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Harlamova29_29 [7]
3 years ago
12

Solve for u. -3(-6u+5) – 7u= 3 (u-4) - 1 Simplify your answer as much as possible.

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

1/4

Step-by-step explanation:

1. 18u−15−7u=3u−12−1

2. 11u−15=3u−12−1  (Simplify  18u-15-7u18u−15−7u  to  11u-1511u−15)

3. 11u−15=3u−13  (Simplify  3u-12-13u−12−1  to  3u-133u−13)

4. 11u=3u−13+15 (Add 1515 to both sides)

5. 11u=3u+2  ( Simplify  3u-13+153u−13+15  to  3u+23u+2)

6. 11u−3u=2  ( Subtract 3u3u from both sides)

7. 8u=2 ( Simplify  11u-3u11u−3u to 8u)

8. u= 2/8  (Divide both sides by 8)

9.u=1/4 (Simplify 2/8 to 1/4)

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Step-by-step explanation:

16856.70x=1.3362

So:  x = 16856.7/1.3362

x=12615.4019

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A rectangular garden measures 21ft by 29ft . Surrounding (and bordering) the garden is a path 3ft wide. Find the area of this pa
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4 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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