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wolverine [178]

2 years ago

15

Diego took a survey to find out how many students plan to go to a school band concert. There are more than 1,000 students in his

school. He selected a sample by asking the first 50 students he met at band practice.

2 answers:

puteri [66]2 years ago

6 0

Answer:

The one above is wrong

The actual answers are A and D

Gennadij [26K]2 years ago

4 0

Answer: N/A

Step-by-step explanation: We need more info to solve this problem

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Juan has a supply of rope that is 9 yards long. He needs pieces of rope that are each yard long. How many pieces of rope of this

FinnZ [79.3K]

It can be cut into nine different pieces that size with no rope left over

4 0

2 years ago

Find the slope of the line that passes through the given points (-1.6) and (8.4) Select the correct choice below and, if necessa

fiasKO [112]

Answer:

The slope is $m=-\frac{2}{9}$

Step-by-step explanation:

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

$A(-1,6)\ B(8,4)$

Substitute the values

$m=\frac{4-6}{8+1}$

$m=\frac{-2}{9}$

$m=-\frac{2}{9}$

6 0

2 years ago

I need help with this. Im so confused

vekshin1

2. .8333333 repeated 3x1\3 is 1 the 1x5\6 is .83
3. Order of operations PMDOS 30\2 first giving you 15 then take 50-15=35
4. 10x1.5=15
5. 20+30 first because of order of operations then multiply by 10 giving you 500
6. 120 \ 12 = 10
7. 1 + 9 = 10
8. 0.15-0.02=.13x100 = 13
9. 4.9/0.07=70

8 0

3 years ago

Which fractions are equivalent to 1/8+5/8? Check all that apply.

dolphi86 [110]

The equivalent fractions are 3/4 & 12/16 hope this helps!!

7 0

2 years ago

Read 2 more answers

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

2 years ago