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2 years ago

Help me find “a” and “b” please!!

Mathematics

1 answer:

sattari [20]2 years ago

7 0

Answer:

36-A

9-B

Step-by-step explanation:

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At winds speeds above 1000 centimeters per second (cm/sec), significant sand-moving events begin to occur. Wind speeds below 100

alisha [4.7K]

Answer:

Lower limit = 1007.95 cm/s

Upper limit = 1142.05 cm/s

Step-by-step explanation:

We have to calculate a 95% CI with population's standard deviation known.

The sample is size n=60, with mean M=1075.

The confidence interval can be expressed as:

$M-z*\sigma/\sqrt{n} \leq\mu\leq M+z*\sigma/\sqrt{n}$

For a 95% CI, the value of z=1.96.

$1075-1.96*265/\sqrt{60} \leq\mu\leq 1075+1.96*265/\sqrt{60} \\\\1075-67.05 \leq\mu\leq 1075+67.05\\\\ 1007.95 \leq\mu\leq 1142.05$

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3 years ago

What is 235 rounded to the nearest ten and explain to me why?

Westkost [7]

There is no decimal place

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2 years ago

Read 2 more answers

Convert y=-(x-3)^2+11 to standard form

Citrus2011 [14]

Y=−x2+6x+2 is the standard form of the equation you provided.

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3 years ago

You are painting a rectangular tabletop. The length of the table top is 6 inches less than 3 times the

dedylja [7]

Answer:

Step-by-step explanation:

Let l represent length and w represent width

We know that l = 3w-6

(3w-6)(w) = the area

Use distributive property

$Area = 3w^2-6w$

A is the correct answer

Hope this helps :)

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2 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

2 years ago