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Rashid [163]
3 years ago
10

The difference of two integers is less than either integer

Mathematics
1 answer:
Marina86 [1]3 years ago
3 0
That is true based on only positive integers.
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-3/10q - 8/5q + 9 = 10<br><br> Solve for q
GuDViN [60]

Answer:

-1 9/10

Step-by-step explanation:

-3/10q - 8/5q + 9 = 10

-3/10q - 16/10q + 9 = 10

-19/10q + 9 = 10

-19/10q = 1

-19 = 10q

-19/10 = 10q/10

-1  9/10 = q

7 0
3 years ago
Simplify the expression. 8+6x3-(20/2)^2
REY [17]
<span>8 + 6 x 3 - (20/2)^2
</span><span>= 8 + 18 -(10)^2
</span>= 8 + 18 - 100
= -74

answer
- 74
8 0
3 years ago
Creating a Function from a Mapping The mapping shows a relationship between input and output values. Which ordered pair could be
Afina-wow [57]

I'm not sure what some of the ordered pairs are in what you wrote, but a function is when each input has one output. So, for every x there is one y. The x values can not repeat. So, remove one of two ordered pairs where the x value is repeating.

6 0
3 years ago
The lifetime of a computer can be modeled by an exponential random variable with an expected lifetime of 900 days. 1. Find the p
weeeeeb [17]

Answer:

1.  0.108368

2. 0.188876

Step-by-step explanation:

Let X be the exponential random variable that represents the lifetime of a computer.

i.e. X\sim Exp(0.001)

The probability that the computer will function more than 2000 days can be computed as follows:

P(X > 2000)

: f_X(x) = \lambda e^{-\lambda x

P(X > 2000) = 1  - P(X< 2000)

P(X > 2000) = exp(-2000/β) = e⁻²²

P(X > 2000) = 0.108368

2.

By applying conditional probability;

P(X>2000 \bigg | X>500) =\dfrac{P(X>2000 \  \cap  \ X>500)}{P(X> 500)}

P(X>2000 \bigg | X>500) =\dfrac{P(X>2000 )}{P(X> 500)}

P(X>2000 \bigg | X>500) =\dfrac{0.108368 }{0.57375}

\mathbf{P(X>2000 \bigg | X>500) =0.188876}

7 0
3 years ago
Round 84,630 to the nearest thousands
Salsk061 [2.6K]
The answer should be 85,000
5 0
3 years ago
Read 2 more answers
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