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2 years ago

15

A florist uses 200 liliesto make 8 identical flower arrangements. How many lilies would the floristuse to create 6identical arra

ngements?

1 answer:

Roman55 [17]2 years ago

3 0

Answer:

150

Step-by-step explanation:

200 divided by 8 = 25 each

25 x 6 = 150

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What's the lowest term of 14/28

rjkz [21]

Answer:

Step-by-step explanation:

IN a fraction it would be 1/2

As a decimal it would be 0.5

As a percent is would be 50%

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3 years ago

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PLS HELP ME ASAL FOR 7, 8, and 9. A and B! (SHOW WORK!!!!) + LOTS IF POINTS!! (BE SPECIFIC ON WHICH your ANSWERING)

nevsk [136]

Ok so
Q.1=B
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3 years ago

If the experiment is consisting of rolling a fair die once, find:

Lerok [7]

question one

1, 2, 3, 4, 5, 6

question two

3/6 or 1/2

question three

3/6 or 1/2

question four

2/6 or 1/3

question five

4/6 or 2/3

you use the list from question one, and count how many fit the criteria and put that number over the total ( which is 6 in this case )

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3 years ago

The lengths of the sides of a triangle are in the extended ratio 7: 9 : 10. The perimeter of the triangle is 52 cm. What are t

Pavel [41]

Answer:

14 cm, 18 cm, 20 cm

Step-by-step explanation:

7x + 9x + 10x = 26x

equation: 26x = 52

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2 years ago

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected

jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0

3 years ago