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Alex
9 months ago
9

A glitter glue recipe takes a mixture of 1 5 cup of glitter to 2 5 cup of glue. For A, how many cups of glitter are used for 1 c

up of glue? For B, how many cups of glitter glue is made with 1 cup of glue?
Mathematics
1 answer:
tatiyna9 months ago
3 0

0.6 cup of gliter is used for 1 cup of glue and 1.6 cup of gliter glue is made with 1 cup of glue.

Part A

Given that,

15 cup of gliter is for 25 cup of glue ,

Therefore, "X" cup of gliter is for 1 cup of glue

X = (1*15)/25

X= 0.6 cup of gliter

Part B,

Given that,

From part A 0.6 cup of gliter is for 1 cup of glue

Thus, 1 cup of glue and 0.6 cup of gliter when mixed will form "Y" cup of gliter glue

Now, Y= 0.6 cup of gliter + 1 cup of glue

Y= 1.6 cup of gliter glue

which approximates to 2 cup of gliter glue.

For more information on cross multiplication kindly visit to

brainly.com/question/11203238

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Two cyclists, 112 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other, and the
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Answer:

Speed of a= 21 miles/hr

r = Speed of b= 7 miles/hr

Speed of a = 3r

Step-by-step explanation:

The cyclist are 112 miles apart

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Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

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