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2 years ago

What is 1/4 of 32 PLEASE HELP ME IM STUCK

Mathematics

2 answers:

exis [7]2 years ago

7 0

$\frac{1}{4} \ of \ 32= \frac{1}{4}*32= \\\\= \frac{1}{1}*8\to\boxed{8}$

salantis [7]2 years ago

4 0

32/4 = 8 1/4 of 32 is 8

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Reika [66]

Answer:

Quantitative: i) The number of students in the school vis - a - vis the number of classes in the school. In order words, simply number of students per class

ii) The rate of graduation of students.

Categorical:i) Courses being taken by students

ii) Region of country of students.

Step-by-step explanation:

To answer this question effectively, it's important for us to know that a quantitative variable usually has numerical values, and also have units of measure as well while categorical variables simply places each student into a specific category.

Now, the two quantitative variables we will consider are;

i) The number of students in the school vis - a - vis the number of classes in the school. In order words, number of students per class

ii) The rate of graduation of students.

Also, the two categorical variables we will consider are;

i) Courses being taken by students

ii) Region of country of students

5 0

3 years ago

Select "Rational" or "Irrational" to classify each number.

xenn [34]

Answer:

rational rational irrational

Step-by-step explanation:

6 0

2 years ago

R(x)=2√x s(x)=√x (r/s)(3)

anygoal [31]

This is a trick question. First just understand that any number over itself is 1 so $\frac{2 \sqrt{x} }{\sqrt{x}}$ is just $2 * \frac{ \sqrt{x} }{ \sqrt{x}} = 2$ so it is just 2

5 0

3 years ago

Read 2 more answers

X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y

andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

$\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}$