Which of the following shapes are polygons with more than 444 sides?

Please help it’s easy I think

Nikitich [7]

The frequency of revision times is given by the product of the frequency

density value and the class width.

Correct response:

The number of students are <u>72 students</u>

<h3>Methods of calculation using a histogram</h3>

The estimate of the number of students who revised for less than 45

minutes is given by the area, <em>A,</em> under the histogram to the left of the 45

minute mark as follows;

Frequency, f = Frequency density × Class width

Number of students = ∑f

Therefore;

A = 5 × 2 + 10 × 2.2 + 20 × 1.6 + 10 × 0.8 = 72

The number of students that revised for less than 45 minutes = <u>72 </u>students

Learn more about histograms here:

brainly.com/question/17139138

4 0

2 years ago

In what real-world situations would you want to round a number?

solmaris [256]

Taxes,paying for things, estimating how much money you have or will get or pay

7 0

3 years ago

find the values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (1, -8)

frutty [35]

Answer:

cosФ = $\frac{1}{\sqrt{65}}$ , sinФ = $-\frac{8}{\sqrt{65}}$ , tanФ = -8, secФ = $\sqrt{65}$ , cscФ = $-\frac{\sqrt{65}}{8}$ , cotФ = $-\frac{1}{8}$

Step-by-step explanation:

If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:

cosФ = $\frac{x}{r}$
sinФ = $\frac{y}{r}$
tanФ = $\frac{y}{x}$
secФ = $\frac{r}{x}$
cscФ = $\frac{r}{y}$
cotФ = $\frac{x}{y}$

Where r = $\sqrt{x^{2}+y^{2} }$ (the length of the terminal side from the origin to point (x, y)

You should find the quadrant of (x, y) to adjust the sign of each function

∵ Point (1, -8) lies on the terminal side of angle Ф in standard position

∵ x is positive and y is negative

→ That means the point lies on the 4th quadrant

∴ Angle Ф is on the 4th quadrant

∵ In the 4th quadrant cosФ and secФ only have positive values

∴ sinФ, secФ, tanФ, and cotФ have negative values

→ let us find r

∵ r = $\sqrt{x^{2}+y^{2} }$

∵ x = 1 and y = -8

∴ r = $\sqrt{x} \sqrt{(1)^{2}+(-8)^{2}}=\sqrt{1+64}=\sqrt{65}$

→ Use the rules above to find the six trigonometric functions of Ф

∵ cosФ = $\frac{x}{r}$

∴ cosФ = $\frac{1}{\sqrt{65}}$

∵ sinФ = $\frac{y}{r}$

∴ sinФ = $-\frac{8}{\sqrt{65}}$

∵ tanФ = $\frac{y}{x}$

∴ tanФ = $-\frac{8}{1}$ = -8

∵ secФ = $\frac{r}{x}$

∴ secФ = $\frac{\sqrt{65}}{1}$ = $\sqrt{65}$

∵ cscФ = $\frac{r}{y}$

∴ cscФ = $-\frac{\sqrt{65}}{8}$

∵ cotФ = $\frac{x}{y}$

∴ cotФ = $-\frac{1}{8}$

8 0

3 years ago

What is There are fourteen whole snack bars on a plate. There are five friends.

AleksandrR [38]

Carl is incorrect. Dave ate a higher fraction of snack bars, by 0.2 snack bars.

Carl had .5 left of a snack bar.

Dave had .3 left of a snack bar.

Tony had .5 left of a snack bar.

Gary had .0 left of a snack bar.

Tryone had .7 left of a snack bar.

If we add the above snack bars, there is a total of two remaining snack bars, meaning they only ate 12 of 14 snack bars.

3 0

3 years ago

At the movie theatre, child admission is $5.60 and adult admission is $9.50. On Saturday, four times as many adult tickets as ch

romanna [79]

Answer:

-6.20x-6.20y=-948.6

add this to equation 2

6.20x-6.20x +9.30y-6.20y=1125.30-948.60

3.10y= 176.70

/3.10

y=176.70/3.10

y = 57 number of adult tickets

153-57=96 child tickets

Step-by-step explanation:

child admission is $6.20 : x numbers

adult admission is $9.30 : y numbers

153 tickets were sold

sales of $1125.30 .

x+y = 153.........................1

6.20x+9.30y= 1125.30..............2

4 0

3 years ago