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2 years ago

8

A jar contains 960 beans. Of them 600 are Azuki beans and the rest are cranberry beans. what is the ratio of the Azuki beans to

all the beans.

2 answers:

ira [324]2 years ago

7 0

600:360 simplify
10:6 simplify
5:3
Hope this helps.

victus00 [196]2 years ago

5 0

600:960 or, you can reduce it to 5:8

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Kipish [7]

Answer:

12cm and 16cm

Step-by-step explanation:

The hypotenuse of the right angle triangle = 20cm

let the other two sides be x and y;

The difference;

x = y - 4

Problem:

Find x and y;

Solution:

According to the pythagoras theorem;

x² + y² = 20² ------ i

x² + y² = 400 ----- i

and x - y = 4 ---- ii

So; x = 4 + y

Now input the value into equation (i);

(y + 4)² + y² = 400

(y+4)(y+4) + y² = 400

y² + y² + 4y + 4y + 16 = 400

2y² + 8y + 16 = 400

2y² + 8y + 16 -400 = 0

2y² + 8y - 384 = 0;

y² + 4y - 192 = 0

Factorize the equation;

y² + 16y - 12y - 192 = 0

y(y + 16) - 12(y + 16) = 0

(y-12)(y + 16) = 0

y -12 = 0 or y+ 16 = 0

y = 12 or -16

It is not realistic for the length of a body to be a negative value, so y= 12;

since;

x - y = 4,

x = 4 + 12

x = 16

5 0

3 years ago

Why was an attack against Vicksburg by river just not possible for Union forces?

Vsevolod [243]

<span>Because the town was built above sheer cliffs that were heavily defended by artillery.</span>

8 0

3 years ago

Harvey the wonder hamster can run 3 1/6 km in 1/4 hour. Harvey runs at a constant rate. Find his average speed in kilometers per

inysia [295]

$3 \frac{1}{6}: \frac{1}{4}= \frac{19}{6}* \frac{4}{1}= \frac{38}{3}= 12 \frac{2}{3}$

<span>his average speed is </span>12 2/3 kilometers per hour

7 0

3 years ago

Read 2 more answers

99. What is the greatest prime factor of 5,355?<br> A. 17<br> B. 51<br> C. 119<br> D. 131<br> E. 153

NemiM [27]

Hello,

5355 is divisble by 3 since 5+3+5+5=18 is divisible by 3
5355=3*1785
1785 is divisble by 3 since 1+7+8+5=21 is divisible by 3
1785=3*595
595 is not divisible by 3 but by 5
595=5*119
119 is divisible by 7 since 1*2+1*3+9*1=14 is divisible by 7.
119=7*17

17 is prime
Answer A

8 0

2 years ago

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.

3 0

2 years ago