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Arisa [49]
3 years ago
7

For Numbers 14a-14d, tell which expressions require you to rename mixed numbers before you can subtract. Find each difference. W

rite each expression and the difference as an equation in the correct box 14a 2 1/3-1 3/4. 14b 1 3/4 -7/8. 14c 5 2/3-2 1/3. 14d 6 1/5-2 1/3
Mathematics
2 answers:
weqwewe [10]3 years ago
8 0
The answer would be 142 degrees
VMariaS [17]3 years ago
6 0

Solution:

14 a, 14 c, 14 d are the Expressions which require to rename mixed numbers before we can subtract.

14 a. →

=2 \frac{1}{3}-1 \frac{3}{4}\\\\= \frac{7}{3}-\frac{7}{4}\\\\= \frac{28-21}{12} \\\\= \frac{7}{12}

14 b.

1 \frac{3}{4} -\frac{7}{8}\\\\ = \frac{7}{4}-\frac{7}{8} \\\\ =\frac{14-7}{8} \\\\ =\frac{7}{8}

14 c.

5 \frac{2}{3}-2 \frac{1}{3}\\\\ = \frac{17}{3}-\frac{7}{3} \\\\ =\frac{17-7}{3}\\\\ =\frac{10}{3}=3 \frac{1}{3}

14 d.

6\frac {1}{5}-2 \frac{1}{3}\\\\ = \frac{31}{5}-\frac{7}{3} \\\\ = \frac{93-35}{15}=\frac{58}{15}=3 \frac{13}{15}


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Answer:

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Answer:

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Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.

The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>

The probability mass function of a Geometric distribution is:

P(Y=y)=(1-p)^{y}p;\ y=0,12...

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) =  P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

              =(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable <em>Y</em> as follows:

\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445

Compute the standard deviation of the random variable <em>Y</em> as follows:

\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88

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                    = P (Y ≥ 3)

                    = 1 - P (Y < 3)

                    = 1 - P (X = 0) - P (X = 1) - P (X = 2)

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Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

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