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3 years ago

One and two step equation error analysis. Please help me.

Mathematics

2 answers:

uysha [10]3 years ago

7 0

1). not a -27
2). -8x = -96 (x = 12)
3). y = -150

olga nikolaevna [1]3 years ago

3 0

X - 45 = -27
+45 = 45
= 18

-8x = -96
12

3 is correct

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A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

$Z = \frac{X - \mu}{s}$

X = 205

$Z = \frac{X - \mu}{s}$

$Z = \frac{205 - 200}{5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{195 - 200}{5}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

$Z = \frac{X - \mu}{s}$

$Z = \frac{210 - 200}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{190 - 200}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0

3 years ago

What is the equation in standard form of the line that passes through the point (1,24) and has a slope of negative 0.6

Mumz [18]

let's first off convert the 0.6 to a fraction, and then let's keep in mind that

Standard Form of a Linear Equation

- variables must be on the left-hand-side, usually sorted in ascending order

- there must not be any fractions, just integers

- the variable "x" must not have a negative coefficient.

$\bf 0.\underline{6}\implies \cfrac{06}{1\underline{0}}\implies \cfrac{3}{5}\impliedby m = slope
~\hspace{12em}
(\stackrel{x_1}{1}~,~\stackrel{y_1}{24})
\\\\[-0.35em]
~\dotfill\\\\
\begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-24=\cfrac{3}{5}(x-1)\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{5y-120=3(x-1)}
\\\\\\
5y-120=3x-3\implies -3x+5y=117\implies \stackrel{\textit{standard form}}{3x-5y=-117}$

3 0

3 years ago

A light plane flew from its home base to an airport 255 miles away. With a head wind, the trip took 1.7 hours. The return

ryzh [129]

The speed from home base to airport was:
$v=255miles/1.7h\ /:1.7\\v=150mph$
The speed from airport to home base was:
$v=255miles/1.5h\ /:1.5\\v=170mph$
Average speed of plane was:
$\frac{170mph+150mph}{2}=160mph$
Average speed of wind was:
$x\ -\ wind\ speed\ (mph)\\150+x=170-x\ /+x\\150+2x=170\ /-150\\2x=20\ /:2\\x=10$
Answer: Average speed of plane was 160mph and average speed of wind was 10mph.

I hope I helped :D

6 0

4 years ago

Read 2 more answers

B/4 = 3.2 help me plz

Fynjy0 [20]

Answer:

Step-by-step explanation:

B/4=3.2

B=12.8

3 0

3 years ago

Can someone help me with my essay

wlad13 [49]

I would use the fact that everything involves math. sleeping, eating, walking, running, brushing your teeth, playing games, the matters are endless because it really just uses angles and how long you’ve been doing an activity.

6 0

3 years ago