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quester [9]
3 years ago
5

Sergio thinks he will need 90 cups of fruit juice to make 210 cups of punch. Explain his method. How many cups of fruit juice wo

uld Sergio need to make 154 cups of punch? Explain your calculations.
Mathematics
1 answer:
Eduardwww [97]3 years ago
3 0

Answer:

66 cups of fruit juice will be required .

Step-by-step explanation:

According to the question ,

90 Cups of fruit juice is required for making of 210 cups of punch .

So,

For making of one cup of punch ,

\frac{90}{210} =  \frac{3}{7} cups of fruit juice will be required ,

Thus,

For making of 154 cups of punch ,

Total number of cups of fruit juice required will be \frac{3}{7}×154.

= 66 cups.

Thus a total of 66 cups of fruit juice will be required.

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mote1985 [20]

Answer:

The correct option is option (c).

Sin \ P= \frac {RQ}{PQ}

Step-by-step explanation:

Right angled triangle:

  • One angle must be 90° and other two angles are acute angle.
  • The hypotenuses is the longest side of the triangle and opposite right angle.
  • It follows the Pythagorean Theorem.

Given that,

∠QRP= 90°, ∠RPQ= 30°, ∠PQR = 60°

we know that,

sin \theta =\frac{Opposite }{Hypotenuse}

for sin P , the opposite is QR.

The hypotenuse is PQ.

Therefore,

Sin \ P= \frac {RQ}{PQ}

5 0
3 years ago
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Ivenika [448]

Answer:

102.00

Step-by-step explanation:

If you want to leave a 20% tip, multiply the cost by 0.20 to get the tip amount or multiply the cost by 1.20 to get the total including tip. If you want to leave a 18% tip, multiply the cost by 0.18 to get the tip amount or multiply the cost by 1.18 to get the total including tip.

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How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide
kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The information provided is:

<em>σ</em> = $60

<em>MOE</em> = $2

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the sample size as follows:

MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

       n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}

          =[\frac{1.96\times 60}{2}]^{2}

          =3457.44\\\approx 3458

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

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