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4 years ago

A middle school took all of its

Mathematics

1 answer:

bearhunter [10]4 years ago

7 0

If there is 1000 seats, and the students took 45% percent of them, then 450 sixth graders went on the trip

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True or False: As long as you pay the minimum payment, you don't have to pay the rest of the bill off

Inga [223]

False. the minimum is just to hold off, it’s not forever. it will keep you in good condition, but not for the long run.

5 0

3 years ago

HailyDenney2 I need your assistance please!

lorasvet [3.4K]

Answer:

x= 3

Step-by-step explanation:

6/4.5= 1.3

4/1.3=3

5 0

3 years ago

In this lesson, you worked with composite functions. Now make up any two functions f(x) and g(x), defined in terms of the variab

yaroslaw [1]

There are g(f(x)) and f(g(x))
do them then evaluate

lets say
f(x)=3x
g(x)=9x^3-4

g(f(x))=9(3x)^3-4=243x^3-4
f(g(x))=3(9x^3-4)=27x^3-12
we can see that g(f(x)) with yield a higher value for x=10

4 0

3 years ago

About 1% of people are allergic to bee stings. What is the probability that exactly 1 person in a class of 25 is allergic to bee

Zanzabum

Answer:

Ok so first of all we know that 1% or 1/100 people are allergic to bee stings. In this class their is 25 people and the change of one of them being alergic so 1/25 this is 4 time 1/00 so

Hope This Helps!!!

8 0

2 years ago

Read 2 more answers

A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres

topjm [15]

Let $U(t)$ denote the amount of uranium-238 in the reactor at time $t$ . As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

$\dfrac{\mathrm dU}{\mathrm dt}=-kU$

where $k>0$ is constant. Separating variables and integrating both sides gives

$\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}$

Suppose we start some amount $u$ . This means that at time $t=0$ we have $U(0)=u$ , so that

$u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}$

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if $t$ is taken to be in years) for some starting amount $u$ ,

$0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}$

The half-life is the time $t_{1/2}$ it takes for the starting amount $u$ to decay to half, $0.5u$ :

$0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}$

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

7 0

3 years ago