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shtirl [24]
2 years ago
7

Solve the system of linear equations by substitution: (SHOW UR WORK) y=5x-6 y=4x-2

Mathematics
2 answers:
MArishka [77]2 years ago
5 0

Answer:

x = 4

y = 2

Step-by-step explanation:

1) y = 5x - 6

2) y = 4x - 2

sub 1 into 2

5x - 6 = 4x - 2

5x - 4x = 6 - 2

x = 4 ---> (3)

sub 3 into 2

y = (4 * 1) - 2

y = 4 - 2

y = 2

suter [353]2 years ago
4 0

Answer:

4

Step-by-step explanation:

Since they both equal y, they are both equivinlent

5x-6=4x-2

5x=4x+4

x=4

CHECK

20-6=16-2

14=14

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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
The sum of 5 consecutive integers is 265. What is the fifth number in this sequence ?
Yanka [14]
Hello,
Let's assume n,n+1,n+2,n+3,n+4 the 5 numbers

n+(n+1)+...+(n+4)=5n+10=265
5n=265-10
5n=255
n=51
The 5th number is 51+4=55

6 0
3 years ago
Please I really need help with this
romanna [79]

Answer:

A, 14

Step-by-step explanation:

Assuming that she has a total of d dollars, and spends half of it, then we have 1/2d. She then spends <em>another</em> 6 dollars, and therefore we would subtract. She is left with one dollar.

1/2d-6=1

Working backwards, we can solve for d, her total amount of money:

1/2d=7

d=14

<em>I hope this helps! :)</em>

6 0
3 years ago
The sum of a number wand 4 is more than -12.
Mazyrski [523]
The answer is w + 4 > -12
4 0
3 years ago
. If x = 7 + 4√3 then find the value of x2<br> – 1/x2
inessss [21]

Answer:

you can use mathpapa it works

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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