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vladimir1956 [14]
3 years ago
14

(4.7 * 108) + (3.35 x 10)

Mathematics
2 answers:
Bond [772]3 years ago
7 0

Answer: 541.1

Step-by-step explanation:

multiply each set of parenthesis and add the two answers

anygoal [31]3 years ago
5 0

Answer:

541.1

Step-by-step explanation:

4.7 * 108 = 507.6 (according to BADMAS)

3.35 x 10 = 33.5

507.6 + 33.5 = 541.1 in fraction 5411/10

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What percent of one dollar is a nickle​
-Dominant- [34]

Answer:

5

Step-by-step explanation:

.05 ×100=5 per cent

a nickel is 5 cent

8 0
3 years ago
Which equation has the same solution as 2x^2+16x-10=0
dmitriy555 [2]
D
Divide the whole thing by 2
2(X^2+8x-5)
X^2+8x. =5
Half of 8 is 4 sq it so you add 16
X^2+8x+16= 5+16
(X+4)^2=21
8 0
3 years ago
Read 2 more answers
A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates
KIM [24]

Answer:

The probability is 0.971032

Step-by-step explanation:

The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.

The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:

P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}

In this case, we have 2000 electronics components with a probability 0.005 of fail during the useful life of the product and a probability 0.995 that each component operates without failure during the useful life of the product. Then, the probability that x components of the 2000 fail is:

P(x)=\frac{2000!}{x!(2000-x)!}*0.005^{x}*(0.995)^{2000-x}     (eq. 1)

So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:

P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)

We can also calculated that as:

P(x ≥ 5) = 1 - P(x ≤ 4)

Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)

Then, if we calculate every probability using eq. 1, we get:

P(x ≤ 4) = 0.000044 + 0.000445 + 0.002235 + 0.007479 + 0.018765

P(x ≤ 4) = 0.028968

Finally, P(x ≥ 5) is:

P(x ≥ 5) = 1 - 0.028968

P(x ≥ 5) = 0.971032

3 0
3 years ago
We want to find the zeros of this polynomial:<br> p(x)= 3x^3 – 3x^2– 18x
bulgar [2K]

Answer:

x=0   x=3   x=-2

Step-by-step explanation:

p(x)= 3x^3 – 3x^2– 18x

Factor out the greatest common factor, 3x

p(x)= 3x (x^2 – x– 6)

Factor inside the parentheses

What 2 numbers multiplies to -6 and adds to -1

-3*2 = -6

-3+2 = -1

p(x)= 3x (x-3)(x+2)

Setting the function equal to zero to find the zeros

0 = 3x (x-3)(x+2)

Using the zero product property

3x = 0  x-3 =0   x+2 =0

x=0   x=3   x=-2

3 0
3 years ago
Read 2 more answers
The probability that a customer of a network operator has a problem about you needing technical staff's help in a month is 0.01.
snow_tiger [21]

Answer:

(a) average calls = 5  

(b) probability that there is exactly one call in 6 consecutive monts = 0.038

Step-by-step explanation:

Let event of a customer requiring help in a particular month = H

and event of a customer not requiring help in a particular month = ~H

Given

p= 0.01,  therefore

Number of households, n = 500.

Binomial distribution:

x = number of households requiring help in a particular month

P(x,n,p) = C(x,n)*p^x*(1-p)^(n-x)

where

C(x,n) = n!/(x!(n-x)!) is the the number of combinations of x objects out of n

(a) Average number of households requiring help = np = 500*0.01 = 5

(b)

Probability that there are no calls requiring help in a particular month

P(0), q= C(0,n)*p^0(1-p)^(n-0)

= 1*1*0.99^500

= 0.006570483

Applying binomial probability over six months,

q = 0.006570483

n = 6

x = 1

P(x,n,q)

= C(x,n)*q^x*(1-q)^(n-x)

= 6!/(1!*5!) * 0.006570483^1 * (1-0.006570483)^5

= 0.038145

Therefore the probability that in 6 consecutive months there is exactly one month that no customer has called for help = 0.038

3 0
2 years ago
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