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atroni [7]
3 years ago
14

What do the following two equations represent? • y-8=1/5(x+5) • y-8=-1/2(x+5)

Mathematics
1 answer:
tangare [24]3 years ago
8 0

Answer:

Intersecting, but not perpendicular lines

When you graph it, it isn't gonna be apart, its gonna be intersecting

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Please help show work?
Romashka-Z-Leto [24]
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So what you have to do is leave the same denominator and just count. Once a fraction has the same numerator and denominator that means it is a whole number.
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4 years ago
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Dracula jumped 5 times every 30 minutes at that rate, how long,in minutes will it take to jump 6 times?
pentagon [3]

Answer:

36 minutes

Step-by-step explanation:

If I read this correctly, Dracula jumps 1 time every 6 minutes. If he jumped 5 times in 30 minutes, then that means when he jumped a sixth time, it was 36 minutes that passed since:

30 / 5 = 6

So, if he were to jump a sixth time:

30 + 6 = 36

Therefore, Dracula jumps 1 time per 6 minutes.

3 0
3 years ago
6. (4.2.12) Of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired. a. F
nata0808 [166]

Answer:

(a) Probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.

Step-by-step explanation:

We are given that of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired.

Let Probability that item are defective = P(D) = 0.20

Also, R = event of item being repaired

Probability of items being repaired from the given defective items = P(R/D) = 0.60

<em>So, Probability of items not being repaired from the given defective items = P(R'/D) = 1 - P(R/D) = 1 - 0.60 = 0.40 </em>

(a) Probability that a randomly chosen item is defective and cannot be repaired = Probability of items being defective \times Probability of items not being repaired from the given defective items

              = 0.20 \times 0.40 = 0.08 or 8%

So, probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Now we have to find the probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired.

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 = 20 items

            r = number of success = exactly 2

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

                  chosen item to be defective and cannot be repaired, i.e; 8%

<em>LET X = Number of items that are defective and cannot be repaired</em>

So, it means X ~ Binom(n=20, p=0.08)

Now, Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is given by = P(X = 2)

   P(X = 2) = \binom{20}{2} \times 0.08^{2} \times  (1-0.08)^{20-2}

                 = 190 \times 0.08^{2}  \times 0.92^{18}

                 = 0.2711

<em>Therefore, probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is </em><em>0.2711.</em>

             

6 0
3 years ago
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