MATHEMATICS MIDDLE SCHOOL

A box contains 100 colored chips; some are yellow and some are green. John chooses a chip at random, records the color, and places it back in the bag. John has recorded 57 yellow chips and 19 green chips. Using these results, what is the predicted number of green chips in the box?

Answers

Answer 1

Answer:

The predicted number of green chips in the box is 25.

We have given that,

A box contains 100 colored chips.

some are yellow and some are green.

John chooses a chip at random, records the color, and places it back in the bag.

John has recorded 57 yellow chips and 19 green chips.

$57+19 = 76$

Use proportion.

$(19)/(76)=(x)/(100)$

$1900 = 76x$

divide both sides by 76

$(1900)/(76)=(76x)/(76) \n \n\n(1900)/(76)=x$

$25 = x$

The predicted number of green chips in the box is 25.

To learn more about the proportion of visits:

brainly.com/question/19994681

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Answer 2

Answer: 57+19 = 76
Use proportion.
19    x
--- = ----
76 100
1900   76x
------- = -----
  76 76
25 = x

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Trumpets and cornets are brass instruments. Fully stretched out, the length of a trumpet is 5 and 1/4 feet and the length of a cornet is 4 and 2/4 feet. The trumpet is how much longer than the cornet?

Answers

s(0.75 feet) longer than the corne

f is a function such that f(x) < 0. The graph of the new function g defined by g(x) = | f(x) | is a reflection of the graph of f.

Answers

The graph Would reflect on the x axis

Reduce the ratio 14:63

Answers

14:63

Divide both numbers to get the most simple ratio.

14/7 = 2
63/7 = 9

2:9 is our final answer

14:63.
14/7 = 2
63/7 = 9
Reduced ratio:- 14:63 = 2:9.

What is the value of the figure 8 in 1,083,096

Answers

80,000. it easy rally we learnt this in 2nd grade

Amy is choosing a 2-letter password from the letters A, B, C, D, E, and F. The password cannot have the same letter repeated in it. How many such passwords are possible?

Answers

Answer:

Amy can choose from 60 possible passwords.

Solution:

Given that the two-letter password is from the letters A, B, C, D, E and F

Total number of letters = 6

The password is a two digit letter and repetition of letters is not allowed.

Method 1:

\therefore The number of ways of obtaining an ordered subset of r elements from a set of n elements is given by

$^(n) P_(r) = (n!)/((n-r)!)$

So, the total possible password = $^(6)P_(r) = (6!)/((6-4)!)$ = $(6!)/(4!)$ = $(6*5*4!)/(4!)$

On cancelling the $4!$ in numerator and denominator we get,

Here we get, $(6*5)/(1)$ = 30 possible passwords

Method 2:

There are 6 possible choices for the first term of the password and 5 choices for the second term (i.e. leaving the selected letter of the $\bold1^(st)\bold$ term)

So, $6* 5 = 30$

Answer:

Step-by-step explanation:

6 choices for the 1st, then 5 choices for the 2nd, so: 6*5=30

or nPk=n!/(n-k)! with n=6 and k=2

Put it simply each letter can be combined with 5 diffrent letters and with 6 total letters 6x5=30

L=P+2G over K Solve for G?

Answers

L = P + 2G / K

Multiply both sides by K to get rid of the denominator

KL = P + 2G

Subtract P from both sides

KL - P = 2G

Divide each side by 2

KL - P / 2 = G

G = KL - P / 2

Hi Brand

L=P+(2G)/k

Multiply both sides by K

KL= KP+2G

Flip the equation

KP+2G=KL

add -KP to both sides

KP+2G-KP=KL-KP

2G=KL-KP

Divide both sides by 2

2G/2= (KL-KP)/2

G= 1/2 KL + -1/2 KP