21 is 7 Times as many as 3. What is the equation?

Mathematics

1 answer:

aalyn [17]3 years ago

5 0

21 = 7 x 3 or 7(3) if you're no longer using x as a multiplication symbol

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Calculate the mean of this data set: 12, 35, 44, 74, 23, 49, 45, 18, 90, 56, 84,

Blababa [14]

Well, here is the mean of ur data collected 48.181818...

4 0

3 years ago

Read 2 more answers

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$

5 0

3 years ago

Can someone please explain how to do these?

Pani-rosa [81]

Answer:

First question answer: The limit is 69

Second question answer: The limit is 5

Step-by-step explanation:

For the first limit, plug in $x=8$ in the expression $(9x-3)$ , that's the answer for linear equations and limits.

So we have:

$9x-3\\9(8)-3\\72-3\\69$

The answer is 69

For the second limit, if we do same thing as the first, we will get division by 0. Also indeterminate form, 0 divided by 0. Thus we would think that the limit does not exist. But if we do some algebra, we can easily simplify it and thus plug in the value $x=1$ into the simplified expression to get the correct answer. Shown below: