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3 years ago

If your annual pay is $36,010, what is your biweekly pay based on 26 biweekly paychecks?

1 answer:

4 0

To get your bi weekly pay you would divide $36,010 by 26, seeing as bi weekly means you get payed every other week.
36,010 / 26 = $1,385
I hope this helps :)

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- A contractor plans to build a rectangular office complex. The floor area on the first floor of the

Neko [114]

Answer:

Width = 30

Step-by-step explanation:

Area = 18x * 10y = 180xy

Length = 6xy

Width = ?

Since the shape of the office complex is a rectangle,

Area of a rectangle = length × width

180xy = 6xy × width

Width = 180xy / 6xy

Width = 30

4 0

3 years ago

Rory built a sand castle that measured 73 inches by 47 inches, and dug a moat around it. afterward he decided to change the cast

navik [9.2K]

Answer

The bolded word is an compound adjective, a compound adjective is a adjective which contains two or more words joined by an hyphen.

Step-by-step explanation:

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3 years ago

This is the function f(x)= x2 − 4x + 6 .<br><br><br> What is the axis of symmetry?

kiruha [24]

Answer:

x = 2

Step-by-step explanation:

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3 years ago

Paul buys a number of large sacks of fertilizer costing $x each

Komok [63]

X times S = $27. X is the cost of the large sacks and S is the amount of sacks. Once they are multiplied it should equal $27. If this is incorrect please let me know so I can try and correct mistakes made.

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3 years ago

In proof testing of circuit boards, the probability that any particular diode will fail is . Suppose a circuit board contains di

ira [324]

Complete Question

In proof testing of circuit boards, the probability that any particular diode will fail is 0.01.

Suppose a circuit board contains 200 diodes. How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail? What is the (approximate) probability that at least four diodes will fail on a randomly selected board?If five boards are shipped to a particular customer, how likely is it that at least four of them will work properly? (A board works properly only if all its diodes work)

Answer:

The number of diode expected to fail is

$\mu =2$

The standard deviation is

$\sigma = 1.407$

The probability that at least four diodes will fail on a randomly selected board is

$P(X \ge 4) = 0.1419$

Step-by-step explanation:

Generally the testing follows a binomial distribution because

1 The number of trials is finite

2 There is only two outcomes( which (a) A diode fail (b) a diode does not fail)

3 The each individual trial is independent of each other.

4 The chances of success is constant for every trial

From the question we are told that

The probability that a diode will fail is $p = 0.01$

The sample size is n = 200

Generally the number of diode expected to fail is mathematically represented as

$\mu = n * p$

=> $\mu = 200 * 0.01$

=> $\mu =2$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{ np * (1 - p)}$

=> $\sigma = \sqrt{ 200* 0.01 * (1 - 0.01 )}$

=> $\sigma = 1.407$

Generally the probability that at least 4 diode will fail is mathematically represented as

$P(X \ge 4) = 1 - P(X < 4 )$

Generally

$P(X < 4 ) = P(X = 0 ) + P(X = 1) + P(X = 2) + P(X = 3)$

=> $P(X < 4 ) = ^nC_0 p^0 * (1-p)^{n-0} +^nC_1 p^1 * (1-p)^{n-1}+^nC_2 p^2 * (1-p)^{n-2}+ ^nC_3 p^3 * (1-p)^{n-3}$

=> $P(X < 4 ) = ^{200}C_0 p^0 * (1-0.01)^{200-0} +^{200}C_1 (0.01)^1 * (1-0.01)^{200-1}+^{200}C_2 (0.01)^2 * (1-0.01)^{200-2}+ ^{200}C_3 (0.01)^3 * (1-0.01)^{200-3}$

=> $P(X < 4 ) = 0.134 + 0.2707 + 0.2727 + 0.1814$

=> $P(X$

$P(X \ge 4) = 1 -0.8581)$

=> $P(X \ge 4) = 0.1419$

3 0

3 years ago