Preparing a glucose solution In a certain medical test

nora has 3/4 as many books as victor. let v represent the number of books victor has. create an expression to represent the numb

xxTIMURxx [149]

Answer:

N = $\frac{3}{4}$ * v

Step-by-step explanation:

Since the value that we are provided is a fraction that is tied to the number of books that Victor has, we would simply need to multiply this fraction by the total number of books that Victor has. This value is represented by the variable v while the total number of books that Nora has is represented by the variable N. Therefore, the expression would be the following

N = $\frac{3}{4}$ * v

4 0

3 years ago

HELP FAST PLEASE!!! PLEASE<br><br> Indentify the interval on which the function is increasing.

Katena32 [7]

Answer: [0,3]

The line is going up from left to right between 0 and 3 on the x-axis. We use a closed bracket because it is a open circle.

5 0

3 years ago

Explain how solving 8+2 can help you solve 80+20.

katrin [286]

Becuase 80 is the same as 8 but has an extra 0 at the end (which makes it 80) and same for 20 to 2. 8 + 2 = 10 (now add a zero at the end). The answer should be 100.

5 0

3 years ago

Read 2 more answers

How to do graphs of quadratic functions

Anastaziya [24]

Answer:

The Graph of a Quadratic Function

A quadratic function is a polynomial function of degree 2 which can be written in the general form.

Step-by-step explanation:

Example: Graph: f(x)=−x2−2x+3.

Solution:

Step 1: Determine the y-intercept. To do this, set x=0 and find f(0).

f(x)f(0)===−x2−2x+3−(0)2−2(0)+33

The y-intercept is (0,3).

Step 2: Determine the x-intercepts if any. To do this, set f(x)=0 and solve for x.

f(x)000x+3x======−x2−2x+3−x2−2x+3x2+2x−3(x+3)(x−1)0−3orx−1x==Set f(x)=0.Multiply both sides by −1.Factor.Set each factor equal to zero.01

Here where f(x)=0, we obtain two solutions. Hence, there are two x-intercepts, (−3,0) and (1,0).

Step 3: Determine the vertex. One way to do this is to first use x=−b2a to find the x-value of the vertex and then substitute this value in the function to find the corresponding y-value. In this example, a=−1 and b=−2.

x====−b2a−(−2)2(−1)2−2−1

Substitute −1 into the original function to find the corresponding y-value.

f(x)f(−1)====−x2−2x+3−(−1)2−2(−1)+3−1+2+34

The vertex is (−1,4).

Step 4: Determine extra points so that we have at least five points to plot. Ensure a good sampling on either side of the line of symmetry. In this example, one other point will suffice. Choose x=−2 and find the corresponding y-value.

x −2 y 3f(−2)=−(−2)2−2(−2)+3=−4+4+3=3Point(−2,3)

Our fifth point is (−2,3).

Step 5: Plot the points and sketch the graph. To recap, the points that we have found are

y−intercept:x−intercepts:Vertex:Extra point:(0,3)(−3,0) and (1,0)(−1,4)(−2,3)

Answer:

4 0

3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago