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salantis [7]
2 years ago
14

A,B and C are coplanar.truefalse

Mathematics
2 answers:
Yuri [45]2 years ago
8 0
Fasle false false false false
12345 [234]2 years ago
5 0

Answer:


Step-by-step explanation:

is TRUE

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Assume a random sample of size n is from a normal population. Assume a single sample t test is used to for hypothesis testing. T
valentinak56 [21]

Answer:

True

Step-by-step explanation:

Type I and Type II are not independent of each other - as one increases, the other decreases.

However, increases in N cause both to decrease, since sampling error is reduced.

A small sample size might lead to frequent Type II errors, i.e. it could be that your (alternative) hypotheses are right, but because your sample is so small, you fail to reject the null even though you should.

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6 1/2 minus 5 3/4 in math
Nata [24]

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0.75, or 3/4

Step-by-step explanation:

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The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2885 and standard deviation
aliina [53]

Answer:

a.  X~N(2,885, 651)

b.  0.086291

c.  0.00058

d.  3213.10 calories

Step-by-step explanation:

a. -A normal distribution is expressed in the form X~N(mean, standard deviation).

-Let X a random variable denoting  the number of calories consumed.

-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.

-This distribution is expressed as X~N(2,885, 651)

b. The probability that less than 2000 calories are consumed is calculated using the formula:

P(X

#substitute the given values in the formula to solve for P:

P(X

Hence, the probability of consuming less than 2000 calories is 0.08691

c. The proportion of customers consuming more than 5000 calories is calculated as:

P(X>x)=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(Z>\frac{5000-2885}{651})\\\\=P(z>3.2488)\\\\=1-0.99942\\\\=0.00058

Hence, the proportion of customers consuming over  5000 calories is 0.00058

d. The least amount of calories to get the award is calculated as:

1% is equivalent to a z value of 0.50399.

-We equate this to the formula to solve for the mean consumption:

0.01=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(z>\frac{\bar X-2885}{651})\\\\\1\%=0.50399 \\\\\frac{\bar X-2885}{651}=0.50399 \\\\\bar X=0.50399\times 651+2885\\\\=3213.09

Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.

8 0
3 years ago
Factor -1/2out of -1/2x+6
Volgvan
-\dfrac{1}{2}x+6=-\dfrac{1}{2}\left(\dfrac{-\frac{1}{2}x}{-\frac{1}{2}}+\dfrac{6}{-\frac{1}{2}}\right)=-\dfrac{1}{2}\left(x-\dfrac{6\cdot2}{1}\right)=\boxed{-\dfrac{1}{2}\left(x-12\right)}
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