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Papessa [141]
3 years ago
7

Pedro simplified four over seven divided by five over eight.; his work is shown below. Identify where he made his error. Origina

l division problem: four over seven divided by five over eight. Step 1: four over seven multiplied by eight over five. Step 2: four times five over seven times eight. Step 3: twenty over fifty-six. Step 4: five over fourteen.
Mathematics
2 answers:
Triss [41]3 years ago
6 0
He  cross multiplied when he should've multiplied straight across...

<span>Original division problem: four over seven divided by five over eight
</span>
step one: f<span>our over seven multiplied by eight over five</span>

step two: 4*8=32    7*5=35

32/35
Vlad [161]3 years ago
5 0
Step 3 is wrong.
It should be 4/7 / 5/8
                    4/7 x 8/5
                    32/ 35

                    
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take 2 points from the graph

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Find the volume of 5.4 inches 2 inches 7.1 inches round to the nearest 10th is necessary
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If it is a cube, the answer would be 76.7 .

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4 years ago
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he port of South Louisiana, located along miles of the Mississippi River between New Orleans and Baton Rouge, is the largest bul
iVinArrow [24]

Answer:

a) P(x < 5) = 0.7291

b) P(x ≥ 3) = 0.9664

c) P(3 < x < 4) = 0.2373

d) 5.35 million tons of cargo in a week will require the port to extend operating hours.

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 4.5 million tons of cargo per week

Standard deviation = σ = 0.82 million

a) The probability that the port handles less than 5 million tons of cargo per week

= P(x < 5)

We first standardize/normalize 5.

The standardized score of any value is the value minus the mean divided by the standard deviation.

z = (x - μ)/σ = (5 - 4.5)/0.82 = 0.61

To determine the probability that the port handles less than 5 million tons of cargo per week

P(x < 5) = P(z < 0.61)

We'll use data from the normal probability table for these probabilities

P(x < 5) = P(z < 0.61) = 0.72907 = 0.7291 to 4 d.p

b) The probability that the port handles 3 or more million tons of cargo per week?

P(x ≥ 3)

We first standardize/normalize 3.

z = (x - μ)/σ = (3 - 4.5)/0.82 = -1.83

To determine the probability that the port handles less than 3 or more million tons of cargo per week

P(x ≥ 3) = P(z ≥ -1.83)

We'll use data from the normal probability table for these probabilities

P(x ≥ 3) = P(z ≥ -1.83)

= 1 - P(z < -1.83)

= 1 - 0.03362

= 0.96638 = 0.9664

c) The probability that the port handles between 3 million and 4 million tons of cargo per week = P(3 < x < 4)

We first standardize/normalize 3 and 4.

For 3 million

z = (x - μ)/σ = (3 - 4.5)/0.82 = -1.83

For 4 million

z = (x - μ)/σ = (4 - 4.5)/0.82 = -0.61

To determine the probability that the port handles between 3 million and 4 million tons of cargo per week

P(3 < x < 4) = P(-1.83 < z < -0.61)

We'll use data from the normal probability table for these probabilities

P(3 < x < 4) = P(-1.83 < z < -0.61)

= P(z < -0.61) - P(z < -1.83)

= 0.27093 - 0.03362

= 0.23731 = 0.2373 to 4 d.p

d) Assume that 85% of the time the port can handle the weekly cargo volume without extending operating hours. What is the number of tons of cargo per week that will require the port to extend its operating hours?

Let x' represent the required number of tons of cargo thay will require the port to extend its operating hours.

Let its z-score be z'

P(x < x') = P(z < z') = 85% = 0.85

Using the normal distribution table,

z' = 1.036

z' = (x' - μ)/σ

1.036 = (x' - 4.5)/0.82

x' - 4.5 = (0.82 × 1.036) = 0.84952

x' = 0.84952 + 4.5 = 5.34952 = 5.35 to 2 d.p

Therefore, 5.35 million tons of cargo in a week will require the port to extend its operating hours.

Hope this Helps!!!

3 0
3 years ago
suppose that a department contains 11 men and 17 women. how many different committees of 6 members are possible if the committee
slega [8]

If a department has 11 male employees and 17 female employees, then 198968 different committees of 6 members are possible with the condition that the committee has strictly more female employees than male.

As per the question statement, a department has 11 male employees and 17 female employees.

We re required to calculate the total number different committees that can be formed with 6 members strictly having more female employees than male.

Now for committee of 6 members to have more women than men, there can be two combinations:

(4 women and 2 men)

Or, (5 Women and 1 man).

That is, we will need to calculate the number of combinations we can have by selecting 4 female employees from a group of 17 and 2 Male employees from a group of 11 and the number of combinations we can have by selecting 5 female employees from a group of 17 and 1 Male employee from a group of 11 and add up the two number of combinations to obtain our required answer.

Then comes the most important thing to know to be able to solve this question, i.e., the formula to calculate combinations, which goes as

nCr=\frac{n!}{r!(n-r)!}

Therefore, the total number different committees that can be formed with 6 members strictly having more female employees than male is

[(17C4)*(11C2)+(17C5)*(11C1)]\\=[(2380*55)+(6188*11)]\\=(130900+68068)\\=198968

  • combination(s): In mathematics, a combination is a way of selecting items from a collection or set, where the order of selection does not matter, i.e., for example, we have a set of three numbers X, Y and Z and then, in how many ways can we select two numbers from each set, is defined by combination.

To learn more about Combinations, click on the link below.brainly.com/question/8044761

#SPJ4

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