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kobusy [5.1K]
3 years ago
15

The ratio of a basketball player's completed free throws to attempted free throws is 4 to 7. If she completed 12 free throws, fi

nd how many free throws she attempted. a. 3 free throws. c. 21 free throws. b. 7 free throws. d. 4 free throws
Mathematics
2 answers:
sammy [17]3 years ago
8 0

Answer:

21 free throws is true.

Step-by-step explanation:

It is given that:

Let the completed free throws be denoted by : C

and attempted free throws be denoted by A.

The ratio of a basketball player's completed free throws to attempted free throws is 4 to 7.

i.e. \dfrac{C}{A}=\dfrac{4}{7}

If she completed 12 free throws, we have to find how many free throws did she attempted.

Now we are given C=12.

\dfrac{12}{A}=\dfrac{4}{7}\\\\A=12\times \dfrac{7}{4}\\\\A=3\times 7\\\\A=21

Hence, total number of free throws attempted= 21


Ghella [55]3 years ago
3 0
The ratio of a basketball player's
completed free throws : attempted free throws
7 : 4

There are 12 free throws.
= 7 / 4 of 12
= 21 free throws.

If she completed 12 free throws, the number of free throws <span>she attempted is 21 free throws.</span>
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Semmy [17]

Answer:

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  • a7 = 65

Step-by-step explanation:

The explicit formula for the n-th term of an arithmetic sequence is ...

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where a1 is the first term and d is the common difference.

The sequence of seat counts has a1=5 and d=10, so the explicit formula is ...

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3 years ago
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Step-by-step explanation:

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7 0
3 years ago
The school store opened on the first day of school with 45 notebooks and 15 pencils. Within two days it sold all of these items.
malfutka [58]

The number of pencils sold on the first day  is 5 pencils.

The number of notebooks sold on the first day  is 10 notebooks.

The number of pencils sold on the second day 10 pencils.

The number of notebooks sold on the second day   35 notebooks.

<u>Step-by-step explanation:</u>

Given data,

  • The total number of notebooks in the store = 45 notebooks.
  • The total number of Pencils in the store = 15 pencils.
  • The number of days that all items were sold = 2 days.

Now, you have to calculate the no.of notebooks and no. of pencils sold each day.

<u>In the first day :</u>

The number of sale of notebooks and pencils are given by,

Twice as many notebooks were sold as pencils.

Let us take, the number of pencils sold on the first day  = x

And,  the number of notebooks sold on the first day  = 2x (Twice as pencils).

<u>In the second day :</u>

The number of sale of notebooks and pencils are given by,

For every 7 notebooks​ sold, 2 pencils were sold.

The number of pencils sold on the second day =   2y

The number of notebooks sold on the second day   = 7y

<u> The equation is framed for number of notebooks sold on each day :</u>

The number of notebooks sold ⇒ 2x + 7y = 45   -------(1)

<u> The equation  is framed for number of pencils sold on each day :</u>

The number of pencils sold: x + 2y = 15  ----------(2)

Solving the equations by multiplying eq(2) by 2 and subtract it from eq(1),

  2x + 7y = 45

-<u> (2x + 4y) = 30</u>

   <u>        3y = 15  </u>

⇒ y = 15/3

⇒ y = 5

The value of y is 5.

Substitute y=5 in eq (2),

⇒ x + 2(5) = 15

⇒ x + 10 = 15

⇒ x = 15 - 10

⇒ x = 5

The value of x is 5.

First day sale,

The number of pencils sold on the first day  = x ⇒ 5 pencils

The number of notebooks sold on the first day  = 2x ⇒ 10 notebooks

Second day sale,

The number of pencils sold on the second day =   2y ⇒ 10 pencils

The number of notebooks sold on the second day   = 7y ⇒ 35 notebooks.

8 0
3 years ago
The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth
Art [367]

Answer:

a) N_0=250\; k=0.15

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies t=0. And N(0)=N_0. The process is:

N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250

b) After 2 days means t=48. So, we just replace and operate:

N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}

c) N(t_1)=4000; \;t_1=?

N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\  \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h

d) t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)

N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h

6 0
3 years ago
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