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3 years ago

Round off to nearnest thousand 5,486 + 8,602

Mathematics

2 answers:

Korolek [52]3 years ago

6 0

5,486 + 8,602 is equal to around 14,000

Mars2501 [29]3 years ago

5 0

5,486 plus 8,602 is equal to around 14,000, give or take.

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Choose the quadrilateral that is not a rhombus of congruent sides and two angle that each measure 120 degrees

Amanda [17]

Answer for the above question is option C

Step-by-step explanation:

Option C - It's a quadrilateral but not a rhombus . As we know rhombus has congruent sides while its two angle can't be measured 120 degrees but in option C it seems two vertically opposite angles of to be 120 degrees. In rhombus opposite angles are same while adjacent angles are supplementary. All options except C are the rhombus of congruent sides.

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Determine the intercepts of the line. x-intercept: ( , ) y-intercept: ( , )

olga2289 [7]

Answer:

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Step-by-step explanation:

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3 years ago

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

Calculation the value from standard normal z table, we have,

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$