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jarptica [38.1K]
3 years ago
15

Jami is keeping track of the weather for a class project she recorded data on 16 days out of 16 days 1/4 were clear blue skies h

ow many days had clear blue skies
Mathematics
1 answer:
katovenus [111]3 years ago
8 0

Answer:

4 days

Step-by-step explanation:

She recorded date for 16 days, and 1/4 of the days were clear blue skies.

One way to find this is to multiply 16 and 1/4

16*1/4=4

Another way is to first convert 1/4 to a decimal

1/4=0.25

Then multiply 16 and 0.25

16*0.25 =4

So, 4 out of 16 days had clear blue skies

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% slope  = m
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The volume of a gas is 10.0 liters and its pressure is 1.5 atm. If the pressure is decreased to 0.75 atm,
JulsSmile [24]

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The new volume is 20litres

Step-by-step explanation:

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3 0
3 years ago
21. Who is closer to Cameron? Explain.
pickupchik [31]

Problem 21

<h3>Answer:  Jamie is closer</h3>

-----------------------

Explanation:

  • A = Arthur's location = (20,35)
  • J = Jamie's location = (45,20)
  • C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

=================================================

Problem 22

<h3>Answer:  Arthur is closest to the ball</h3>

-----------------------

Explanation:

We have these key locations:

  • A = Arthur's location = (20,35)
  • J = Jamie's location = (45,20)
  • C = Cameron's location = (65,40)
  • B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

  • AB
  • JB
  • CB

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

-------------

So in summary, we have these approximate segment lengths

  • AB = 29.1547595
  • JB = 41.2310563
  • CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

3 0
3 years ago
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