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KengaRu [80]
4 years ago
6

Solve the system using elimination.

Mathematics
1 answer:
Leona [35]4 years ago
6 0

Answer:

the answer is A

Step-by-step explanation:

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Independent random samples from two regions in the same area gave the following chemical measurements (ppm). Assume the populati
Basile [38]

Answer:

d. The interval contains only negative numbers. We cannot say at the required confidence level that one region is more interesting than the other.

Step-by-step explanation:

Hello!

You have the data of the chemical measurements in two independent regions. The chemical concentration in both regions has a Gaussian distribution.

Be X₁: Chemical measurement in region 1 (ppm)

Sample 1

n= 12

981 726 686 496 657 627 815 504 950 605 570 520

μ₁= 678

σ₁= 164

Sample mean X[bar]₁= 678.08

X₂: Chemical measurement in region 2 (ppm)

Sample 2

n₂= 16

1024 830 526 502 539 373 888 685 868 1093 1132 792 1081 722 1092 844

μ₂= 812

σ₂= 239

Sample mean X[bar]₂= 811.94

Using the information of both samples you have to determina a 90% CI for μ₁ - μ₂.

Since both populations are normal and the population variances are known, you can use a pooled standard normal to estimate the difference between the two population means.

[(X[bar]₁-X[bar]₂)±Z_{1-\alpha /2}* \sqrt{\frac{Sigma^2_1}{n1}+\frac{Sigma^2_2}{n_2}  }]

Z_{1-\alpha /2}= Z_{0.95}= 1.648

[(678.08-811.94)±1.648*\sqrt{\frac{164^2}{12}+\frac{239^2}{16}  }]

[-259.49;-8.23]ppm

Both bonds of the interval are negative, this means that with a 90% confidence level the difference between the population means of the chemical measurements of region 1 and region 2 may be included in the calculated interval.

You cannot be sure without doing a hypothesis test but it may seem that the chemical measurements in region 1 are lower than the chemical measurements in region 2.

I hope it helps!

6 0
4 years ago
For geometry:(<br><br> please help, will give brainist
mash [69]

Answer:

5x+16=6x+7

16=X+7

11=X

Step-by-step explanation:

WAIT I PUT THE WRONG SYMBOL

5X+16=6X-7

16=X-7

23=X

7 0
3 years ago
Rewrite each fraction with a denominator of 24.<br> 3/8 <br> 7/6
Illusion [34]

9/24

28/24

If you need an explanation just ask

6 0
4 years ago
Read 2 more answers
Classify each number by indicating in which set or sets it belongs 8 and 0.25
Daniel [21]

we are given with two numbers: 8 and 0.25. In this case, we are asked in the problem to determine the set in which the numbers belong to. 8 belongs to whole numbers and integers, as well as natural numbers. 0.25 belongs to decimal numbers and whole numbers
6 0
4 years ago
If logb2=x and logb3=y, evaluate the following in terms of x and y:
Alja [10]

log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x

<em><u>Solution:</u></em>

Given that,

log_b2 = x\\\\log_b3 = y --------(i)

<em><u>Use the following log rules</u></em>

Rule 1: log_b(ac) = log_ba + log_bc

Rule 2: log_b\frac{a}{c} = log_ba - log_bc

Rule 3: log_ba^c = clog_ba

a) log_b{162}

Break 162 down to primes:

162 = 2^1 \times 3^4

log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y

Thus we get,

log_b162 = x + 4y

Next

b) log_b 324

Break 324 down to primes:

324 = 2^2 \times 3^4

log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y

Next

c) log_b\frac{8}{9}

By rule 2

log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} =  3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} =  3x - 2y

Next

d) \frac{log_b27}{log_b16}

By rule 2

\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} =  3y - 4x

Thus the given are evaluated in terms of x and y

3 0
3 years ago
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