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3 years ago

10

PLZ HURRY IM TIMEED!! ＞﹏＜

1 answer:

Mumz [18]3 years ago

4 0

B) Rotterdam. Because sea level is 0 and Rotterdam is the only one below sea level

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The population of a city is 45,000 and decreases 2% each year. If the trend continues, what will the population be after 15 year

Sloan [31]

The answer is 31,500.
Let's clarify some things :
1% of 45,000 is 450
Maybe you're wondering why , it's actually very simple .
Just move the decimal to the left once if you want to find 10% , move the decimal twice if you want to find 1%.
We now that 1% equals 450. Then 2% equals to....
Well just add another 1% .
450 + 450 or 450 times 2
Which equals 900.
Now we want to find what's the population after 15 years.
900 times 15= 13,500
But that's not the answer ....
That's what is decreasing.
45,000 - 13,500= 31,500
That's why the answer is 31,500.

4 0

2 years ago

Help I don’t know the steps

Korvikt [17]

Answer:

down below

Step-by-step explanation:

straight lines are 180 degrees and you have 2 angles on a line. one of which is 72 and the other is unknown. It is solved for below.

$180=72+x\\180-72=72+x-72\\108=x$

your missing degree is 108.

i may not be right, if i'm not i am very sorry.

7 0

2 years ago

Use the digits 5,4 and 8 once each to write an odd number greater than seven hundred

natali 33 [55]

845 because 800 is greater than 700 and 845 is an odd number because it ends in 5 (which is an odd number).

4 0

2 years ago

Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and

Irina18 [472]

Answer:

0.918 is the probability that the sample average sediment density is at most 3.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.80

Standard Deviation, σ = 0.85

Sample size,n = 35

We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.

Formula:

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

Standard error due to sampling:

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437$

P(sample average sediment density is at most 3.00)

$P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)$

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

$P(x \leq 3.00) = 0.918$

0.918 is the probability that the sample average sediment density is at most 3.00

4 0

2 years ago

Please help me solve this and what are the answers

mixas84 [53]

Answer:

5y=-2x - 3x

5y = -5x

5y + 5x = 0

5(y + x ) = 0

maybe

7 0

2 years ago