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Oduvanchick [21]

3 years ago

5

A random variable is not normally distributed, but it is mound shaped. It has a mean of 14 and a standard deviation of 3. If you

take a sample of size 10, can you say what the shape of the sampling distribution for the sample mean is

1 answer:

Over [174]3 years ago

5 0

Answer:

Step-by-step explanation:

from the question,

the mean 14

the standard deviation is 3

and sample size is 10.

since the n which is the sample size is 10, then the distribution is mound shaped.

why?

this is due to the fact that the random variable from which we took the sample is mound shaped.

The sampling distribution of the mean is normally distributed although the question says the random variable is not normally distributed. so the shape is bell shaped and normally distributed.

the standard deviation of the mean is

3/√10

= 0.948

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steposvetlana [31]

Answer:

1.) 12x-10

2.)18x+10

Step-by-step explanation:

8 0

2 years ago

A piece of plywood was cut so it's length with 8 ft by 4 ft what is the area of the wood

Elena L [17]

The area of an object is found by multiplying the length by the width, so to solve your problem, multiply 8*4 to get 32.

8 0

3 years ago

A survey of students at a university shows that 4 out of 5 drink coffee. Of the students who drink coffee, 1 out of 6 adds crea

Andru [333]

The number who drink coffee is 3/4 of 8000, so ...
... (3/4)×8000 = 6000.
Of those, 1/5 add cream, so the number of coffee drinkers who add cream to their coffee is ...
... (1/5)×6000 = 1200.
_____
Alternate solutions
We have computed ...
... (1/5)×((3/4)×8000) = 1200
Since multiplication is associative, you can multiply the fractions first, if you wish:
... ((1/5)×(3/4))×8000 = (3/20)×8000
... = 24000/20 = 1200
or
... = 3×(8000/20) = 3×400 = 1200

6 0

2 years ago

Bonita has a standard of 2 hours of labor per unit, at $12 per hour. In producing 2600 units, Bonita used 5050 hours of labor at

Eduardwww [97]

Answer:

1800

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3 0

3 years ago

The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

Alchen [17]

Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
$a_2=a(r)$
$a_2=a(r)^2$

the sum is -7/10
$\frac{-7}{10}=a+ar+ar^2$
and their product is -1/125
$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$

from the 2nd equation we can take the cube root of both sides to get
$\frac{-1}{5}=ar$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$
subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
$r=\frac{5+3}{4}=\frac{8}{4}=2$ or $r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

6 0

3 years ago