Question 1:

The table gives the mass of liquids with a volume of 5 cm?.
Liquid
Mass (g)
Water
5
Glycerin
6.3
Milk
5.15
Olive oil
4.9
Density is the ratio of mass to volume.
Density = mass/volume
The density of milk is ___9/m^2

Question 2:
Did you know that the outdoor temperature can be found by counting the chirps made by crickets?
According to the Farmer's Almanac, the formula c = = + 4 relates to degrees
Celsius (c) to the number of cricket chirps (n) in 25 seconds. One night while camping, Charlie counts 51 cricket chirps in 25 seconds.
Note: Enter the number of degrees in the blank in order for this question to be scored correctly.
(1 point)
The temperature in degrees Celsius is____

Question 3:
Logan is solving this problem:
The cheetah ran 0.05 hours at a rate of 70 mph. How far did the cheetah run?
Logan solves the problem by writing:
d = rt
d = 70.05 miles
What is wrong with Logan's work?
(1 point)
• Logan's work is correct.
• Logan did not multiply the numbers.
• Logan used the wrong formula. He should have used A = bh.
• Logan labeled his answer incorrectly. The correct label is feet.

1 answer

Question 1:

To find the density of milk, we use the formula:

\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]

From the table, the mass of milk is 5.15 g, and the volume is 5 cm³.

So,

\[ \text{Density of milk} = \frac{5.15 \text{ g}}{5 \text{ cm}^3} = 1.03 \text{ g/cm}^3 \]

The answer is: 1.03 g/cm³ (Note: You have mentioned "___9/m^2" but the correct density should be in g/cm³ for appropriate units.)


Question 2:

According to the formula provided:

\[ c = \frac{n - 40}{4} \]

Where \( n \) is the number of cricket chirps in 25 seconds. Charlie counted 51 chirps.

So we plug 51 into the equation:

\[ c = \frac{51 - 40}{4} = \frac{11}{4} = 2.75 \]

The temperature in degrees Celsius is: 2.75


Question 3:

Logan's work uses the formula for distance:

\[ d = rt \]

Where \( r \) is the rate and \( t \) is the time. However, Logan wrote:

\[ d = 70.05 \text{ miles} \]

He incorrectly wrote "70.05" instead of calculating using the given time of 0.05 hours.

To find the correct distance:

\[ d = r \times t = 70 \text{ mph} \times 0.05 \text{ hours} = 3.5 \text{ miles} \]

The issue is that Logan did not multiply the numbers correctly to find the distance.

So the correct answer is: Logan did not multiply the numbers.