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Hitman42 [59]
3 years ago
11

Mariah wanted to make cookies for her family but needed to reduce the recipe. The original recipe required five cups of flour an

d 2 cups of sugar. For the reduced recipe, she will use three cups of flour.
How much sugar should Mariah use? Explain why your answer is correct.
Mathematics
2 answers:
OleMash [197]3 years ago
5 0

Answer:

1 1/5 cups of sugar

Step-by-step explanation:

Evgesh-ka [11]3 years ago
3 0

Answer:

1⅕ or 1.2 cups of sugar

Step-by-step explanation:

Flour : Sugar

5 : 2

3 : X

5/2 = 3/X

X = 3×2/5

X = 6/5 cups

1⅕ or 1.2 cups of sugar

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Listed below are the annual tuition amounts of the 10 most expensive colleges in a country for a recent year.
erastovalidia [21]

Answer:

\bar X = \frac{52384 +54079 +52762 +54364 +52584+53987 +53108 +51588 +50554 +53987}{10}=52939.7

Mode= 53987

Median = \frac{52762+53108}{2}=52935

Midrange=\frac{50554+54364}{2}=52459

Step-by-step explanation:

We can calculate the mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace the values given we got:

\bar X = \frac{52384 +54079 +52762 +54364 +52584+53987 +53108 +51588 +50554 +53987}{10}=52939.7

The mode is the most repeated value and for this case with a frequency of 2 the mode is:

Mode= 53987

In order to find the median we need to order the dataset on increasing way like this:

50554, 51588, 52384  ,52584, 52762

53108, 53987 , 53987 , 54079, 54364

Since we have 10 values an even number the median is calculated from the average between positions 5 and 6 on the data ordered, and we got:

Median = \frac{52762+53108}{2}=52935

The mid range is defined like this:

Midrange = \frac{Max +Min}{2}

And if we replace we got:

Midrange=\frac{50554+54364}{2}=52459

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The one you have highlighted is correct
7 0
3 years ago
A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 ki
shepuryov [24]

Answer

given,

mean = 12 Kg

standard deviation = 0.5 Kg

assume the observed statistic is = 11.1

 now, \bar{X}=11.1 , \mu = 12 , \sigma = 0.5

assuming the number of sample = 4

n = 4

Hypothesis test:

H₀ : μ≥ 12

Ha : μ < 12

now,

significant level α = 0.05

z* = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

z* = \dfrac{11.1-12}{\dfrac{0.5}{\sqrt{4}}}

     z* = -3.60

Test statistics, Z* = -3.60

P-value

P(Z<-3.60) = 0.002 (from z- table)

P- value = 0.002

now,

reject the value of H₀ when P-value < α

    0.002 < 0.05

since, it is less P-value < α , we have to reject the null hypothesis

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3 years ago
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