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2 years ago

A hi school is conducting a survey to see how many students have part time jobs. Would this sample be biased or unbiased?

1 answer:

8 0

It would be biased because one would assume that students can get to school by means other than those that put them 8n the parking lor. (Parking lot = car drivers who would be the only students sampled.). Others may ride to school by either parent or school bus who could drop them off somewhere else or possibly walk to school. The freshmen/sophomores would be excluded generally as they would be too young to drive altogether

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NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.

slava [35]

Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

$\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2$

<u>Total Surface Area</u>

Area of the isosceles triangle:

$\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2$

Total surface area:

$\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)$

<u>Volume</u>

$\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)$

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

$\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2$

<u>Total Surface Area</u>

Area of a pentagon:

$\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2$

where a is the side length.

Therefore:

$\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)$

Total surface area:

$\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)$

<u>Volume</u>

$\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)$

8 0

1 year ago

Children's tickets to a local play cost $\$1.50$ less than adult tickets. At one performance 325 children tickets and 175 adult

svetoff [14.1K]

Answer:

$8.00

Step-by-step explanation:

The problem statement gives two relations between the prices of two kinds of tickets. These can be used to write a system of equations for the ticket prices.

<h3>setup</h3>

Let 'a' and 'c' represent the prices of adult and children's tickets, respectively. The given relations can be expressed as ...

a - c = 1.50 . . . . . . . adult tickets are $1.50 more

175a +325c = 3512.5 . . . . . total revenue from ticket sales.

<h3>solution</h3>

We are only interested in the price of an adult ticket, so we can eliminate c to give one equation we can solve for 'a'. Using the first equation, an expression for c is ...

c = a -1.50

Substituting that into the second equation, we have ...

175a +325(a -1.50) = 3512.50

500a -487.50 = 3512.50 . . . . . . simplify

500a = 4000 . . . . . . add 487.50

a = 8 . . . . . . . . . divide by 500

An adult ticket costs $8.

5 0

1 year ago

HELP i need the answers to these! explain if possible:)

IrinaVladis [17]

so one: the middle Angles are all 90 degrees triangles equal 180, the angle opposite of "x" is equal to 34 so 180-90-34=x

two: both angles are equal set it up like an equation and solve for n

three: the corner angle of the right triangle is equal to 61 as well again do 180-(61×2)=x

4: do this equation (<ktu)2+ x = 180 then solve for x

5: not exactly.sure what its asking

hope this helps

5 0

2 years ago

Ricky made 86 phone calls in 3 months. In the first month she made 21 calls, and in the second month she made 19 calls. How many

IrinaVladis [17]

She made 46 calls the third month

3 0

2 years ago

Read 2 more answers

Please ,help hurry up,will give u brainliest.

Nataliya [291]

Answer:

Take a look at the picture!

5 0

1 year ago