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Shalnov [3]
3 years ago
12

?? I'm stuck and I have no idea where to go from here

Mathematics
1 answer:
Alenkasestr [34]3 years ago
4 0
\cfrac{4(3(-2)-6)}{1+(-2)}= \cfrac{4(-6-6)}{1-2}= \cfrac{4(-12)}{-1}= \cfrac{-48}{-1}  =48
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Which is an equation of a line through (7,1) and perpendicular to y=-x+3
zysi [14]

Answer:

The answer is y=-1x+8 or you can write it this way y=-x+8 either way will work fine.

Step-by-step explanation:

y=mx+b

m=-1

y-y=m(x-x¹)

y-1=-1(x-7)

y-1=-1x+7

+1 +1

y=-x+8

I hope this helps:)

6 0
2 years ago
I need help on this question
Alex777 [14]
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6 0
2 years ago
"enter the values for m, n, and p that complete the difference:"
Keith_Richards [23]

m is 3

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8 0
3 years ago
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Which of the following is best description of the number 1.381432...?
gizmo_the_mogwai [7]

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Step-by-step explanation:

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4 0
3 years ago
A computer can be classified as either cutting dash edge or ancient. Suppose that 94​% of computers are classified as ancient. ​
taurus [48]

Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94​% of computers are classified as ancient.

(a) <u>Two computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 2 computers

            r = number of success = both 2

           p = probability of success which in our question is % of computers

                  that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient​</em>

So, it means X ~ Binom(n=2, p=0.94)

Now, Probability that both computers are ancient is given by = P(X = 2)

       P(X = 2)  = \binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}

                      = 1 \times 0.94^{2} \times 1

                      = 0.8836

(b) <u>Eight computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 8 computers

            r = number of success = all 8

           p = probability of success which in our question is % of computers

                  that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient</em>

So, it means X ~ Binom(n=8, p=0.94)

Now, Probability that all eight computers are ancient is given by = P(X = 8)

       P(X = 8)  = \binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}

                      = 1 \times 0.94^{8} \times 1

                      = 0.6096

(c) <u>Here, also 8 computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 8 computers

            r = number of success = at least one

           p = probability of success which is now the % of computers

                  that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

<em>LET X = Number of computers classified as cutting dash edge</em>

So, it means X ~ Binom(n=8, p=0.06)

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X \geq 1)

       P(X \geq 1)  = 1 - P(X = 0)

                      =  1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}

                      = 1 - [1 \times 1 \times 0.94^{8}]

                      = 1 - 0.94^{8} = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge​ is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

7 0
2 years ago
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