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ivanzaharov [21]
2 years ago
11

CAN SOMEONE PLEASE HELP

Mathematics
2 answers:
Allushta [10]2 years ago
4 0
4 is the right answer
galina1969 [7]2 years ago
4 0

i think the right answer is 4

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Antonio is also working on a new spreadsheet program. The program takes up 14 bytes of memory. If each byte is equal to 8 bits,
Tanya [424]
14 x 8 = 112

The answer is 112
5 0
3 years ago
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Can anyone plz help me in this plz step by step​
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Step-by-step explanation:

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3 years ago
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How to simplify (x+6)(x-6)
SCORPION-xisa [38]
Using (a-b)(a+b)= a2-b2
Simply the product
x2-62
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6 0
4 years ago
Calculate s5 for the sequence defined by
pantera1 [17]
ANSWER

S_5= 42.5

EXPLANATION

The general term of the arithmetic sequence is

a_n=1+\frac{5}{2}n

For the first term,we substitute n=1 to get,

a_1=1+\frac{5}{2}\times 1

a_1=1+2.5=3.5

Similarly

a_5=1+\frac{5}{2}\times 5

a_5=13.5

The sum of the first n-terms is given by the formula,

S_n=\frac{n}{2}(a_1+l)

where l is the last term.
In this case

l=a_5=13.5

We substitute these values to get,

S_5= \frac{5}{2} (3.5 + 13.5)

S_5= \frac{5}{2} (17)
S_5=42.5



We could have also used the formula,

S_n=\frac{n}{2}(2a_1+(n-1)d)

S_5=\frac{5}{2}(2(3.5)+(5-1)2.5)

S_5=42.5
3 0
4 years ago
Read 2 more answers
A probability distribution for a random variable Y is given by P(Y=y)=cy, for y=1,2,4,5 and c is a constant. Find E(Y), the expe
Dahasolnce [82]

Answer:

c(1+2+4+5) =1

c =\frac{1}{12}

Now we can find the expected value with this formula:

E(Y) =\sum_{i=1}^n Y_i P(Y_i) =\sum_{i=1}^n cy_i *y_i = cy^2_i

And replacing we got:

E(Y) = \frac{1}{12} (1^2 + 2^2 + 4^2 +5^2) = \frac{46}{12}= \frac{23}{6}

Step-by-step explanation:

For this case we know the following probability mass function given:

P(y) = cy , y= 1,2,4,5

For this case we need to satisfy the following condition in order to have a probability distribution function:

\sum_{i=1}^n P(y_i) =1

And we have this:

c(1+2+4+5) =1

c =\frac{1}{12}

Now we can find the expected value with this formula:

E(Y) =\sum_{i=1}^n Y_i P(Y_i) =\sum_{i=1}^n cy_i *y_i = cy^2_i

And replacing we got:

E(Y) = \frac{1}{12} (1^2 + 2^2 + 4^2 +5^2) = \frac{46}{12}= \frac{23}{6}

6 0
3 years ago
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