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maks197457 [2]
3 years ago
10

A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15, respectively. If some of the nuts of one of the three ty

pes are removed, which of the following could be the ratio of pecans to cashews to almonds remaining in the bowl?
i. 1 : 2 : 3
ii. 2 : 3 : 4
iii. 4 : 7 : 10
A. I only
B. II only
C. III only
D. I and III only
E. II and III only
Mathematics
1 answer:
kirza4 [7]3 years ago
5 0

Answer:

iii

Step-by-step explanation:

because of the amount taken from the cashews.and nuts and 1 of 3 were taken away

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Plz help worth 50 points
serious [3.7K]

Answer:

The answer is A

Step-by-step explanation:

Starting from -3 in the Y values of option A. If you subtract three from each value, you will get the next value to the right.

  • -3 minus -3 = -6
  • -6 minus -6 = -9

       

6 0
3 years ago
Read 2 more answers
Which fraction, decimal number, and percent are equivalent?
uranmaximum [27]

Answer:

A. 2/5, 0.40, 40%

Step-by-step explanation:

To turn a fractions into a decimal divided the numerator by the denominator. 2/5=0.40.

8 0
3 years ago
Juanita is measuring a table for her dining room. Table area is 46 7/8 square feet. Table length is 7 1/2 feet. She does not wan
Anastaziya [24]
No. If you still want the  46 7/8 square foot table and less than 6ft wide than the table length must be greater than 7 1/2 feet long. For the math to work out. Because, the 7 1/2 foot long table would need a 6 1/4 wide table to fit the 46 7/8 square foot.
5 0
3 years ago
Solve 8x2 + 6x + 5 = 0
MariettaO [177]
8 x 2 + 6x + 5=0
16 + 6x + 5=0
6x+21=0
6x+21-21=0-21
6x=-21
6x/6=-21/6
x=-7/2
5 0
3 years ago
How do you do this question?
Alex Ar [27]

Step-by-step explanation:

(a) dP/dt = kP (1 − P/L)

L is the carrying capacity (20 billion = 20,000 million).

Since P₀ is small compared to L, we can approximate the initial rate as:

(dP/dt)₀ ≈ kP₀

Using the maximum birth rate and death rate, the initial growth rate is 40 mil/year − 20 mil/year = 20 mil/year.

20 = k (6,100)

k = 1/305

dP/dt = 1/305 P (1 − (P/20,000))

(b) P(t) = 20,000 / (1 + Ce^(-t/305))

6,100 = 20,000 / (1 + C)

C = 2.279

P(t) = 20,000 / (1 + 2.279e^(-t/305))

P(10) = 20,000 / (1 + 2.279e^(-10/305))

P(10) = 6240 million

P(10) = 6.24 billion

This is less than the actual population of 6.9 billion.

(c) P(100) = 20,000 / (1 + 2.279e^(-100/305))

P(100) = 7570 million = 7.57 billion

P(600) = 20,000 / (1 + 2.279e^(-600/305))

P(600) = 15170 million = 15.17 billion

7 0
2 years ago
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