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3 years ago

Use the equation f=6g to find the value of f when g=5

Mathematics

2 answers:

sashaice [31]3 years ago

8 0

Answer:

Step-by-step explanation:

F= 30

kompoz [17]3 years ago

5 0

Answer:

f=30

Step-by-step explanation:

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What is the range of the function in this table?

marin [14]

Hello from MrBillDoesMath!

Answer:

Choice B: {1,2,4}

Discussion:

The range of the function is the set of "y" values shown in the table, which is 1,4,4,2. As sets do not contain duplicate elements, the range is {1,2,4). This is Choice B

Thank you,

MrB

8 0

3 years ago

Read 2 more answers

There was a 6 square foot piece of wrapping paper for a birthday present. It takes 3 3/8 feet of square feet of the paper to wra

Citrus2011 [14]

Answer: However, in real life it depends on the shape of the actual presents. If two sheets are used for the 2 presents. There should be enough left from those sheets to use for the third present. (They should be taped together.)

Step-by-step explanation: 6/3 then 3 over 8 ––> 3/8

6/ 27/8

= 6/1 / divide 27/8

= 6^2 /1 x 8/27 ^9 (^) –––> this mean square root of 2 or 9

6 cancels and 27 cancel

= 16/9

=1 7/9

However you will still need to buy 2 sheets

3 0

3 years ago

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:

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

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