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tresset_1 [31]
3 years ago
5

Dina wants to make 15 3/4 cups of strawberry drink by mixing water and strawberry syrup with a ratio of 2 1/4 cup of water for e

very 3/4 cup of syrup. how much water and syrup will she need ro use?
Mathematics
1 answer:
bekas [8.4K]3 years ago
4 0

Answer:

The water she need is 7\frac{7}{8} \ cups. and strawberry syrup 11\frac{13}{16}\ cups.

Step-by-step explanation:

Given:

Dina wants to make 15 3/4 cups of strawberry drink by mixing water and strawberry syrup with a ratio of 2 1/4 cup of water for every 3/4 cup of syrup.

Now, to find the quantity of water and syrup she need to use.

As given in question ratio so:

Strawberry syrup = 2 1/4 = 9/4.

Water = 3/4.

Total cups of strawberry drink = 15 3/4 = 63/4.

Let the strawberry syrup be \frac{9}{4} x.

And let the water be \frac{3}{4} x.

According to question:

\frac{9x}{4} + \frac{3x}{4}=\frac{63}{4}.

<em>On adding the fractions:</em>

⇒\frac{9x+3x}{4} =\frac{63}{4}

⇒\frac{12x}{4} =\frac{63}{4}

<em>Multiplying both sides by 4 we get:</em>

⇒12x=63

<em>Dividing both sides by 12 we get:</em>

⇒x=\frac{63}{12}

<em>Dividing numerator and denominator by 3 on R.H.S we get:</em>

⇒x=\frac{21}{4}

Now, putting the value of x on ratios:

Strawberry syrup =  \frac{9}{4}\times x=\frac{9}{4}\times\frac{21}{4}

                             =  \frac{189}{16}

                             =  11\frac{13}{16}\ cups  

Water = \frac{3}{4}\times x =\frac{3}{4} \times\frac{21}{4}

          = \frac{63}{8}

          = 7\frac{7}{8} \ cups.

Therefore, the water she need is 7\frac{7}{8} \ cups. and strawberry syrup 11\frac{13}{16}\ cups.

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Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that X \sim Poisson(\lambda=4)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

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Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183

P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733

P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465

P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

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c. Compute P(8≤ X).

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d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

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3 years ago
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