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2 years ago

13

During a harvest, a farmer filled 70 containers each with 46 ears of corn. When he was repacking the ears for shipment, he place

d them in 5 crates.
How many ears of corn did he put in each crate?
A.
634
B.
644
C.
654
D.
649

1 answer:

Alexxx [7]2 years ago

8 0

Answer:

B ( hope this was helpful) :)

Step-by-step explanation:

You do 70x46=3220 then u divide 3220 by 5 to get 644.

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How does the measure of the pink inscribed angle compare with the measure of the blue intercepted arc? How do you know?

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Answer:

The pink angle is half the angle subtended by the blue arc.

Step-by-step explanation:

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See attachment for circle

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2 years ago

Need help with this question please!

nevsk [136]

I would say it is A because if you subtract <em>p,</em> the original price by $2.50, you would get <em>d, </em>the discounted price. Look at B u see that you're adding the discount which doesn't make sense. Looking at C, the discounted price of different prices can't always be the same. And finally, D, the discounted price is greater than the original. Also, if you subtract you would get different discounts.

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2 years ago

A banks loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviati

Alona [7]

Answer:

2.87%

Step-by-step explanation:

We have the following information:

mean (m) = 200

standard deviation (sd) = 50

sample size = n = 40

the probability that their mean is above 21.5 is determined as follows:

P (x> 21.5) = P [(x - m) / (sd / n ^ (1/2))> (21.5 - 200) / (50/40 ^ (1/2))]

P (x> 21.5) = P (z> -22.57)

this value is very strange, therefore I suggest that it is not 21.5 but 215, therefore it would be:

P (x> 215) = P [(x - m) / (sd / n ^ (1/2))> (215 - 200) / (50/40 ^ (1/2))]

P (x> 215) = P (z> 1.897)

P (x> 215) = 1 - P (z <1.897)

We look for this value in the attached table of z and we have to:

P (x> 215) = 1 - 0.9713 (attached table)

P (x> 215) =.0287

Therefore the probability is approximately 2.87%

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3 years ago

Sec (of theta)=25/24<br> find cot(of theta)

Len [333]

If secant of theta = <span> <span> <span> 1.0416666667 </span> </span> </span> then
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3 years ago

Read 2 more answers

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

<em>A</em> = a student attended public high school

<em>B</em> = a student attended private high school

<em>C</em> = a student was home schooled

<em>D</em> = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

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2 years ago