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2 years ago

If you spend 780 minutes per year waiting at traffic lights, how many hours is this?

Mathematics

2 answers:

rjkz [21]2 years ago

8 0

Answer:

13 hours

Step-by-step explanation:

minutes spent waiting on light/60= hours spent waiting on light

Tanzania [10]2 years ago

3 0

Answer:
That will me 13 hours

You might be interested in

The area of a small pond is 78.5 yds. What is the diameter of the pond?

Zolol [24]

Answer:

Diameter of pound = 10 yd

Step-by-step explanation:

Given area of circular pond = 78.5 square yd

To find the diameter of the pond.

Solution:

The pond being circular in shape, the area of pond can be given as:

⇒ $\pi\ r^2$

where $r$ represents radius of the pond.

Thus, we have:

⇒ $\pi\ r^2=78.5$

Using $\pi=3.14$

⇒ $3.14r^2=78.5$

Solving for $r$

Dividing both sides by 3.14

⇒ $\frac{3.14r^2}{3.14}=\frac{78.5}{3.14}$

⇒ $r^2=25$

Taking square root both sides.

⇒ $\sqrt{r^2}=\sqrt{25}$

⇒ $r=\pm5$

So, radius = 5 yd, as distances are always positive.

Diameter = $2\times radius=2\times 5\ yd= 10\ yd$

Thus, diameter of pound = 10 yd

7 0

2 years ago

Kayla has 24 yellow beads and 36 green beads a. what is the greatest number of necklaces she could make? b. how many yellow bead

creativ13 [48]

She can make 6 necklaces and these will be 4 yellow and 6 green on each necklace

7 0

1 year ago

PI-3.

Fudgin [204]

Answer:

A. $301

B. $721

Step-by-step explanation:

Let $x be the amount of money they raised.

Rowena tried to put the $1 bills into two equal piles and found one left over at the end, then

$x=2q_1+1$

Polly tried to put the $1 bills into three equal piles and found one left over at the end, then

$x=3q_2+1$

Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then

$x=4q_3+1\\ \\x=5q_4+1\\ \\x=6q_5+1$

Finally Rowena put all the bills evenly into 7 equal piles, and none were left over, then

$x=7q_6$

This means $x-1$ is divisible by 2, 3, 4, 5 and 6 without remainder, so

$x-1=2\cdot 3\cdot 2\cdot 5n=60n$

Hence,

$x=60n+1, \ n\in N$

The smallest amount of money they could have raised is $301, because

$x=60\cdot 5+1=301$ is divisible by 7.

Now, the number $x=60n+1$ should be divisible by 7 and must be greater than 500.

So,

$60n+1>500\\ \\60n>499\\ \\n>8$

When n = 9,

$x=60\cdot 9+1=541$ is not divisible by 7.

When n = 10,

$x=60\cdot 10+1=601$ is not divisible by 7.

When n = 11,

$x=60\cdot 11+1=661$ is not divisible by 7.

When n = 12,

$x=60\cdot 12+1=721$ is divisible by 7.

B. The least amount of money they could have raised is $721

7 0

2 years ago

Read 2 more answers

True or false: the regression sum of squares (ssr) can never be greater than the total sum of squares (sst).

Valentin [98]

. . . . . . .. . . . . . . . . . . . . . . . . . . . ./ . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8 0

2 years ago

Tim and Jim are working on a school project together. If he had to work by himself, it would take

Vladimir [108]

Answer: 6 hours

Step-by-step explanation:

Given, Tim and Jim are working on a school project together.

Tim takes 10 hours to complete the school project.

If Jim worked alone, it would take him 15 hours

Let t be the time they both take if they work together.

Then, $\dfrac{1}{t}=\dfrac{1}{\text{Time taken by Tim}}+\dfrac{1}{\text{Time taken by jim}}$

$\Rightarrow\ \dfrac{1}{t}=\dfrac{1}{10}+\dfrac{1}{15}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{3+2}{30}=\dfrac{5}{30}\\\\\Rightarrow\dfrac{1}{t}=\dfrac{1}{6}\\\\\Rightarrow t=6$

So, it will take<u> 6 hours</u> to complete the school project together .

4 0

1 year ago