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3 years ago

How do I answer this question

Mathematics

1 answer:

viva [34]3 years ago

4 0

$\text{The formula of a circle with a center (h; k) and a radius r}\\\\(x-h)^2+(y-k)^2=r^2\\\\x^2+y^2=4\to (x-0)^2+(y-0)^2=2^2\\\\\text{therefore we have the circle with a center (0; 0) and a radius 2}\\\\x-y=1\ \ \ |-x\\-y=1-x\ \ \ |\cdot(-1)\\y=x-1\\\\for\ x=0\to y=0-1=-1\to(0;-1)\\for\ x=1\to y=1-1=0\to(1;\ 0)\\\\\text{Look at the picture}$

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-3x-(-2-4x) in simplest form

gogolik [260]

Answer:

-3+(2+4x)

combine like terms

-1+4x that's the answer

5 0

3 years ago

Make r the subject of t = 6r^2 + 6

SpyIntel [72]

First step, divide 6 from both sides.

t/6=r^2+1

second step, subtract 1 from both sides

t/6 -1=r^2

Now take the square root of each side.

$\sqrt{\frac{t}{6}-1}=r$

Done!

3 0

3 years ago

What is the first step in solving the equation?<br><br> 14 = -2m

ollegr [7]

Answer:

first you need to divide each side by -2

Step-by-step explanation:

start with dividing by -2, then you are left with -7 = m which is the answer.

7 0

2 years ago

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand

Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

Part b

We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

4 0

3 years ago

Whitch fraction is equivalent to 5/20

hichkok12 [17]

1/4 is 5/20 simplified

8 0

3 years ago

Read 2 more answers