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borishaifa [10]
3 years ago
5

Not sure how to answer or figure out

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
3 0

Answer:

112

Step-by-step explanation:

Hey There!

So remember opposite the sides given are corresponding with its opposite angle

So the corresponding angle of segment QO is angle P and the corresponding angle of segment PQ is angle O

Thus meaning that angle P and angle O are equal to 34

Now we want to find angle Q

So remember all of the angles in a triangle add up to equal 180

so we subtract the given angles (34 and 34) from 180 to find the missing angle (angle Q)

180-34-34=112

So we can conclude that angle Q = 112

~<em>TheMathWiz</em>

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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

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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
4 years ago
Also need help with this
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