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3 years ago

15

if buyer purchases 25% of a lot in the spring and then purchases 50% more in the fall what percentage of the lot is still availa

ble to purchase

1 answer:

brilliants [131]3 years ago

8 0

Answer:

25%

Step-by-step explanation:

Total purchase = 25%+50%

Total purchase = 75%

Percentage left = 100% - 75%

Percentage left = 25%

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Which data set COULD NOT be represented by the histogram shown?

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D. Only has 3 numbers in the 5-9 range, while the histogram shows 4. Hope this helps :)

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3 years ago

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Perform the indicated operation. Be sure the answer is reduced.

avanturin [10]

<h3>Given Equation:-</h3>

$\boxed{ \rm \frac{4x^{2}y^{3}z}{9} \times \frac{45y}{8 {x}^{5} {z}^{5} }}$

<h3>Step by step expansion:</h3>

$\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times \dfrac{45y}{8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }$

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$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times \dfrac{45y}{2{x}^{5} {z}^{3} }$

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$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }$

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$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5} {z}^{3} }$

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$\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }$

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$\dashrightarrow \sf\dfrac{y^{3}z}{1} \times \dfrac{5y}{2{x}^{3} {z}^{3} }$

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$\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }$

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$\dashrightarrow \sf\dfrac{y^{3}}{1} \times \dfrac{5y}{2{x}^{3} {z}^{2} }$

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$\dashrightarrow \sf \dfrac{5y \times {y}^{3} }{2{x}^{3} {z}^{2} }$

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$\dashrightarrow \sf \dfrac{5 \times {y}^{4} }{2{x}^{3} {z}^{2} }$

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$\dashrightarrow \bf \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }$

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$\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}$

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2 years ago

What is 25 percent of 100

Yanka [14]

25% of 100 equals 25.

4 0

3 years ago

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One garden had 5 times as many raspberry bushes as a second garden. After 22 bushes are transplanted from the first garden to th

Aleonysh [2.5K]

Let the number of raspberry bushes in one garden = x

And the number of raspberry bushes in second garden = y

Garden one has 5 times as many raspberry bushes as second garden,

So the equation will be,

x = 5y -------(1)

If 22 bushes were transplanted from garden one to the second, number of bushes in both the garden becomes same,

Therefore, (x - 22) = (y + 22)

x - y = 22 + 22

x - y = 44 ------(2)

Substitute the value of x from equation (1) to equation (2)

5y - y = 44

4y = 44

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Substitute the value of 'y' in equation (1),

x = 5(11)

x = 55

Therefore, Number of bushes in garden one were 55 and in second garden 11 originally.

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3 years ago

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Answer:

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3 years ago