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2 years ago

I need help please I don't get this question

Mathematics

2 answers:

SpyIntel [72]2 years ago

8 0

Answer:

70.5%

Step-by-step explanation:

$\frac{1 + \frac{41}{100} }{2} = \frac{x}{100}$

I wish I could help

Norma-Jean [14]2 years ago

5 0

<h3>Answer: 159</h3>

No need to type in the percent symbol as that's already taken care of.

==============================================================

Explanation:

The square on the left is fully shaded in to represent 100% so far.

The square on the left is partially shaded in to represent an additional percentage on top of that 100%.

Note that we have 10 rows and 4 columns of white squares, so 10*4 = 40 white squares. Plus the additional white square to the right to get 41 white. That means there are 100-41 = 59 blue squares shaded in. This means that 59% of the second square is shaded blue.

That leads us to 100% + 59% = 159%

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2 years ago

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2 years ago

Find possible zeroes<br> f(x)=3x^6+4x^3-2x^2+4

larisa [96]

The possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are $\mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}$

<h3>How to determine the possible zeros?</h3>

The function is given as:

f(x) = 3x^6 + 4x^3 -2x^2 + 4

The leading coefficient of the function is:

p = 3

The constant term is

q = 4

Take the factors of the above terms

p = 1 and 3

q = 1, 2 and 4

The possible zeros are then calculated as:

$\mathbf{Zeros = \pm\frac{Factors\ of\ q}{Factors\ of\ p}}$

So, we have:

$\mathbf{Zeros = \pm\frac{1,2,4}{1,3}}$

Expand

$\mathbf{Zeros = \pm\frac{1,2,4}{1},\pm\frac{1,2,4}{3}}$

Solve

$\mathbf{Zeros = \pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}$

Hence, the possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are $\mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}$

Read more about rational root theorem at:

brainly.com/question/9353378

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1 year ago

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2 years ago

Of the 32 students in a class, 18 play the violin, 16 play the piano, and 7 play neither.

german

Answer:

(a)9

(b)I. 0.28125

II. 0.46875

III. 0.25

Step-by-step explanation:

There are a total of 32 students, therefore the number of elements in the Universal set, n(U)=32

$1$8 play the violin, n(V)=18\\16 play the piano, n(P)=16\\ 7 play neither, n(P \cup V)' =7$

(a)The Venn diagram is attached below.

$n(U)=n(V)+n(P)-n(V \cap P)+ n(P \cup V)'\\32=18+16+7-n(V \cap P)\\32=41-n(V \cap P)\\n(V \cap P)=41-32\\n(V \cap P)=9\\$

Therefore, 9 students play both the violin and piano.

(b)

I. Probability that the student plays the violin but not the piano

Number of Students who play violin only =18-x=18-9=9

$P(V \cap P') = \dfrac{n(V \cap P')}{n(U)}\\= \dfrac{9}{32}\\=0.28125$

ii.Probability that the student does not play the violin

Number of Students who does not play violin only =17-x+7=17-9+7=15

P(does not play violin only)

$= \dfrac{15}{32}\\=0.46875$

iii.Probability that the student plays the piano but not the violin

Number of Students who play piano only =17-x=17-9=8

$P(V' \cap P) = \dfrac{n(V' \cap P)}{n(U)}\\= \dfrac{8}{32}\\\\=0.25$

3 0

2 years ago

I need help please I don't get this question​

I need help please I don't get this question