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trapecia [35]
3 years ago
13

Pastries made out of filo dough and brushed with either olive oil or butter (but not both). Pastries made out of shortcrust doug

h are not brushed with anything. Rashid and Mikhail submitted a total of x pastries to a baking competition. Mikhail used filo dough for all of his pastries, Rashid used shortcrust dough for all of his pastries, and each pastry was made using only one kind of dough. If Rashid made 2323 as many pastries as Mikhail, and 5858 of the filo dough pastries were brushed with olive oil, then how many of the pastries submitted by Rashid and Mikhail, in terms of x, were brushed with butter?
A. 3X203X20
B. 9X409X40
C. 1X41X4
D. 3X83X8
E. 5X12
Mathematics
1 answer:
Genrish500 [490]3 years ago
6 0

Answer:

\frac{9x}{40} is the answer.

Step-by-step explanation:

The question has some typo errors.

Rashid and Mikhail submitted a total of x pastries to a baking competition.

Suppose say :

Mikhail made x pastries .

Hence Rashid made \frac{2x}{3} pastries.

And in total they made \frac{5x}{3} pastries.

Now we have that out of x, \frac{5}{8} of x were brushed with olive oil.

So, \frac{3}{8} of x that is \frac{3x}{8} are brushed by butter.

So, it becomes (\frac{3}{8}) / ( \frac{5x}{3} )

= \frac{3}{8} \times\frac{3x}{5} = \frac{9x}{40}

Hence, answer is option B.

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3 years ago
The sodium content of a popular sports drink is listed as 200 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample me
julia-pushkina [17]

Answer:

The sample does not contradicts the manufacturer's claim.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 200 mg

Sample mean, \bar{x} = 211.5 mg

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 18.5 mg

a) First, we design the null and the alternate hypothesis  for a two tailed test

H_{0}: \mu = 200\\H_A: \mu \neq 200

We use Two-tailed t test to perform this hypothesis.

b) Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} } Putting all the values, we have

t_{stat} = \displaystyle\frac{211.5 - 200}{\frac{18.5}{\sqrt{20}} } = 2.7799

c) Now,

t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = \pm 2.0930

Since, the calculated t-statistic does not lie in the range of the acceptance region(-2.0930,2.0930), we reject the null hypothesis.

Thus, the sample does not contradicts the manufacturer's claim.

d) P-value = 0.011934

Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis.

Yes, both approach the critical value and the p-value approach gave the same results.

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3 years ago
If (a, -5) is a solution to the equation 3a = -2B - 7, what is a?
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I think what you need to do is make B = -5
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3 years ago
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Answer:

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Step-by-step explanation:

Given;

original price of the smart phone = $679

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total discount of the smart phone if you use the store credit card

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P = 0.5 X $679

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