Carlisle is testing different types of planting soil to determine which type is most effective in growing blueberry bushes. He purchases 2 different brands of planting soil from a local store. Carlisle applies Brand A to an area of the yard that receives full sunlight and Brand B to another area of the yard that is in partial shade. He waters both areas daily. At the end of the study, Carlisle concludes Brand A is more effective in growing blueberry bushes. Why is his conclusion not valid?

Answer:

Its D i just took the test

Step-by-step explanation:

Answer:

D. g=7 and h=3

Step-by-step explanation:

Final answer:

To solve the expression 2(√(80/5) - 5), we simplify the square root first, then subtract 5, and finally multiply by 2, which results in the solution -2.

Explanation:

To solve the expression 2(√(80/5) - 5), let's first simplify the square root and the division inside it.

We have √(80/5) = √16 = 4, since 80 divided by 5 is 16, and the square root of 16 is 4.

Next, we subtract 5 from 4: 4 - 5 = -1.

Finally, we multiply the result by 2: 2 * (-1) = -2.

Therefore, the solution to the expression 2(√(80/5) - 5) is -2.

The answer to the given question is: [tex]\[ 2\left(\frac{\sqrt{80}}{5} - 5\right) = \frac{2\sqrt{80}}{5} - 10 \][/tex]

To solve the expression step by step:

 1. First, simplify the square root within the parentheses. The number 80 can be factored into [tex]\( 16 \times 5 \)[/tex], and since [tex]\( \sqrt{16} = 4 \),[/tex] we can rewrite[tex]\( \sqrt{80} \) as \( 4\sqrt{5} \).[/tex]

[tex]\[ 2\left(\frac{4\sqrt{5}}{5} - 5\right) \][/tex]

2. Next, distribute the 2 across the terms inside the parentheses:

[tex]\[ 2 \times \frac{4\sqrt{5}}{5} - 2 \times 5 \][/tex]

 3. Simplify each term:

[tex]\[ \frac{8\sqrt{5}}{5} - 10 \][/tex]

 4. At this point, we have the simplified form of the expression. However, if we want to combine the terms into a single fraction, we need a common denominator. The second term, -10, can be written as [tex]\( \frac{-10 \times 5}{5}[/tex]) to get the common denominator of 5:

[tex]\[ \frac{8\sqrt{5}}{5} - \frac{50}{5} \][/tex]

 5. Now, subtract the two fractions:

[tex]\[ \frac{8\sqrt{5} - 50}{5} \][/tex]

 6. This is the final simplified form of the expression. It cannot be simplified further algebraically because [tex]\( 8\sqrt{5} \)[/tex] and 50 do not have common factors other than 1.

Therefore, the final answer is: [tex]\[ \boxed{\frac{8\sqrt{5} - 50}{5}} \][/tex]

[tex]\[ 2\left(\frac{\sqrt{80}}{5} - 5\right) \][/tex]

And the final simplified answer is: [tex]\[ 2\left(\frac{\sqrt{80}}{5} - 5\right) = \frac{2\sqrt{80}}{5} - 10 \][/tex]

The new yearly salary is $28,140.

Amount of raise.
5%=0.05 
Multiply the original salary times the percentage of the raise.

$26,800×0.05=$1,340

New salary .

$26,800+1,340=$28,140

HOPE THIS HELPS! :)

This question is based on the concept of percentage. Therefore new salary when there is 5% raise in 26,800 is 28,140. There are many uses of calculating percent.

Percent is a proportionate of a number. Percent is a dimensionless quantity that is it has no unit. Percent is always out of hundred whereas fraction is always from 1. It is represented with the sign %. There are many uses of calculating percent. It is used in calculating interest in bank. It used in calculating inflation rate.

Mathematically,

The new yearly salary  of a person is 28,140.

percentage of raise in salary is

5%=0.05

Multiply the original salary and  the percentage of the raise in salary

26,800×0.05=1,340

New salary =26,800+1,340=28,140

Therefore new salary when there is 5% raise in 26,800 is 28,140

To know more about percent, here:

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The area of the two designs are given as 52 square inches and 61.375 square inches.

In these the second design 61.375 square inches is having the greater area.

If 500 envelops are made then area of paper = 61.375x500 =30687.5 square inches.

If envelops having 52 square inches are made out of 30687.5 square inches of paper ,number of envelops made =30687.5 ÷ 52=590.144

590 small envelops can be made having area as 52 square inches.

It is 590-500=90 more than the larger envelops made.

At 4:00 pm, the distance between the ships is changing at a rate of 30 km/h.

To find how fast the distance between the two ships is changing, you can use the Pythagorean theorem, as the two ships are moving at right angles to each other (north and south). Let's denote the distance between the two ships as "D," the speed of ship A as "vA," and the speed of ship B as "vB."

At any given time, you have:

D² = (Distance ship A travels)² + (Distance ship B travels)²

Now, we can differentiate both sides of this equation with respect to time "t" to find how the distance "D" is changing:

2D * dD/dt = 2(vA * dA/dt) + 2(vB * dB/dt)

Here, dD/dt represents the rate of change of the distance between the ships, dA/dt is the speed of ship A (10 km/h), and dB/dt is the speed of ship B (20 km/h).

Plug in the values:

2 * (dD/dt) = 2 * (10 km/h) + 2 * (20 km/h)

Now, solve for dD/dt:

2 * (dD/dt) = 20 km/h + 40 km/h

2 * (dD/dt) = 60 km/h

Now, divide by 2:

dD/dt = 60 km/h / 2 = 30 km/h

So, at 4:00 pm, the distance between the two ships is changing at a rate of 30 km/h.

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The volume of the space not filled by the sphere is the difference between the volume of a cube with edge length 6 inches and the volume of a sphere with radius 3 inches.

The volume of a cube of edge length s is

... V = s³

When the edge length is 6 in, the volume is

... V = (6 in)³ = 216 in³

The volume of a sphere with radius r is

... V = (4/3)π·r³

When the radius is 3 in, the volume is

... V = (4/3)π·(3 in)³ = 36π in³

Then the volume of the space between the cube and the sphere is

... Vcube - Vsphere = 216 in³ - 36π in³ ≈ 102.9 in³ . . . . corresponding to choice C

In Juanita's science experiment, the attribute measured is the plant's height, a key phenotypic trait in studying plant growth, which she records in centimeters, using consistent and precise units for scientific reliability.

The attribute measured in Juanita's science experiment is the height of a plant, which is a phenotypic trait important in plant biology. Juanita is collecting data related to plant growth by observing and recording the plant's height in centimeters daily. This meticulous tracking allows for analysis of plant growth patterns over time and the effect of various treatments, such as different fertilizers or amounts of water. The typical unit of measurement for the height of a plant is in centimeters or meters, which are part of the metric system used in scientific research.

During a laboratory experiment like the one Juanita is conducting, it is essential to measure such variables with precision and consistency. Measurements should be recorded using uniform units and methods to ensure reliability in the results, which may include the plant height, recorded in centimeters, at regular intervals. A graph representing this data might portray the number of days since the start of the experiment along the X-axis and the plant's height along the Y-axis, thereby allowing a visual assessment of the plant's growth over time.

Answer:

They just have 2 factors :/

Step-by-step explanation:

Bc yes

To calculate the percentage of players weighing 194 pounds or more from a box plot, we need to know where 194 pounds falls in relation to the quartiles on the plot. Without the box plot details, we cannot perform the needed analysis.

The student's question appears to concern the analysis of a box plot depicted in their materials to find the percentage of players weighing 194 pounds or more on a football team. However, to accurately answer this question, we need the specific details from the box plot, including the median (Q2), first quartile (Q1), third quartile (Q3), and any outliers if provided. From the box plot, we would determine where the weight of 194 pounds falls in relation to the quartiles. If 194 pounds is at or above Q3, then you would expect a lower percentage of players above this weight; if it is below Q3, then the percentage could be higher.

The approach typically involves identifying the number of players within each quartile and calculating the corresponding percentages. Without the specific box plot data, we cannot perform these calculations. Should you provide the box plot details, we can review the quartiles and calculate the exact percentage of players weighing 194 pounds or more.