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3 years ago

A cruise ship can cover 21 nautical miles in 483 minutes. How many nautical miles will it travel in 230 minutes?

Mathematics

2 answers:

never [62]3 years ago

8 0

The answer is probably 2 in my opinion

vaieri [72.5K]3 years ago

6 0

Answer: About 10.95 nautical miles

Step-by-step explanation: First, lets find how many minutes the cruise ship can cover in 1 nautical mile.

In order to do that divide 21 nautical miles by 21 and 483 minutes by 21.

21/21 = 1 nautical mile 483/21 = 21 minutes

Now, divide 230 minutes by 21 minutes, which equals about 10.95.

So, in 230 minutes the cruise ship will travel about 10.95 nautical miles.

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Alan buys 5.3 ounces of sour patch kids candy. After sharing with his friends he returns to buy an additional 6.75 ounces. How m

Natasha_Volkova [10]

Answer:

Alan buys 12.05 ounces of sour patch kids candy in all.

Step-by-step explanation:

We are given that Alan buys 5.3 ounces of sour patch kids candy. After sharing with his friends he returns to buy an additional 6.75 ounces.

And we have to find the total ounces he buys in all.

As we know that for finding the total quantity or amount, we will use addition for calculating it.

Firstly, Alan buys = 5.3 ounces of sour patch kids candy

Additional ounces of sour patch kids candy Alan buys = 6.75

So, the total ounces of sour patch kids candy he buys in all = 5.3 + 6.75

= 12.05 ounces

Hence, he buys 12.05 ounces in all.

7 0

3 years ago

Drag the tiles to list the sides of △MNO from shortest to longest.

sweet [91]

The smaller the angle subtended by a side, the smaller the length of the

side.

The correct responses are;

Question 1: The list of sides from shortest to longest are;

MO/Shortest MO/Medium and MO/Longest

a) Friday

b) 70 minutes

c) 40%

d) Yes, the sum of the mean number of minutes spent on aerobic training and the mean number of minutes spent on strength training is equal to the mean total number of minutes spent training.

From the given diagram, we have, the measure of the third angle, ∠O, is

found as follows;

∠O = 180° - 54° - 61° = 65°

Therefore, ∠O = The largest angle

We get;

The longest side is opposite the largest angle, which gives;

The shortest side is the side opposite ∠N (54°)= $\frac{}{MO}$

The next shortest side is the side opposite ∠M(61°) = $\frac{}{NO}$

The longest side is the side opposite ∠O(65°) = $\frac{}{MN}$

a) The time spent training on Tuesday = 60 + 10 = 70 minutes

The time spent training on Thursday = 50 + 30 = 80 minutes

The time spent training on Friday = 45 + 40 = 85 minutes

Therefore, the day the athlete spent the longest total amount of time training is on Friday

b) The time spent training on Monday = 10 + 20 = 30 minutes

The time spent training on Wednesday = 20 + 15 = 35 minutes

Therefore, we get;

30, 35, 70, 80, and 85

The median total number of minutes the athlete spent training each day = 70 minutes

c) The time spent strength training = 20 + 10 + 15 + 30 + 45 = 120

The total number of minutes the athlete spent training = 70 + 80 + 85 + 30 + 35 = 300

$The$ $percentage$ $spent$ $on$ $strength$ $training$ = $\frac{120}{300}$ × $100 =$ $\frac{40}$ %

d) The mean number of minutes spent on strength training is found as follows;

$Mean_{strength} =\frac{120}{5} =24$

The mean number of minutes spent on aerobic training is found as follows;

$Mean_{aerobic} =\frac{10+60+20+50+40}{5} =36$

$Mean_{strength} +Mean_{aerobic} =24+36=60$

The mean total number of minutes spent training, $Mean_{total}$ = $\frac{300}{5}$ = 60

Therefore;

$Mean_{strength}+Mean_{aerobic} = Mean_{total} \\$

Learn more here:

brainly.com/question/2962546

4 0

3 years ago

In 2005 there were 1352 students in 2006 there were 1014 students work out the decrease in the number of students

Eduardwww [97]

I’m. It sure but I need points so sorry

6 0

3 years ago

How many inches (in.) are in 1 foot (ft.)?

velikii [3]

Answer:

12 inches are in 1 foot.

7 0

3 years ago

Read 2 more answers

Complete the tables of values.

zlopas [31]

Answer with Step-by-step explanation:

We are given two tables and we are to complete the missing values.

For table one, we are to substitute the given values of x in the term $4^{-x}$ :

when x = 0, $4^{-0} = 1$

when x = 2, $4^{-2} = \frac{1}{16}$

when x = 4, $4^{-4} = \frac{1}{256}$

For table two, we are to substitute the given values of x in the term $(\frac{2}{3} )^x$ :

when x = 0, $(\frac{2}{3} )^0 = 1$

when x = 2, $(\frac{2}{3} )^2 = \frac{4}{9}$

when x = 4, $(\frac{2}{3} )^4 = \frac{16}{81}$

6 0

4 years ago

Read 2 more answers