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saw5 [17]
3 years ago
9

The events committee buys 100 flowers for a school dance. The flowers are a

Mathematics
1 answer:
iogann1982 [59]3 years ago
6 0

Answer:

D) 68 carnations & 32 roses

Step-by-step explanation:

First of all, we need to identify the parts of the problem that we already know.

1) The events committee bought a total of 100 flowers.

2) The events committee spent a total of $139 on the carnations and roses, combined.

3) One carnation costs $0.75

4) One rose costs $2.75

We can create a formula to find the price of the flowers:

(x times $0.75) + (y times $2.75)= $139

If we calculate it out, the only option that fits the equation is 68 carnations and 32 roses.

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marusya05 [52]
18/27 can be reduced to 2/3 so the second option should be the correct answer
5 0
4 years ago
Read 2 more answers
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melamori03 [73]
58 is 50 plus 8. So its 50 and 8.That is for the first one.
3 0
3 years ago
Find the determinant of the following matrix. [1 -2] [3 4]​
Andrew [12]

Answer:

10

Step-by-step explanation:

[1 -2]

[3 4]

We can obtain the determinant of the above matrix by doing the following:

Determinant =(1 × 4) – (3 × –2)

Determinant = 4 – – 6

Determinant = 4 + 6

Determinant = 10

Thus, the determinant of the above matrix is 10

8 0
3 years ago
Jeanine is twice as old as her brother Marc. If the sum of their ages is 24. How old is Jeanine?
Tanya [424]
Let Marc's age be x. If Jeanine is twice as old as Marc, then her age will be 2x. If their ages added together are 24, then you can create the following equation:

2x + x = 24
3x = 24
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Remember that x is Marc's age. So if Marc is 8, and Jeanine is twice as old, she will be 16. 
4 0
4 years ago
Which of the following statements is not true? a) The standard deviation of the sampling distribution of sample mean = σ/√n b) T
I am Lyosha [343]

Answer:

e) The mean of the sampling distribution of sample mean is always the same as that of X, the distribution from which the sample is taken.

Step-by-step explanation:

The central limit theorem states that

"Given a population with a finite mean μ and a finite non-zero variance σ2, the sampling distribution of the mean approaches a normal distribution with a mean of μ and a variance of σ2/N as N, the sample size, increases."

This means that as the sample size increases, the sample mean of the sampling distribution of means approaches the population mean.  This does not state that the sample mean will always be the same as the population mean.

6 0
4 years ago
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