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Jet001 [13]
4 years ago
6

(07.05 MC)

Mathematics
1 answer:
inysia [295]4 years ago
7 0

Answer:

The X-intercepts are 2,3, and -5.

Step-by-step explanation:

X-2=0

X=2

X-3=0

X=3

X+5=0

X=-5

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Simplify: 70x/2x.<br> A. 32<br> B. 33<br> C. 34<br> D. 35
Tresset [83]
The answer is D. 35. The x’s cancel out because x divided by x is 1 leaving you with 70/2 which is then 35.
Hope this helped.
6 0
3 years ago
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I need help plz help
ivanzaharov [21]

Answer:

B. 70%

Step-by-step explanation:

7/10 of the numbers are more than 5

7/10= 70%

5 0
3 years ago
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The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4096, 0.4096, 0.1536, 0.0
Zinaida [17]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i)

And if we replace we got:

E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8

So we expect about 0.8 defective computes in a batch of 4 selected.

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

Solution to the problem

For this case we have the following distribution given:

X           0             1               2               3             4

P(X)  0.4096    0.4096    0.1536    0.0256    0.0016

And we satisfy that P(X_i) \geq 0 and \sum P(X_i) =1 so we have a probability distribution. And we can find the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And if we replace we got:

E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8

So we expect about 0.8 defective computes in a batch of 4 selected.

5 0
4 years ago
Help I will give brainliest
sineoko [7]

Answer:

Step-by-step explanation:

C

3 0
3 years ago
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Will give brainliest
natka813 [3]

Answer:

9 and -45

Step-by-step explanation:

they are whole numbers and integers are positive/negative whole numbers

4 0
3 years ago
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