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siniylev [52]
3 years ago
8

This should be the last question for the day...hopefully

Mathematics
1 answer:
andrew11 [14]3 years ago
5 0
I’d think it’s equated to the one, why I think that? Idek.
You might be interested in
Juan is working two summer jobs, making $12 per hour babysitting and making $16 per hour lifeguarding. In a given week, he can w
Semenov [28]

\text{Let }b=

Let b=

\,\,\text{the number of hours babysitting}

the number of hours babysitting

\text{Let }l=

Let l=

\,\,\text{the number of hours lifeguarding}

the number of hours lifeguarding

\text{\textquotedblleft at most 17 hours"}\rightarrow \text{17 or fewer hours}

“at most 17 hours"→17 or fewer hours

Use a \le≤ symbol

Therefore the total number of hours worked in both jobs, b+lb+l, must be less than or equal to 17:17:

b+l\le 17

b+l≤17

\text{\textquotedblleft at least \$240"}\rightarrow \text{\$240 or more}

“at least $240"→$240 or more

Use a \ge≥ symbol

Juan makes $12 per hour babysitting, so in bb hours he will make 12b12b dollars. Juan makes $16 per hour lifeguarding, so in ll hours he will make 16l16l dollars. The total amount earned 12b+16l12b+16l must be greater than or equal to \$240:$240:

12b+16l\ge 240

12b+16l≥240

\text{Plug in }\color{green}{3}\text{ for }b\text{ and solve each inequality:}

Plug in 3 for b and solve each inequality:

Juan worked 3 hours babysitting

\begin{aligned}b+l\le 17\hspace{10px}\text{and}\hspace{10px}&12b+16l\ge 240 \\ \color{green}{3}+l\le 17\hspace{10px}\text{and}\hspace{10px}&12\left(\color{green}{3}\right)+16l\ge 240 \\ l\le 14\hspace{10px}\text{and}\hspace{10px}&36+16l\ge 240 \\ \hspace{10px}&16l\ge 204 \\ \hspace{10px}&l\ge 12.75 \\ \end{aligned}

b+l≤17and

3+l≤17and

l≤14and

​  

 

12b+16l≥240

12(3)+16l≥240

36+16l≥240

16l≥204

l≥12.75

​  

 

\text{The values of }l\text{ that make BOTH inequalities true are:}

The values of l that make BOTH inequalities true are:

\{13,\ 14\}

{13, 14}

\text{(the final answer is this entire list)}

(the final answer is this entire list)

6 0
3 years ago
WORKSHEET-2
Margarita [4]

Answer:

1. \frac{C}{D} = 3.143

2. Circumference

3. Area of the circle is 3850 cm^{2}.

   ii. Circumference of the circle is 220 cm.

Step-by-step explanation:

The area, A, and the circumference, c, of a circle can be determined respectively by:

A = \pir^{2}

C = 2\pir

where r is the radius of the circle.

1. the ratio of the circumference and diameter, D, of a circle is:

r = \frac{D}{2}

so that,

C = 2\pi\frac{D}{2}

C = \piD

\frac{C}{D} = \pi

   = \frac{22}{7}

   = 3.143

2. The distance around a circular region is known as it's circumference.

3. Given a circle of radius 35 cm, then:

A = \pir^{2}

   = \frac{22}{7} x (35)^{2}

   = \frac{22}{7} x 1225

   = 22 x 175

A = 3850

Area of the circle is 3850 cm^{2}.

C = 2\pir

   = 2 x \frac{22}{7} x 35

   = 2 x 22 x 5

   = 220

C = 220 cm

Circumference of the circle is 220 cm.

8 0
3 years ago
Someone please solve x for me please
Eva8 [605]

Answer:

x = 40

Step-by-step explanation:

50 + 150 + 2x + x + x

200 + 4x

360 - 200 = 160

160 / 4 = 40

x = 40

50 + 80 + 40 + 150 + 40

8 0
3 years ago
Please help I need this for a math pre-test.
mestny [16]

Answer:

612 I think well how this helps

6 0
4 years ago
Read 2 more answers
In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position wh
Lady_Fox [76]

Answer: = ( 0.411, 0.409)

Therefore at 95% confidence interval (a,b) = (0.411, 0.409)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean gain x = 0.41

Standard deviation r = 0.016

Number of samples n = 1000

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

0.41+/-1.96(0.016/√1000)

0.41+/-1.96(0.000506)

0.41+/-0.00099

0.41+/-0.001

= ( 0.411, 0.409)

Therefore at 95% confidence interval (a,b) = (0.411, 0.409)

4 0
4 years ago
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