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babymother [125]
3 years ago
14

Of 24 tulips 1/3 are red. how many tulips are red?

Mathematics
2 answers:
Soloha48 [4]3 years ago
3 0
6 tulips are red out of 24
hram777 [196]3 years ago
3 0
8 tulips are red, the others are yellow
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Use the data below.
LuckyWell [14K]

Step-by-step explanation:

mean = sum of all data points / number of data points

(22+16+18+14+16+34+20)/7 = 140/7 = 20

median is the number, where half of the data points are smaller, and the other half are larger.

first we need to sort the data points

14, 16, 16, 18, 20, 22, 34

median = 18

mode is the data point occurring most often.

mode = 16 (occurring 2 times, while the others occur only once).

range is the difference between the largest and the smallest data point.

range = 34 - 14 = 20

5 0
1 year ago
Suppose you have $16,000 to invest in three stocks, A, B, and C. Stock A is a low-risk stock that has expected returns of 4%. St
podryga [215]

To solve this problem, let us say that:

money invested in stock A = A

money invested in stock B = B

money invested in stock C = C

The given problem states that:

C = A * (1 / 4) = 0.25 A

B = A * (1 / 2) = 0.50 A

It was stated that we only have $16,000 to invest. Therefore:

A + B + C = 16,000

Substituting values of C and B in terms of A:

A + 0.50 A + 0.25 A = 16,000

1.75 A = 16,000

A = $9,142.86

So C and B is:

C = 0.25 (9142.86)
C = $2285.71

 

B = 0.50 (9142.86)

B = $4571.43 

5 0
3 years ago
Read 2 more answers
What is the simplify of this expression -2(2b-3)
patriot [66]

-2(2b-3)

Multiply the bracket by -2

Whenever multiplying a -negative number with a -negative number= + positive number.

(-2)(2b)(-2)(-3)

-4b+6

Answer: -4b+6

6 0
3 years ago
Read 2 more answers
If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.
Pie
Given that f^{(n)}(0)=(n+1)!, we have for f(x) the Taylor series expansion about 0 as

f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n

Replace n+1 with n, so that the series is equivalent to

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}

and notice that

\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}

Recall that for |x|, we have

\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}

which means

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}
\implies f(x)=\dfrac1{(1-x)^2}
5 0
3 years ago
Help Plz!
Len [333]

Answer:

10 for each side so 40 tiles

Step-by-step explanation:

6 0
3 years ago
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