Hewo people of da wowld

The cost of parking is an initial cost plus an hourly cost.
The first hour costs $7.
You need a function for the cost of more than 1 hour,
meaning 2, 3, 4, etc. hours.
Each hour after the first hour costs $5.

1 hour: $7
2 hours: $7 + $5                  = 7 + 5 * 1          = 12
3 hours: $7 + $5 + $5          = 7 + 5 * 2          = 17
4 hours: $7 + $5 + $5 + $5 = 7 + 5 * 3          = 22

Notice the pattern above in the middle column.
The number of $5 charges you add is one less than the number of hours.
For 2 hours, you only add one $5 charge.
For 3 hours, you add two $5 charges.

Since the number of hours is x, according to the problem, 1 hour less than the number of hours is x - 1.

The fixed charge is the $7 for the first hour.
Each additional hour is $5, so you multiply 1 less than the number of hours,
x - 1, by 5 and add to 7.

C(x) = 7 + 5(x - 1)

This can be left as it is, or it can be simplified as

C(x) = 7 + 5x - 5

C(x) = 5x + 2

Answer: C(x) = 5x + 2

Check:
For 2 hours: C(2) = 5(2) + 2 = 10 + 2 = 12
For 3 hours: C(3) = 5(3) + 2 = 15 + 2 = 17
For 4 hours: C(3) = 5(4) + 2 = 20 + 2 = 22
Notice that the totals for 2, 3, 4 hours here
are the same as the right column in the table above.

6 0

3 years ago

according to "Black Hole Beginnings," the destruction of some stars results in inbursts of light. an intense gravitational pull.

Colt1911 [192]

According to "Black Hole Beginnings," the destruction of some stars results in an an intense gravitational pull.

To answer the question, we need to know what Black Holes are.

<h3>What are Black Holes?</h3>

Blackholes are regions of space in which no matter can escape

<h3>Characteristics of black holes</h3>

Black holes are formed from the remnants of large star that dies in a supernova
As long as the star is three times or more larger than the sun, it will collapse
The gravitational force on the star causes it to collapse into a black hole

So, when black holes are formed, some stars are destroyed and collapse due to intense gravitational forces.

So, according to "Black Hole Beginnings," the destruction of some stars results in an an intense gravitational pull.

Learn more about black holes here:

brainly.com/question/6037502

#SPJ1

5 0

1 year ago

Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa

xxMikexx [17]

Answer:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins. Yes No Total

Small 32 18 50

Medium 68 7 75

Large 89 11 100

_____________________________________

Total 189 36 225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1: NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{50*189}{225}=42$

$E_{2} =\frac{50*36}{225}=8$

$E_{3} =\frac{75*189}{225}=63$

$E_{4} =\frac{75*36}{225}=12$

$E_{5} =\frac{100*189}{225}=84$

$E_{6} =\frac{100*36}{225}=16$

And the expected values are given by: