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Sonja [21]
3 years ago
13

describe the distribution of sample means (shape, expected value, and standard error) for samples of n=100 selected from a popul

ation with a mean of mu=40 and a standard deviation of 10
Mathematics
1 answer:
motikmotik3 years ago
8 0
If you multiply 40 by 100 and subtract by 54 and add 2 divided by 3 and theres your answer!
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Micah is planning a road trip. He estimates that he will drive a total of of 8 1/4 hours. He wants to stop for gas after 1/3 of
tigry1 [53]

8.25 hours is our base number here. 1/3 of it is taken up when he stops at the gas station.

8.25, 4.125 is half of it, 2.75 is our answer for this,

8.25/3 gives you 2.75 due to 2.75 making it 5.5 when doubled and so on

so, the answer is 2 3/4 hours, D

3 0
4 years ago
A person’s theoretical maximum heart rate (in heartbeats per minute) is 220 - x where x is the person’s age in years (20 ≤ x ≤ 6
Lyrx [107]

Let f(x) represents the heart beat of the person.

Now, we have been given that person’s theoretical maximum heart rate is given by  220 - x.

Also, we have been given that person strive for a heart rate that is at least 50% of the maximum. At least means greater than equal to.

Thus, we have

f(x)\geq (220-x)0.5\\
f(x)\geq 110-0.5x

Also, we have been given that person strive for a heart rate that is at most 75% of the maximum. At most means less than equal to.

f(x)\leq (220-x)0.75\\
\\
f(x)\leq 165-0.75x

Therefore, the system of inequality is given by

f(x)\geq 110-0.5x\\
f(x)\leq 165-0.75x

6 0
3 years ago
3^5 . write each power as a product of the same factor . then find the value
8090 [49]

Answer:

3*3*3*3*3=243

Step-by-step explanation:

It is self explanatory

6 0
4 years ago
Read 2 more answers
After standardizing a NaOH solution, you use it to titrate an HCl solution known to have a concentration of 0.205 M. You perform
Gnoma [55]

Answer:

The right answer is:

(a) 0.205

(b) 0.005425

Step-by-step explanation:

The given data is:

0.213, 0.204, 0.208, 0.200, and 0.198

(a)

The mean will be:

= \frac{Sum \ of \ all \ data}{No. \ of \ data}

= \frac{0.213+ 0.204+ 0.208+ 0.200+0.198}{5}

= \frac{1.023}{5}

= 0.2046

or,

= 0.205

(b)

The standard deviation will be:

\sigma^2=\Sigma\frac{(x_i-\mu)^2}{N}

    =\frac{(0.213-0.2046)^2+...+(0.198-0.2046)^2}{5}

    =\frac{0.0001472}{5}

    =2.994\times 10^{-5}

or,

    =0.005425

7 0
3 years ago
Use the drop-down menus to choose steps in order to correctly solve<br> 4k−6=−2k−16−2 for k.
stich3 [128]

Answer:

<em>down menus to choose steps in order to correctly solve</em>

<em>down menus to choose steps in order to correctly solve4k−6=−2k−16−2 for k.</em>

5 0
3 years ago
Read 2 more answers
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