find the mean of the data set. if necessary, round to the nearest tenth. 15.8, 14.9, 16.7, 17.6, 13.4 15 points

A snail climbs to the top of a coconut tree 18 meters high. In an hour it will climb 2 meters. But if it go 2 meters, it will sl

kumpel [21]

Answer:

The snail will reach the top of the coconut in 18 hours.

Step-by-step explanation:

If the snail will climb 2 meters in 1 hour but it will slip down one meter for every 2 meters climbed, then the distance traveled will be:

$d = 2 m - 1 m = 1 m$

Since in 1 hour, it climbs 2 meters minus 1 meter, the time in which the snail will reach the top is:

$t = \frac{1 h}{1 m}*18 m = 18 h$

Therefore, the snail will reach the top of the coconut in 18 hours.

I hope it helps you!

5 0

2 years ago

Justin and michelle are playing a card game. In the first round, caitlin scored 100 points and justin scored 75 points. In each

Arada [10]

Answer:

Step-by-step explanation: first you have to add up the point that each scored altogether and Caitlin scored 300 and Justin scored 250

5 0

2 years ago

Read 2 more answers

Solve the Inequality: m+16>8m+2

Setler [38]

Answer:

m<2

Step-by-step explanation:

To get to this, you first need to move the variable to one side, so you could subtract m from both sides to make the equation 16>7m+2. Next, you subtract 2 from both sides to make the equation into 14>7m. Finally, you would divide 7 from both sides to get 2>m, or m<2.

Hope this helps!

5 0

3 years ago

ANYONE KNOW THIS NEED ASAP

Dmitry [639]

B, it’s the point that the lines cross

6 0

1 year ago

Read 2 more answers

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

The angle subtended by the distance of the buoy from the port, C is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ 307.35°

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0

2 years ago