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Murljashka [212]
3 years ago
6

Convert to to decimal : 83/10​

Mathematics
2 answers:
Damm [24]3 years ago
8 0
Answer: 8.3



Explanation:
Verizon [17]3 years ago
4 0

Answer:

8.3

Step-by-step explanation:

just move the decimal in 83 over one digit since your already just dividing by 10

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There are 50 pennies in a roll. If you have 130 rolls of pennies, how many pennies do you have?
riadik2000 [5.3K]

Answer:

6,500 pennies or $65

Step-by-step explanation:

5 0
3 years ago
CD is a perpendicular bisector of . Which of the following could be the slopes of the two lines if plotted on a coordinate grid?
NISA [10]

Step-by-step explanation:

A is the correct answer of this question

8 0
3 years ago
For the function F(x) = 1/x+2, which of these could be a value of F(x) when xis
hram777 [196]

Answer:

B 2

Step-by-step explanation:

The computation of the value of f(x) in the case when x = 2 is shown below

As per the question, following function is given

F(x) = (1 ÷ x) + 2

Based on this, the x = 2

Now put the x value in the above equation

So,

= (1 ÷ 2) + 2

= (1 + 4) ÷ 2

= 5 ÷ 2

= 2.5

Hence the closet number is 2

Therefore the value of f(x) in the case when x = 2 is 2

Hence, the correct option is b.

5 0
3 years ago
Read 2 more answers
Is (-3,0)a solution to the equation 2x+6y=-6?
sineoko [7]

Answer:

Yes

Step-by-step explanation:

 x , y

(-3 , 0)

plug in the numbers

2(-3) + 6(0) = -6    

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4 0
3 years ago
One hundred items are simultaneously put on a life test. Suppose the lifetimes
romanna [79]

Answer:

a) \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b) \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( i-1 )st and the ith failures. Then, the T_{i} are independent with \mathrm{T}_{\mathrm{i}} being exponential with rate \frac{(101-i)}{200} .

Therefore,

a) E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]

=\sum_{i=1}^{5} \frac{200}{101-i}

\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b)

The variance is given by, \mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]

\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

7 0
4 years ago
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