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Basile [38]
3 years ago
12

More 5th grade work c’mon

Mathematics
1 answer:
fgiga [73]3 years ago
6 0

Answer: honestly for the first one, i think the answer would be 2.

they both have 5 at the end so you know its divisible :)

Step-by-step explanation:

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What is the answer to this problem?
Viktor [21]

Answer:

Step-by-step explanation:

The 75th percentile is the 3rd line, the median (50th percentile) is the second line, and the 25th percentile is the 1st line.

So, 10 is the answer.

3 0
3 years ago
A granola bar weighs 2 ounces. Do 15 granola bars weigh more than 3 pounds?
GarryVolchara [31]
No that is not correct. granola bars don't weigh that much.
5 0
3 years ago
Read 2 more answers
Find the value of the expression. Show your work.
Kay [80]

Answer:

n=16

Step-by-step explanation:

m=9

solve for now where

n=25 -m

n=25-9

n=16

4 0
3 years ago
The population of a city is modeled by the exponential function ƒ (x​) = 689,254 · 2.40.014x​, where x is the number of years af
Alex Ar [27]

Answer:

<h3>2036</h3>

Step-by-step explanation:

Given the population of a city is modeled by the exponential function

ƒ (x​) = 689,254 · 2.4e^0.014x​, where x is the number of years after 2000

In order to determine the year that the city's population will exceed one million, we will substitute f(x) = 1,000,000 and find x first as shown:

1,000,000 = 689,254  2.4e^0.014x​,

1000000/689,254  = 2.4e^0.014x​,

1.4508 = 2.4e^0.014x​,

1.4508 /2.4 = e^0.014x​,

0.6045 = e^0.014x​,

Apply ln to both sides

ln 0.6045 = lne^0.014x​,

-0.5033 = 0.014x

x = -35.95

Since x cannot be negative, lets say x ≈ 36years

<em>This means that the population will exceed one million after about 36 years. The year it will exceed this amount will be 2000+36 = 2036</em>

<em>Note that the exponential equation used was not correctly written but the same calculation can be employed for any exponential functions you might have.</em>

3 0
3 years ago
16. Solve for x<br> 20°<br> 7x+7
jonny [76]

Answer:

any figure or more info given??

4 0
3 years ago
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