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3 years ago

13

Allison used 2.5 half cups of flour to make a cake, and 3 over 4of a cup of flour to make a pie. If she started with 4 cups of f

lour, how many cups of flour does she have left?

1 answer:

Gennadij [26K]3 years ago

3 0

Answer: Allison has 3/4 cups of flour left.

Step-by-step explanation:

Allison used flour twice, the first time she used 2.5 cups of flour, and the second time she used 3/4 cups of flour. Let's convert these two the the same unit.

2.5= 2 1/2 or 5/2

Let's make this have the same base so it will be easier to add 3/4 to it.

5/2=10/4

We know she used both of these amounts of flour so let's add them. 10/4+3/4=13/4

We know she used 13/4 cups of flour. Let's convert 4 cups of flour to have the denominator of 4.

4=16/4

Now, we just have to subtract how much flour she used to make the pie and cake.

16/4-13/4=3/4

Allison has 3/4 cups of flour left.

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2 years ago

Read 2 more answers

Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mar

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Answer:

a

The 95% confidence interval is

$0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

c

$n = 125 \ stores$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The number of stores that had more than 2 items price incorrectly is k = 52

Generally the sample proportion is mathematically represented as

$\^ p = \frac{ k }{ n }$

=> $\^ p = \frac{ 52 }{ 60 }$

=> $\^ p = 0.867$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} }$

=> $E = 0.0859$

Generally 95% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.867 - 0.0859 < p < 0.867 + 0.0859$

=> $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Considering question b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

Considering question c

From the question we are told that

The margin of error is E = 0.05

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n= [\frac{1.645 }}{0.05} ]^2 * 0.867 (1 - 0.867 )$

=> $n = 125 \ stores$

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