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pantera1 [17]
3 years ago
7

In a school with 500 students,90% are female. Select all expressions that can be used to find the total number of female student

s. A. 90/100 x 500 B.90/10 x 500 C.0.90 x 500
Mathematics
2 answers:
mr Goodwill [35]3 years ago
7 0

Answer:

A)90/100 x 500

C) C.0.90 x 500

Step-by-step explanation:

You can't with B since 90/10 doesn't account for percentage

Brainliest pls

Masja [62]3 years ago
7 0

Answer:

A. 90/100 x 500

C. 0.90 x 500

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Evaluate 2 + 4/5 n for n = 1/4
Sophie [7]

Answer:

2\frac{1}{5}

Step-by-step explanation:

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3 years ago
Brianna bought snacks for her team's practice. She bought a bag of oranges for $2.72 and a 20-pack of juice bottles. The total c
posledela

Answer:

each bottle was $2.12 :P

Step-by-step explanation:

➼start by making an equation to find the price of the juice pack, in this case x would be the price of the juice pack because its unknown, the juice pack price(x) added to the price of the oranges(2.72) and equal the total(45.12), the before tax part is not needed info

2.72 + x = 45.12

➼now our aim to get x alone to find out its worth so we will have to do the opposite and subtract 2.72 from everything, since its positive

2.72-2.72=0 crosses out

45.12-2.72=42.4

➼which gets us x = 42.4 so one 20 pack is 42.40 dollars but we arent done yet, to find out the price of EACH bottle we must divide 42.4 by 20!

42.4/20=2.12

➼so to sum it up, each bottle is $2.12 :)

4 0
3 years ago
Read 2 more answers
Uninhibited growth can be modeled by exponential functions other than​ A(t) ​=Upper A 0 e Superscript kt. For ​ example, if an i
laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form A(t)=A_{0}e^{kt}.

Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

3 0
3 years ago
Find the sum of the constants a, h, and k such that
ra1l [238]

Answer:

  3

Step-by-step explanation:

The value of "a" is the coefficient of x^2, so we know that is 2.

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<u>Solve for h</u>

Now, we have ...

  2x^2 -8x +7 = 2(x -h)^2 +k

Expanding the right side gives us ...

  = 2(x^2 -2hx +h^2) +k

  = 2x^2 -4hx +2h^2 +k

Comparing x-terms, we see ...

  -4hx = -8x

  h = (-8x)/(-4x) = 2

__

<u>Solve for k</u>

Now, we're left with ...

  2h^2 +k = 7 = 2(2^2) +k = 8 +k

Subtracting 8 we find k to be ...

  k = 7 -8 = -1

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And the sum of constants a, h, and k is ...

  a +h +k = 2 +2 -1 = 3

The sum of the constants is 3.

6 0
4 years ago
Equation of a Circle: Mastery Test
xenn [34]

Answer:

(x-1)^2+(y+1)^2=3

Step-by-step explanation:

x² + y2 - 2x + 2y - 1 = 0

add the 1 to get it to the other side of the equation

x² + y2 - 2x + 2y  = +1

group the x's and y's

(x² -2x) + (y2+2y) = +1

then you'll complete the  square on the -2x and + 2y. that just means divide by two and then raise it to the 2nd power.

so (-2/2)^2 and (+2/2)^2

(x²-2x+1)+(y2+2y+1) = 1+1+1

you add the one's to the other side because whatever is done to one side must be done to the other

you'll then need to factor again.

(x-1)^2+(y+1)^2=3

to factor it take one of your squared x's, the sign of the middle term within the parentheses , then the square root of the last term within the parentheses. remeber to put your ^2 (raised to the 2nd power) outside of the parentheses  when you finish.

7 0
3 years ago
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