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Snezhnost [94]
2 years ago
11

I AM GIVING OUT BRAINLIST

Mathematics
2 answers:
Ede4ka [16]2 years ago
8 0
Tenth place is the answer!! :))
Shkiper50 [21]2 years ago
8 0

Step-by-step explanation:

18.706 → 19

18.<u>7</u>06 → 18.71

18.706 = 19

18.706 = 18.71

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Suppose that a random sample of size 10 from the N(µ, 9) distribution resulted in a sample mean of x = 2.7. Give an 85% condiden
stellarik [79]

Answer:

a) CI: 1.3

b) CI: 2.5

c) CI: 2.0

Step-by-step explanation:

a) We have a sample of size n=10 and a sample mean of x=2.7.

The σ² of the population is known and is σ²=9 or σ=3.

For a CI of 85%, the z-value is 1.44.

Then the lower and upper limits of the CI are:

LL=M-z*\frac{\sigma}{\sqrt{n} } =2.7-1.44*\frac{3}{\sqrt{10} }=2.7-1.44*\frac{3}{3.162} =2.7-1.4=1.3\\\\UL=M+z*\frac{\sigma}{\sqrt{n} } =2.7+1.4=4.1

The 85% confidence interval for the mean is defined by the values LL=1.3 and UL=4.1.

CI: 1.3

b) In this case, the variance of the population is unknown.

We have to estimate the variance of the population from the variance of the sample.

Sample data:

n = 100

x(mean) = 2.7

s² = 1.1

The CI is defined as

x-t*\frac{s}{\sqrt{n} }

The value of t depends of the degrees of freedom and the percentage of confidence.

In this case, the degrees of freedom are n-1=100-1=99 and the CI is of 90%.

We look up in a t-table and the t-value for this conditions is 1.6604.

We can now calculate the CI

LL: x-t*\frac{s}{\sqrt{n} }=2.7-1.6604*\frac{\sqrt{1.1} }{\sqrt{100} } =2.7-0.2=2.5\\\\UL: x+t*\frac{s}{\sqrt{n} }=2.7+0.2=2.9

The 90% confidence interval for the mean is defined by the values LL=2.5 and UL=2.9.

CI: 2.5

c) In this case, the variance of the population is unknown.

We have to estimate the variance of the population from the variance of the sample.

Sample data:

n = 10

x(mean) = 2.7

s² = 1.1

The CI is defined as

x-t*\frac{s}{\sqrt{n} }

The value of t depends of the degrees of freedom and the percentage of confidence.

In this case, the degrees of freedom are n-1=10-1=9 and the CI is of 95%.

We look up in a t-table and the t-value for this conditions is 2.2622.

We can now calculate the CI

LL: x-t*\frac{s}{\sqrt{n} }=2.7-2.2622*\frac{\sqrt{1.1} }{\sqrt{10} } =2.7-0.7=2.0\\\\UL: x+t*\frac{s}{\sqrt{n} }=2.7+0.7=3.4

The 95% confidence interval for the mean is defined by the values LL=2.0 and UL=3.4.

CI: 2.0

5 0
2 years ago
How do I find simple interest rates?
butalik [34]

Answer:

Final amount = initial balance ( 1 + ( intrest rate * time ) )

4 0
3 years ago
I need help with number 2!! please help!
Maurinko [17]

Answer:

12.68

Step-by-step explanation:

76 degrees /2 = 38

Half the width of 15m forms the lower side of the triangle: 7.5 m

sin(38) = 7.5/h

=12.1820

12.18+.5

12.68

5 0
3 years ago
How do you do this and what are the answers
scoray [572]
I can answer 1 to 4...
The area of a circle can be found by \pi r^{2}. Reminder: \pi = 3.14 (usually used when finding out the area of a circle)

1. 4^{2} = 16
\pi *16 or 3.14 * 16 = 50.24
50.24/4=12.56

2. 6^{2} = 36
\pi *36 or 3.14 * 36 = 113.04
113.04/10=11.304

(Not really sure if this is correct) 3. 2^{2}=4
\pi *4 or 3.14 * 4
12.56/2.66666667=11.70999999

4. 6^{2}=36
\pi *36= 113.04
113.04/6=18.84
5 0
3 years ago
Question 1
enyata [817]

Answer:

x = 1/8

Step-by-step explanation:

Solve for x by simplifying both sides of the equation, then isolating the variable.

8 0
2 years ago
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