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Bad White [126]

2 months ago

14

There are 10 girls and 6 boys in a class. The teacher randomly draws one student's name and then draws another. What is the prob

ability that the first name drawn will be a boy and the second will be a girl?
60%

23. 4%

16%

25%

1 answer:

marshall27 [118]2 months ago

4 0

The Probability of name drawn will be a boy is 3/8 ways and the girl will be 5/8 ways.

Given that,

In a class, there are 10 girls and 6 boys. After selecting one kid at random, the teacher selects another.

We have to find what probability is it that a boy will be the first name picked and a girl will be the second.

We know that,

Probability is number of favorable outcomes by total number of outcomes.

Total number of outcomes is 16.

For boys the number of favorable outcome is 6

For girls the number of favorable outcome is 10.

For selecting boy is

Probability is 6/16=3/8

For selecting girl is

Probability is 10/16=5/8

Therefore, The Probability of name drawn will be a boy is 3/8 ways and the girl will be 5/8 ways.

To learn more about probability visit: brainly.com/question/11234923

#SPJ4

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Calculate the average rate of change of f(x) = 1 x - x2 - 2 for 3 ≤ x ≤ 6. A) -163 18 B) -18 163 C) 163 18 D) 18 163

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To solve this we are going to use the average rate of change formula: $A(x)= \frac{f(b)-f(a)}{b-a}$
where
$A(x)$ is the average rate of change of the function
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$f(b)$ is the position function evaluated at $b$
$a$ is the first point in the interval
$b$ is the second point in the interval

We can infer for our problem that the first point is 3 and the second point is 6, so $a=3$ and $b=6$ . Lets replace those values in our formula:
$A(x)= \frac{f(b)-f(a)}{b-a}$
$A(x)= \frac{f(6)-f(3)}{6-3}$
$A(x)= \frac{6-6^2-2-(3-3^2-2)}{3}$
$A(x)= \frac{-32-(-8)}{3}$
$A(x)= \frac{-32+8}{3}$
$A(x)= \frac{-24}{3}$
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We can conclude that the average rate of change of the function f(x) = 1 x - x2 - 2 for <span>3 ≤ x ≤ 6 is -8</span>

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Find the reference angle given: t= -5π/3.

Elina [12.6K]

Answer:

The reference angle is $\frac{\pi}{3}$

Step-by-step explanation:

The given angle is $t=-\frac{5\pi}{3}$

The reference angle is the acute angle, that a given angle makes with the x-axis.

First let us make a sketch of the given angle. <em>See diagram in attachment.</em>

Let $\theta$ be the reference angle of $t=-\frac{5\pi}{3}$

Then, from the diagram, we need to subtract $\theta$ from $-\frac{5\pi}{3}$ to get to $-2\pi$ .

We can therefore write the following equation and solve for $\theta$ .

$\Rightarrow -\frac{5\pi}{3}-\theta=-2\pi$

We group like terms to get,

$\Rightarrow -\theta=-2\pi +\frac{5\pi}{3}$

We collect LCM to get,

$\Rightarrow -\theta=\frac{-6\pi+5\pi}{3}$

We simplify to get,

$\Rightarrow -\theta=-\frac{\pi}{3}$

We divide through by $-1$

$\theta=\frac{\pi}{3}$ to get,

The correct answer is D.

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