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cupoosta [38]
2 years ago
14

Consider a study collecting data from a population with an unknown mean and standard deviation. If the sample mean and sample st

andard deviation are the same, what is the effect of increasing the sample size on the following measures? The measure can increase, decrease, not change, or more information may be needed.
Mathematics
1 answer:
Levart [38]2 years ago
8 0

Answer:

Step-by-step explanation:

Increasing the sample size in a particular study will bring about a decrease in the standard deviations of the means.

A decrease in the sample size for a particular study will bring about an increase in the standard deviations of the means.

Whereas the population mean of the distribution of sample means is the same as the population mean of the distribution being sampled from meaning that the population of the distribution of sample means never changes.

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What is the density of sample 3, if the Mass(g) was 5 1/4 and the Volume(mL) was 16.9?
iogann1982 [59]
5 1/4 divided by 16.9 is 0.31
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3 years ago
Tributive Property<br>each of the following. Show for what val<br>x - 2(x - 2) =<br>X​
levacccp [35]

Answer:

Step-by-step explanation:

Subtraction is not commutative. For example, 4 − 7 does not have the same difference as 7 − 4. The − sign here means subtraction.

 

However, recall that 4 − 7 can be rewritten as 4 + (−7), since subtracting a number is the same as adding its opposite. Applying the commutative property for addition here, you can say that 4 + (−7) is the same as (−7) + 4. Notice how this expression is very different than 7 – 4.

4 0
2 years ago
On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

5 0
3 years ago
Can someone help me I'm doing my packet and the quest is hard
CaHeK987 [17]
The answer would be y=-3/4x-3
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2 years ago
I need help with this math
enot [183]

Answer:

question? haven't learn this yet but try asking desmos that app helps me a lot!

3 0
2 years ago
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