The elements least likely to form bonds are found in which group? 15, 16, 17, 18

The perimeter of a parallelogram is 72 meters. The width of the parallelogram is 4 meters less than it length. Find the length a

lesya [120]

P = L + L + w + w

72 = 2(L) + 2(L-4)

Distribute

72 = 2(L) + 2(L) - 8

Add like terms

72 = 4(L) - 8

Subtract 8 from both sides

64 = 4(L)

Divide both sides by 4

16 = L

The length is 16. Now substitute that in the equation for finding w.

w = L - 4

w =

4 0

3 years ago

what formulas could be used to determine the surface area of a cylinder sandwiched between one cone and half of a sphere?

skelet666 [1.2K]

Answer: Rectangle around the cylinder, Lateral surface of the cone, Half a sphere

To find this surface area, you have 3 different parts. First, you need the rectangular portion of the cylinder, because the circles will be on the inside of the composite shape. Then, you will need the lateral side of the cone, that's everything but the circular base. Again, the circle will be on the interior. Finally, you should find the surface area of half of the sphere.

4 0

3 years ago

PLEASE HELP <br><br> WILL GIVE BRAINLIEST<br> AND 5.0 RATING

Vlad [161]

Answer: The coordinates are (2.5, -1.5).

Explanation: The explanation is in the image.

7 0

2 years ago

Read 2 more answers

What is the prime factorization for 945

Scilla [17]

Answer: 3•3•3•5•7

Step-by-step explanation:

Create a factor tree. See photo attached. (:

8 0

2 years ago

Read 2 more answers

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien

diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = $\frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}$

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) = $\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)}$ = $\frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)}$ = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0

3 years ago