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Anni [7]
3 years ago
7

Which expression is equivalent to StartRoot 200 EndRoot?

Mathematics
1 answer:
Xelga [282]3 years ago
8 0

Answer:

Answer

is B

10√2

Step-by-step explanation:

is: √200

Simplifying the given expression, we get:

200 = 2 x 2 x 2 x 5 x 5

⇒ 200 = (2)² x (5) ² x 2

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Sketch the area under the standard normal curve over the indicated interval and find the specified area. (You can use the probab
mafiozo [28]

The area to the right of z = 1.35 is 0.0885 and the area to the left of -0.47 is 0.3192.

<h3>How to compute the values?.</h3>

Given z = 1.35

= 1- P(z < 1.35)

= 1- 0.9115

= 0.0885

The area to the left of -0.47 will be:

= 1 - P(z < 0.47)

= 1 - 0.6808

= 0.3192

Learn more about normal curve on:

brainly.com/question/6758792

#SPJ1

8 0
2 years ago
1. If f(x) = 2x - 3, find the following:<br> a. f(-2) =<br> b. f(7) =<br> c. f(-4) =
Pavlova-9 [17]
A. 2(-2) - 3
-4 - 3
-7

b. 2(7) - 3
14 - 3
11

2(-4) - 3
-8 - 3
- 11
5 0
3 years ago
If the diagonals of a quadrilateral bisect the angles, is the quadrilateral always a parallelogram? Explain your answer.
Liula [17]
Yes. If the diagonals bisect the angles, the quadrilateral is always a parallelogram, specifically, a rhombus.


Consider quadrilateral ABCD. If diagonal AC bisects angles A and C, then ΔACB is congruent to ΔACD (ASA). Hence AB=AD and BC=CD (CPCTC).

Likewise, if diagonal BD bisects angles B and D, triangles BDA and BDC are congruent, thus AB=BC and AD=CD. (CPCTC again). Now, we have AB=BC=CD=AD, so the figure is a rhombus, hence a parallelogram.
6 0
3 years ago
carlos is using a hose to fill a bucket. the path of the water can be represented by the function f(t) where f(t) represents the
Illusion [34]

Answer:

The function represents the height of the water in the bucket

Step-by-step explanation:

Given

f(t) = -t^2 + 4t

Required

State what the function represents

From the question, we understand that t represents the time spent while the function illustrates the water level in the bucket.

This implies that f(t) calculates the height of the water in the bucket, given the value of time (t)

5 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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