If you conducted a survey of favorite foods of class members, what would be the size of the sample

Coach Emanuel is trying to lose weight. the stars 2017 weighing 296 pounds. if he plans to burn enough calories to lose 3 pounds

Vlad1618 [11]

Answer:

ooooooppppppppppp2345vb

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Does anyone know the answer to this??

Kipish [7]

Answer:

very good

Step-by-step explanation:

very good

8 0

3 years ago

1-58. Mr. Wright was making a table to figure out how much it costs to send a certain numt

Drupady [299]

A table of values can be used to represent variables that are directly proportional.

The complete table of proportions is:

$\left[\begin{array}{ccccccccc}Letters&10&2&[150 ]&7&1&500&[420] \\Cost&0.45&0.90&6.75&[0.315]&[0.045 ]&[22.5 ] & 18.90\end{array}\right]$

Given that

$\left[\begin{array}{ccccccccc}Letters&10&2&[\ ]&7&1&500&[\ ] \\Cost&0.45&0.90&6.75&[\ ]&[\ ]&[\ ] & 18.90\end{array}\right]$

Let:

$L \to$ Letters

$C \to$ Cost

Using proportional reasoning, we have:

$C = kL$

Where

$k \to$ ratio of proportion

For the first values of C and L, we have:

$C = kL$

$0.45 = k \times 10$

Divide both sides by 10

$k = 0.045$

So, the equation of proportion is:

$C = 0.045L$

When C = 6.75, we have:

$C = 0.045L$

$6.75 = 0.045L$

Solve for L

$L = \frac{6.75}{0.045}$

$L = 150$

When L = 7, we have:

$C = 0.045L$

$C = 0.045 \times 7$

$C = 0.315$

When L = 1, we have:

$C = 0.045L$

$C = 0.045 \times 1$

$C = 0.045$

When L = 500, we have:

$C =0.045L\\$

$C =0.045 \times 500$

$C =22.5$

When C = 18.90, we have:

$C = 0.045L$

$18.90 = 0.045L$

Solve for L

$L=\frac{18.90}{0.045}$

$L=420$

Hence, the complete table is:

$\left[\begin{array}{ccccccccc}Letters&10&2&[150 ]&7&1&500&[420] \\Cost&0.45&0.90&6.75&[0.315]&[0.045 ]&[22.5 ] & 18.90\end{array}\right]$

Read more about proportions at:

brainly.com/question/21126582

8 0

2 years ago

A club is choosing 2 members to serve on a committee. The club has nominated 2 women and 4 men. Based on chance alone, what is t

tekilochka [14]

Answer:

53.33% probability that one woman and one man will be chosen to be on the committee

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the members are chosen is not important, so we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

What is the probability that one woman and one man will be chosen to be on the committee?

Desired outcomes:

One woman, from a set of 2, and one man, from a set of 4. So

$D = C_{2,1}*C_{4,1} = \frac{2!}{1!1!}*\frac{4!}{1!3!} = 8$

Total outcomes:

Two members from a set of 2 + 4 = 6. So

$T = C_{6,2} = \frac{6!}{2!4!} = 15$

Probability:

$p = \frac{D}{T} = \frac{8}{15} = 0.5333$

53.33% probability that one woman and one man will be chosen to be on the committee

5 0

3 years ago

Use the raw data below to create a table that can be used to create a histogram with 5

schepotkina [342]

Answer:

Five classes

Step-by-step explanation:

1. Sort the data

You get

1.1, 1.3, 1.6, 1.7, 2, 2.6, 2.7, 2.9, 3.2, 3.2, 3.5, 3.5, 3.9, 4.6, 4.7, 4.8, 4.8, 4.9, 4.9, 5.3, 5.7, 6.4, 6.5, 7.1, 7.5, 7.6, 8.1, 8.2, 9.2, 9.4

2. Calculate the range

Range = Max - Min = 9.4 - 1.1 = 8.3

3. Calculate the class width

Divide the range by the number of classes

8.3/5 = 1.7

Round this up to 2.

4. Decide where to start the histogram

You could use classes: 1 - 3, 3 - 5, 5 - 7, 7 - 9, 9 -11

However, even numbers are easier to read.

It would be preferable to use classes: 0 - 2, 2 - 4, 4 - 6, 6 - 8, 8 - 10

5. Prepare a frequency distribution table

Note: Each class does not include its largest possible value. Thus, a value of 2 goes into Class 2 - 4, (not Class 0 - 2).

$\begin{array}{cc}\textbf{Miles run} & \textbf{No. of students} \\0 - 2 & 4 \\2 - 4 & 9 \\4 - 6 & 8 \\6 - 8 & 5 \\8 - 10 & 4 \\\end{array}$

4 0

3 years ago