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3 years ago

PLZ HELP!!!!!! (WILL GIVE BRAINLIEST!!!!)

Mathematics

1 answer:

nasty-shy [4]3 years ago

3 0

I think you times all of the inchs together. 5x3x2=30.

I dont know if i am right tho. Hope this help ya.

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List all the multiples of 4 up to 20

emmainna [20.7K]

4 , 8 , 12 , 16 , 20

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3 years ago

The nth term of a sequence is n^2 + 20.

Olegator [25]

Answer:

$\large\boxed{a)\ 21,\ 24,\ 29}\\\boxed{b)\ \text{five terms}}$

Step-by-step explanation:

$a_n=n^2+20\\\\a)\\\\\text{Substitute}\ n=1,\ n=2\ \text{and}\ n=3\ \text{to the }\ a_n:\\\\a_1=1^2+20=1+20=21\\\\a_2=2^2+20=4+20=24\\\\a_3=3^2+20=9+20=29$

$b)\\\\\text{You have to solve the inequality:}\\\\n^2+20$

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3 years ago

Use angle relationships to find the value of x.

Juli2301 [7.4K]

Answer:

x=12.

Step-by-step explanation:

Let's get this straight out of the way, the 3 angles of a triangle ALWAYS add up to 180. Now that that's said, lets begin. Since the opposing side of an angle is the same as the regular side, we simplify to get: 5x+120=180. 180-120=60. so 5x=60. 60/5=12. so x=12.

7 0

2 years ago

Read 2 more answers

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

What is the simplified polynomial and the value

Elanso [62]

The simplified polynomial is; xy(x²y - xy² +1) and; the value when x=-3 and y=2 is; -186

From the polynomial given;

x²yxy -xy²xy +xy.

The polynomial can be factorised to yield a simpler form as follows;

xy(x²y - xy² +1).

This is so because, xy is a common factor.

The value is therefore;

(-3×2)((-3)²2 - (-3×2²) +1)

-6(18+12+1)

= -186