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Serjik [45]
3 years ago
15

Which of the following are roots of the polynomial function? Check all that apply.

Mathematics
1 answer:
lys-0071 [83]3 years ago
3 0

Answer:

B D and E are the correct answers, according to apex.

Step-by-step explanation:


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Suppose you and a friend each choose at random an integer between 1 and 8, inclusive. For example, some possibilities are (3,7),
Bezzdna [24]

Answer and explanation:

Given : Suppose you and a friend each choose at random an integer between 1 and 8, inclusive.

The sample space is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)  (1,7) (1,8)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)   (2,7) (2,8)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)   (3,7) (3,8)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)   (4,7) (4,8)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)   (5,7) (5,8)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)   (6,7) (6,8)

(7,1) (7,2) (7,3) (7,4) (7,5) (7,6)   (7,7) (7,8)

(8,1) (8,2) (8,3) (8,4) (8,5) (8,6)   (8,7) (8,8)

Total number of outcome = 64

To find : The following probabilities ?

Solution :

The probability is given by,

\text{Probability}=\frac{\text{Favorable outcome }}{\text{Total outcome}}

a) p(you pick 5 and your friend picks 8)

The favorable outcome is (5,8)= 1

\text{Probability}=\frac{1}{64}

b) p(sum of the two numbers picked is < 4)

The favorable outcome is (1,1), (1,2), (2,1)= 3

\text{Probability}=\frac{3}{64}

c) p(both numbers match)

The favorable outcome is (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8) = 8

\text{Probability}=\frac{8}{64}

\text{Probability}=\frac{1}{8}

7 0
3 years ago
Part A: Victor is studying for his Finais. He spends 1/3 of an hour studying for History, 2/3 an hour
Elan Coil [88]

Answer:

3 3/4

Step-by-step explanation: all that you have to do is add 1/3 +2/3+11/4 which gives you 3 3/4

4 0
2 years ago
Help please and thanks! I always give brainliest!
Alika [10]

Answer:

X = 18

y = 9

Step-by-step explanation:

Basically the 2 diagnols bisect each other making them congruent. so that makes

x-3 = 15 and y + 3 = 12

simplify and

the. you simplify and get x = 18 and y =9

3 0
2 years ago
Is cos 314° positive,negative,or zero?
Oliga [24]
Maybe positive not sure
4 0
3 years ago
Helppp me plsssssssss<br><br>​
Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

7 0
3 years ago
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