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Fudgin [204]
3 years ago
9

A storekeeper has two kinds of flour, one selling for 65 cents per pound, and the other selling for 95 cents per pound. How many

pounds of each flour should the storekeeper use, to make 100 pounds of flours which will be sold for 83 cents per pound?
Mathematics
2 answers:
bixtya [17]3 years ago
6 0

The amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 poundsThe amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 poundsThe amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 poundsThe amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 poundsThe amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 pounds

Murrr4er [49]3 years ago
4 0

Answer:

The amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 pounds

Step-by-step explanation:

Let

x ----> amount of 65 cents per pound flour to use

y----> amount of 95 cents per pound flour to use

we know that

x+y=100 ----> y=100-x ----> equation A

0.65x+0.95y=0.83(100) ----> equation B

substitute equation A in equation B

0.65x+0.95(100-x)=0.83(100)

Solve for x

0.65x+95-0.95x=83

0.95-0.65x=95-83

0.30x=12

x=40\ lb

Find the value of y

y=100-40=60\ lb

therefore

The amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 pounds

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What is the equation to the expression -2-3(-4)?​
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Answer:

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Step-by-step explanation:

-2-3-(-4)

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3 years ago
The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a s
Andrew [12]

Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that \mu = 1262, \sigma = 118

a) Determine the 26th percentile for the number of chocolate chips in a bag. ​

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

Z = \frac{X - \mu}{\sigma}

-0.643 = \frac{X - 1262}{118}

X - 1262 = -0.643*118

X = 1186

(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

Z = \frac{X - \mu}{\sigma}

-1.96 = \frac{X - 1262}{118}

X - 1262 = -1.96*118

X = 1031

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

Z = \frac{X - \mu}{\sigma}

1.96 = \frac{X - 1262}{118}

X - 1262 = 1.96*118

X = 1493

Between 1031 and 1493.

​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

Z = \frac{X - \mu}{\sigma}

-0.675 = \frac{X - 1262}{118}

X - 1262 = -0.675*118

X = 1182

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

Z = \frac{X - \mu}{\sigma}

0.675 = \frac{X - 1262}{118}

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X = 1342

IQR:

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3 years ago
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Read 2 more answers
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee
IrinaVladis [17]

Answer:

720 ways

Step-by-step explanation:

Generally, combination is expressed as;

                                  ^{n} C_{r} = \frac{n!}{r!(n-r)!}

The question consists of 9 multiple-choice questions and examinee must answer 7 of the multiple-choice questions.

                                   ⇒ ⁹C₇ =\frac{9!}{7!(9-7)!}

                                        =\frac{9!}{7!(2)!}

                                        = 36

The question consists of 6 open-ended problems and examinee must answer 3 of the open-ended problems.

                                    ⇒ ⁶C₃ =\frac{6!}{3!(6-3)!}

                                         =\frac{6!}{3!(3)!}

                                         = 20

Combining the two combinations to determine the number of ways the questions and problems be chosen if an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problem.

                                        ⁹C₇ × ⁶C₃

                                       = 36 × 20

                                       = 720 ways  

3 0
3 years ago
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