Ask question

Ask question

All categories

4 years ago

10

A news station in Oregon recorded that the low temperatures for 5 days were –3, –2, 2, 2, and 6. What was the average temperatur

e for those days?

2 answers:

Stells [14]4 years ago

6 0

Answer:

Average = 1

Step-by-step explanation:

Let us define the average first:

Average is calculated by adding up all the values and then dividing the sum by total number of values.

The formula for average may be written as:

$Average = \frac{Sum}{count}$

In the following case,

Sum of numbers = -3-2+2+2+6 = 5

Count = 5

So,

Average = 5/5

=> Average = 1

MrRissso [65]4 years ago

4 0

Answer:

the answer is 1

Step-by-step explanation:

Sum of numbers = -3-2+2+2+6 = 5

Count = 5

So,

Average = 5/5

=> Average = 1

You might be interested in

Increase £120 by 35%

Natalka [10]

Answer:

£162

Step-by-step explanation:

Let's first find 35% of £120.

35% here is our multiplier, so we can turn it into a decimal (0.35) in order to use it in our sum.

£120 × 35% = £42

Now add £42 to £120.

£120 + £42 = £162

Super simple!

7 0

3 years ago

Read 2 more answers

Keith ran in a marathon. it took him 3 hours and 2 minutes to run 26 miles. if he ran at the same rate for the whole race, how m

stiks02 [169]

First, you have to find how many minutes he ran which would be 182 minutes. Then, to find out how many minutes he spent running each mile, you divide by 26. 182/26=7. 7 minutes per mile.

7 0

4 years ago

How can you find 10% of a number in your head

Strike441 [17]

Answer: multiply it by .1 or just move the decimal point once to the left ex :

10% of 40 is 4

6 0

3 years ago

How much of a radioactive kind of calcium will be left after 10 days if the half-life is 5 days

Sophie [7]

Answer: 16 g

Step-by-step explanation:

We can solve this with the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$

Where:

$A$ is the final amount of radioactive calcium

$A_{o}=64 g$ is the initial amount of radioactive calcium

$t=10 days$ is the time elapsed

$h=5 days$ is the half life of radioactive calcium

Solving with the given data:

$A=(64 g)2^{\frac{-10 days}{5 days}}$

$A=16 g$ This is the amount of radioactive calcium left

7 0

4 years ago

The quantities x and y are oropottional find the contast of proportional of(r) y=rx

ASHA 777 [7]

Answer:

idkskdlkdogwocogdjemnwmmcweiikrkwkfjww

7 0

3 years ago