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vovikov84 [41]
3 years ago
14

Determine whether the statement is true or false. If it is​ false, rewrite it as a true statement. A sampling distribution is no

rmal only if the population is normal. Choose the correct answer below. A. The statement is false. A sampling distribution is normal only if n30. B. The statement is false. A sampling distribution is normal if either n30 or the population is normal. C. The statement is true. D. The statement is false. A sampling distribution is never normal.
Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
5 0
<h3>Answer: Choice B</h3>

The statement is false. A sampling distribution is normal if either n > 30 or the population is normal.

==========================================

Explanation:

If the underlying population is normally distributed, then so is the sample distribution (such as the distribution of sample means, aka xbar distribution).

Even if the population isn't normally distributed, the xbar distribution is approximately normal if n > 30 due to the central limit theorem. Some textbooks may use a higher value than 30, but after some threshold is met is when the xbar distribution is effectively "normal".

Choice A is close, but is missing the part about the population being normal. If we know the population is normal, then n > 30 doesn't have to be required.

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The first three terms of a geometric sequence are as follows.
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3 years ago
Assignment: Compound Interest Investigation
sukhopar [10]
<span>To help Tyler better understand how his money will increase in an account that uses simple interest and one that uses compound interest, we are going to use two formulas: a simple interest formula for the accounts that use simple interest, and a compound interest formula for the accounts that use compound interest.
- Simple interest formula: </span>A=P(1+rt)
where:
A is the final investment value 
P is the initial investment 
r is the interest rate in decimal form 
t is number of years
- Compound interest formula: A=P(1+ \frac{r}{n} )^{nt}
where: 
A is the final investment value 
P is the initial investment 
r is the interest rate in decimal form
t is he number of years 
n is the number of times the interest is compounded per year

<span>1. 
a. This is a compound interest account, so we are going to use our compound interest formula. We now that </span>P=1500, t=5, and since the interest is compounded annually (1 time a year), n=1. To find the interest rate in decimal form, we are going to divide it by 100%: r= \frac{4}{100} =0.04. Now that we have all the values lets replace them in our compound interest formula:
A=1500(1+ \frac{0.04}{1}) ^{(1)(5)}
A=1824.98
<span>We can conclude that after 5 years he will have $1824.98 in this account.
b. Here we will use our simple interest formula. We know that </span>P=1500, t=5, and r= \frac{4}{100} =0.04. Lets replace those values in our simple interest formula:
A=1500(1+(0.04)(5))
A=1800
We can conclude that after 5 years he will have $1800 in this account.
c. The compound interest account from point a will yield more money than the simple account one from point b. The difference between the tow amounts is 1824.98-1800=24.98

2.
a. Here we are going to use our compound interest formula. We know that P=2000, t=1 and r= \frac{8}{100} =0.08. We also know that the interest is compounded Quaternary (4 times per year), so n=4. Now that we have all our values lets replace them into our formula:
A=2000(1+ \frac{0.08}{4} )^{(4)(1)}
A=2164.86
We can conclude that after 1 year he will have $2164.86 in this account.
b. Here we are going to use our simple interest formula. We know that P=2000, t=1, and r= \frac{8}{100} =0.08. Once again, lets replace those values in our formula:
A=2000(1+(0.08)(1))
A=2160
We can conclude that after 1 year he will have $2160 in this account.
c. The compound interest account from point a will yield more money than the simple account one from point b. The difference between the tow amounts is 2164.86-2160=4.86

3.
a. Since Bank A offers an account with a simple interest, we are going to use our simple interest formula. From the question we know that P=3200, t=3, and r= \frac{3.5}{100} =0.035. Now we can replace those values into our formula to get:
A=3200(1+(0.035)(3))
A=3536
Now, to find the interest earned for Bank A we are going to subtract P from A
InterestEarned=3536-3200=336
We can conclude that <span>the interest earned for Bank A is $336
b. 
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A=3200(1+ \frac{0.034}{1} )^{(1)(3)}
A=3537.62
Now, to find the interest earned for Bank A we are going to subtract P from A:
InterestEarned=3537.62-3200=337.62
We can conclude that the interest earned for Bank B is $337.62
c. Even tough the interest returns between the tow Banks are very similar, Bank B offers a slightly better interest over a period of time, which can make a big difference in the long run. If <span>Tyler wants the earn more money, he definitively should deposit his money in Bank B.
d. </span>The compound interest account from Bank B will yield more money than the simple account one from Bank A The difference between the tow amounts is 3537.62-3536=1.62
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4 years ago
Line v passes through point [6.6] and is perpendicular to the graph of y= 3/4x - 11 line w is parallel to line v and passes thro
lbvjy [14]

Given:

The equation of a line is

y=\dfrac{3}{4}x-11

Line v passes through point (6,6) and it is perpendicular to the given line.

Line w passes through point (-6,10) and it is parallel to the line v.

To find:

The equation in slope intercept form of line w.

Solution:

Slope intercept form of a line is

y=mx+b         ...(i)

where, m is slope and b is y-intercept.

We have,

y=\dfrac{3}{4}x-11      ...(ii)

On comparing (i) and (ii), we get

m=\dfrac{3}{4}

So, slope of given line is \dfrac{3}{4}.

Product of slopes of two perpendicular lines is -1.

m_1\times m_2=-1

\dfrac{3}{4}\times m_2=-1

m_2=-\dfrac{4}{3}

Line w is perpendicular to the given line. So, the slope of line w is -\dfrac{4}{3}.

Slopes of parallel line are equal.

Line v is parallel to line w. So, slope of line v is also -\dfrac{4}{3}.

Slope of line v is -\dfrac{4}{3} and it passes thorugh (-6,10). So, the equation of line v is

y-y_1=m(x-x_1)

where, m is slope.

y-10=-\dfrac{4}{3}(x-(-6))

y-10=-\dfrac{4}{3}(x+6)

y-10=-\dfrac{4}{3}x-\dfrac{4}{3}(6)

y-10=-\dfrac{4}{3}x-8

Adding 10 on both sides, we get

y=-\dfrac{4}{3}x-8+10

y=-\dfrac{4}{3}x+2

Therefore the equation of line v is y=-\dfrac{4}{3}x+2.

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3 years ago
Is it possible to have a function f defined on [ 4 , 5 ] and meets the given conditions? f is not continuous on [ 4 ,5 ] and tak
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Yes. 
       
f(x)={x if  4</=X</=5 
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