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Semmy [17]
3 years ago
12

25pts awarded and brainliest awarded, plz help asap!!!!!!

Mathematics
1 answer:
ivann1987 [24]3 years ago
4 0

Answer:

B. f(-1)=6

C.  The domain of f(x) is the set {-2.-1,0,1,2,3,4,5,6}

Step-by-step explanation:

To find f(5) from the table means, the y-value that corresponds to x=5.

This value is 12.

This implies that:

f(5)=12

Also the y-value that corresponds to -1 is 6.

Hence f(-1)=6

The domain of f(x) are the set of all the x-values.

The domain is : {-2.-1,0,1,2,3,4,5,6}

The range for f(x) is the set of all the corresponding y-values.

From the table, the range is: {5,6,7,8,9,10,11,12,13}

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Among cases of heart pacemaker malfunctions, were found to be caused by firmware, which is software programmed into the device.
alex41 [277]

Complete question is;

Among 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware, which is software programmed into the device. If the firmware is tested in three different pacemakers randomly selected from this batch of 8834 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted? Is this procedure likely to result in the entire batch being accepted?

Answer:

P(All three are not caused by firmware) = 83.84%

Probability that the entire batch will be accepted = 0.8384

Step-by-step explanation:

We are told that out of the 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware.

Thus,

Cases not caused by firmware = 8834 - 504 = 8330

So, probability of the first case not being affected by firmware is;

P(first case not caused by firmware) = 8330/8834

Also,

Probability of second case not being affected by firmware is given as;

P(second case not caused by firmware|first case not affected by firmware) = 8329/8833

Similarly,

Probability of third case not being affected by firmware is given as;

P(third case not caused by firmware|first and second not caused by firmware) = 8328/8832

Now, looking at the 3 Probabilities gotten, it is obvious that the events are not independent because the probability of occurence of one event depends on the probability of occurence of the other event.

Thus, we will make use of the general multiplication rule which is;

P(A & B) = P(B) × P(A|B)

Thus;

P(All three not caused by firmware) = P(first case not caused by firmware) × P(second case not caused by firmware|first case not affected by firmware) × P(third case not caused by firmware|first and second not caused by firmware)

Plugging in the relevant values, we have;

P(All three not caused by firmware) = (8330/8834) × (8329/8833) × (8328/8832)

P(All three are not caused by firmware) = 0.83840506679 ≈ 83.84%

6 0
3 years ago
50 apple's cost $25. How much would 75 apples cost?
8_murik_8 [283]

Answer:

$37.5

Step-by-step explanation:

(25/50) x 75 = 37.5

$25/50  .... each apple cost $0.5

0.5 x 75 = 37.5

8 0
3 years ago
Please help<br><br> 20% of 3km = ?
Lorico [155]

20% of 3km is 0.6 Kilometer

8 0
3 years ago
If u get 2 points every 5 minutes, how long will it take to get 1,000 points?
marshall27 [118]

Answer:
2500 Minutes

Step-by-step explanation:

If you get 2 points every 5 minutes, how long will it take it to get 1000 points? It's simple

Divide the Number of Points by the Number Of Points you get every 5 minutes then times it by the number of minutes you get 2 points making 2,500 Minutes

I hope it helped ya!

3 0
2 years ago
Read 2 more answers
Evaluate the integral ∫2−1|x−1|dx
defon

I think you might be referring to the definite integral,

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx

Recall the definition of absolute value:

|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

Then |x-1|=x-1 if x\ge1, and |x-1|=1-x is x. So spliting up the integral at <em>x</em> = 1, we have

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx

The rest is simple:

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}

5 0
3 years ago
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