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andrey2020 [161]
3 years ago
12

Adrian recipe

Mathematics
2 answers:
sweet [91]3 years ago
4 0
1 3/4 for each batch to make 2 1/2 batches you need to take the amount of raisins for one batch 1 3/4 and multiply it by 2 1/2 the amount of batches.

1 3/4*2 1/2= 4 3/8 cups of raisins
ludmilkaskok [199]3 years ago
3 0
Adrian will need: 1 and 3/7 cups of raisins 

1)Divide 2 and 1/2 by 1 and 3/4 
2)You make both fractions improper fractions 
=2 and 1/2= 5/2
=one and 3/4= 7/4
3)Use the rule KCF K=Keep C=Change F=Flip 
4)Keep= 5/2
5)Change= * becomes a divide sign
6)Flip=7/4 becomes 4/7
7)Multiply across 
=5*4=20
=2*7=14
=20 over 14
8)Change 20 over 14 into a mixed fraction 
9)Divide 10 by 7
=1 with the remainder of 3 
10)1 is the whole number 3 is the numerator and 7 is the denominator 
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Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).
masya89 [10]

Answer:

40\pi \ m^{2}

Step-by-step explanation:

we know that

The area of a circle is equal to

A=\pi r^{2}

In this problem we have

r=10\ m

Substitute and find the area

A=\pi (10)^{2}=100 \pi\ m^{2}

Remember that

360\° subtends the area of complete circle

so

by proportion

Find the area of the shaded regions

The central angle of the shaded regions is equal to 2*72\°=144\°

\frac{100\pi }{360} \frac{m^{2}}{degrees} =\frac{x }{144} \frac{m^{2}}{degrees} \\ \\x=144*100\pi /360\\ \\ x=40\pi \ m^{2}

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2 years ago
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7 0
3 years ago
The quadrilaterals ABBCD<br> and PQRS<br> are similar. <br> Find the length <br> of X<br> .
choli [55]
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6 0
3 years ago
Suppose your boss wants you to obtain a sample to estimate a population mean. Based on previous analyses, you estimate that 49 i
VashaNatasha [74]

Answer:

the minimum sample size n = 11.03

Step-by-step explanation:

Given that:

approximate value of the population standard deviation \sigma = 49

level of significance ∝ = 0.01

population mean = 38

the minimum sample size n = ?

The minimum sample size required can be determined by calculating the margin of error which can be re[resented by the equation ;

Margin of error = Z_{ \alpha /2}} \times \dfrac{\sigma}{\sqrt{n}}

38 = \dfrac{2.576 \times 49}{\sqrt{n}}

\sqrt{n} = \dfrac{2.576 \times 49}{38}

\sqrt{n} = \dfrac{126.224}{38}

\sqrt{n} = 3.321684211

n= (3.321684211)^2

n ≅ 11.03

Thus; the minimum sample size n = 11.03

5 0
3 years ago
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