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3 years ago

Which interval has the most students? 0-4 5-9 10-14 15-19

Mathematics

2 answers:

-Dominant- [34]3 years ago

8 0

The answer is 0-4 it has more

Sonja [21]3 years ago

8 0

Answer:

0-4

Step-by-step explanation:

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The population of Williston is currently 21,200 people. If the population

Pachacha [2.7K]

$21200 + 415(y)$

$27000 = 21200 + 415(y)$

$27000 - 21200 = 21200 - 21200 + 415(y)$

$5800 = 415(y)$

$\frac{5800}{415} = \frac{415(y)}{415}$

$13.9759..... = y$

$it \: will \: take \: about \: 14.0 \: years$

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2 years ago

What type of polynomial is: 3x+x^2+4 A.quadratic B. quartic C. linear D. cubic

Svetradugi [14.3K]

Answer:

A.quadratic

Step-by-step explanation:

3x+x^2+4

Put in order from largest to smallest power of x

x^2 +3x+4

The highest power is 2

A.quadratic highest power 2

B. quartic highest power 4

C. linear highest power 1

D. cubic highest power 3

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2 years ago

HELP I'M SUFFERING DA HECK IS THIS QUESTION!!!!

erica [24]

Answer:

Answer choices A and D. E would also be correct but it's a perfect correlation and the question said "strong but not perfect"

Step-by-step explanation:

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3 years ago

Read 2 more answers

Point R on the coordinate grid below shows the location of a car in a parking lot. A second car needs to park 7 units above R to

Levart [38]

Answer:

it's (2,3)

Step-by-step explanation:

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2 years ago

The concentration of dichlorodiphenyltrichloroethane (DDT - an infamous pesticide) in Lake Michigan has been declining exponenti

mojhsa [17]

Answer: A) $y=13e^{-0.1359t}$

B) H = 5.10

C) Yes

Step-by-step explanation: Exponential Decay function is a model that describes the reducing of an amount by a constant rate over time. Generally, it is written in the form: $y(t)=Ce^{rt}$

A) C is initial quantity, in this case, the initial concentration of DDT. To determine r, using the data given:

$y(t)=Ce^{rt}$

$2.22=13e^{13r}$

$e^{13r}=0.1708$

Using a natural logarithm property called power rule:

$13r=ln(0.1708)$

$r=\frac{ln(0.1708)}{13}$

$r=-0.1359$

The decay function for concentration of DDT through the years is $y(t)=13e^{-0.1359t}$

B) The value of H is calculated by $y=C(0.5)^{\frac{t}{H} }$

$2.22=13(0.5)^{\frac{13}{H} }$

$(0.5)^{\frac{13}{H} }=0.1708$

Again, using power rule for logarithm:

$\frac{13}{H} log(0.5)=log(0.1708)$

$\frac{13}{H} =\frac{log(0.1708)}{log(0.5)}$

$\frac{13}{H} =2.55$

H = 5.10

Constant H in the half-life formula is H=5.10

C) Using model $y(t)=13e^{-0.1359t}$ to determine concentration of DDT in 1995:

$y(24)=13e^{-0.1359.24}$

y(24) = 0.5

By 1995, the concentration of DDT is 0.5 ppm, so using this model is possible to reduce such amount and more of DDT.

8 0

2 years ago