the mean

skad [1K]3 years ago

4 0

Answer: it’s 83%

Step-by-step explanation: you add them up and divide by 3

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2000 soldiers stand in a row. Beginning from the left, each soldier calls out a number, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, and

Cloud [144]

Answer:

The number of soldiers that call out a number 3 is 666 soldiers

Step-by-step explanation:

The parameters given are;

Number of soldiers = 2000

Soldiers calling from the left = 1, 2, 3,.....

Soldiers calling from the right = 1, 2, 3,.....

Therefore, since the number soldiers calling out the number 3 are 1 in 3 from the left and 1 in 3 from the right, we split the soldiers into 2 groups of 1000, with one group calling from left and the other group calling from the right;

The number of 3s in called out in the first group from the left is therefore;

1000/3 = 333.33 which is 333 soldiers or from the 3rd soldier to the 999th soldier

Similarly the number of 3s called out from the second group = 333

Hence the total number of soldiers that call out a number 3 = 333 + 333 = 666 soldiers.

8 0

3 years ago

Which relation is a function?

zloy xaker [14]

<span>{(–1, 3), (5, 3), (–6, 7), (9, 0)} is the correct answer.
I hope this helped!

</span>

3 0

3 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$