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3 years ago

3rd grade math explain how you know if a number can be divided by 10 evenly

2 answers:

4 0

if you can divide it has a 0 at the end of a number you can divide it and also if you can skip count by tens you can divide and multiply by the number.

kap26 [50]3 years ago

3 0

Look at the last digit. If the last digit is 0, then the number can be divided by 10 <span> evenly (without a remainder).</span>

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The total cost to pay for parking at a state park for the day and rent a paddleboat are: 1 hour for $17, 2 Hours for $29, 3 hour

kolezko [41]

$412.8 for 24 hours
Hope this help

7 0

3 years ago

In the middle school debate club,30% of the members are in sixth grade.if there are 12 sixth graders in the debate club,how many

kondaur [170]

Answer:

Step-by-step explanation:

8 0

3 years ago

Andre’s school orders some new supplies for the chemistry lab. The online store shows a pack of 10 test tubes costs $4 less than

navik [9.2K]

10t = b - 4

12b+8t = $348

This is a system of equations. I’ll be solving through substitution.

In the first equation. solving for b (the easier variable to isolate) gives you:

b = 10t + 4

Substitute this into the second equation:

12(10t+4) +8t = 348

120t+48+8t = 348

128t = 300

t = 2.34375 —> round it to the nearest cent to get 2.34 dollars

b = 10t+4

b = 10(2.34)+4

b = 27.4 dollars

4 0

3 years ago

First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x

e-lub [12.9K]

Answer:

$(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

$\int x ln(5+x)dx$

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

$\int (U-5) ln(U)dU$

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

$\int (pq')=pq-\int qp'$

so we must define p, q, p' and q':

p=ln U

$p'=\frac{1}{U}dU$

$q=\frac{U^{2}}{2}-5U$

q'=U-5

and now we plug these into the formula:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU$

Which simplifies to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU$

Which solves to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C$

so we can substitute U back, so we get:

$\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C$

and now we can simplify:

$\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

notice how all the constants were combined into one big constant C.

7 0

3 years ago

Use the given values of n and p to find the minimum usual value μ−2σ and the maximum usual value μ+2σ. Round to the nearest hund

lbvjy [14]

Answer:

μ−2σ = 1,089.26

μ+2σ = 1,097.62

Step-by-step explanation:

The standard deviation of a sample of size 'n' and proportion 'p' is:

$\sigma=\sqrt{\frac{p*(1-p)}{n} }$

If n=1139 and p =0.96, the standard deviation is:

$\sigma=\sqrt{\frac{p*(1-p)}{n}}\\\sigma = 0.001836$

The minimum and maximum usual values are:

$\mu-2\sigma = (p-2\sigma)*n\\\mu+2\sigma = (p+2\sigma)*n$

$\mu-2\sigma = (0.96-2*0.001836)*1139\\\mu-2\sigma = 1,089.26\\\mu+2\sigma = (0.96+2*0.001836)*1139\\\mu+2\sigma = 1,097.62$

5 0

3 years ago