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Anna007 [38]
3 years ago
5

C-7.6=-4 what is the value of c

Mathematics
2 answers:
OLEGan [10]3 years ago
7 0
C = 3.6................
9966 [12]3 years ago
4 0
C - 7.6 = -4
c = -4 + 7.6
c = 3.6
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There are six girls and seven boys in the class a team of 10 players is to be selected from the class how many different combina
sweet [91]

Answer:

In 286 different ways 10 players can be selected.

Step-by-step explanation:

There are 6 girls and 7 boys in a class. So in total there are 6+7 = 13 number of students in the class.

A team of 10 players is to be selected from the class.

As there is no other conditions are given, we can pick any 10 students from 13 students.

The way we can select 10 players from 13 students is;

= (13 10)

= 13!/10!(13-10)!

= 13!/10! 3!

= (13 × 12 × 11 × 10!)/ 10! 3!

= ( 13 × 12 × 11)/6

= 286

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3 years ago
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LenKa [72]

Step-by-step explanation:

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2 years ago
B+F=2.20<br> B+S=1.40<br> F+S=2.60
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Answer:

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Step-by-step explanation:

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5 0
3 years ago
Put a check by all the prime numbers.
PolarNik [594]

Answer: 3, 5, and 11

Step-by-step explanation: a prime number is a number greater than 1 that is not a product of two smaller natural numbers.

8 0
3 years ago
Read 2 more answers
The population of a certain town was 10,000 in 1990. The rate of change of the population, measured in people per year, is model
Katena32 [7]

The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

Solution :

According to the question,

The rate of change of population is given as :

$\frac{dP(t)}{dt}=200e^{0.02t}$  in 1990.

Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

                    $=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$

                   $=10,000[e^{0.4}-1]$

                    $=10,000[0.49]$

                    =4900

$\frac{dP(t)}{dt}=200e^{0.02t}$

$\int1.dP(t)=200e^{0.02t}dt$

$P=\frac{200}{0.02}e^{0.02t}$

$P=10,000e^{0.02t}$

$P=P_0e^{kt}$

This is initial population.

k is change in population.

So in 1995,

$P=P_0e^{kt}$

   $=10,000e^{0.02(5)}$

   $=11051$

In 2000,

$P=10,000e^{0.02(10)}$

   =12,214

Therefore, the change in the population between 1995 and 2000 = 1,163.

8 0
3 years ago
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