1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alexgriva [62]
3 years ago
11

Please help! thanks!

Mathematics
2 answers:
Marrrta [24]3 years ago
8 0
B because 6 + 5 = 11 and there's 6 guys out of 11. 330 divided by 11 is 30, 6 x 30 is 180.
NikAS [45]3 years ago
7 0

Answer:

180 are men.

Step-by-step explanation:

it makes the most sense

You might be interested in
Write 0.4*0.4*0.4*0.4*0.4 using exponentail nation.
zepelin [54]
0.4 is multiplying itself a number of times.
0.4 is the base.
The number of equal factors of 0.4 is 5, so the exponent is 5.

0.4^5

0.4^{5}
8 0
3 years ago
If a+b+c=8 and x+y=7 what is -8y-2c-2b-8x-2a
sergey [27]

Answer:

-40

Step-by-step explanation:

Try rearranging and factoring -8y-2c-2b-8x-2a:

This is equal to -8(x + y) -2(a + b + c).

Since x + y = 7, we have:

                            -8(7) -2(a + b + c)

and since a + b + c = 8, we end up with:

                             -56 - 2(8), or

                               -56 - 16    = -40

6 0
3 years ago
Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou
mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

\frac{dQ}{dt} = -kQ

In which k is the decay rate.

The solution is:

Q(t) = Q(0)e^{-kt}

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that Q(0) = 8000

500 counts per minute 13 days later.

This means that Q(13) = 500. We use this to find k.

Q(t) = Q(0)e^{-kt}

500 = 8000e^{-13k}

e^{-13k} = \frac{500}{8000}

\ln{e^{-13k}} = \ln{\frac{500}{8000}}

-13k = \ln{\frac{500}{8000}}

k = -\frac{\ln{\frac{500}{8000}}}{13}

k = 0.2133

So

Q(t) = Q(0)e^{-0.2133t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.2133t}

0.5Q(0) = Q(0)e^{-0.2133t}

e^{-0.2133t} = 0.5

\ln{e^{-0.2133t}} = \ln{0.5}

-0.2133t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.2133}

t = 3.25

The half-life of the radioactive substance is of 3.25 days.

7 0
3 years ago
Can you show me how to solve tax rates in this problem?
inn [45]

Answer:

7.49 + 12.52 +0.7=

Step-by-step explanation:

20.07 so B correct me if im wrong

4 0
3 years ago
Mass of the prism ?
Anna35 [415]

Answer:

I think the answer is density

Step-by-step explanation:


5 0
4 years ago
Other questions:
  • The measures of the legs of a right triangle are 15 m and 20 m. What is the length of the hypotenuse?
    10·1 answer
  • A mathematical sentence which contains an inequality symbol and one variable raised to the first power is called a _ inequality
    9·1 answer
  • What is the approximate volume of a cylinder with a radius of 9cm and height of 22 cm? Use 3.14 to approximate rr?
    7·2 answers
  • What is the <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B40.23%7D%20" id="TexFormula1" title=" \sqrt{40.23} " alt=" \sqrt{
    5·2 answers
  • Which describes the scatter plot of a car’s value compared to the age of the car ?
    6·2 answers
  • What are the similarities and differences between partial products and regrouping
    14·1 answer
  • All of the following fractions are equivalent to 1/2 , except:
    8·1 answer
  • If a force of 24N moves a paper airplane with an acceleration rate equal to 8m\s/s what was its mass
    8·1 answer
  • Anybody know the answer to this
    10·1 answer
  • Your family ate at a diner and left a 15% tip that was<br><br> $7.35. How much was the bill?
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!