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mars1129 [50]
2 years ago
10

Giuliana has 22 quarters and dollar coins worth a total of $10.75.

Mathematics
2 answers:
sertanlavr [38]2 years ago
7 0
Use the letters q and d to represent # of quarters and # of dollars.

q + d = 22 coins.

$0.25q + $1.00d = $10.75        Mult. this by 100 to eliminate fractions:
                                                 25q + 100d = 1075

Subst. q = 22 - d into the 2nd equation:

25 (22  - d) + 100 d = 1075

550 - 25d + 100d = 1075
550 + 75d = 1075
           75 d = 525
                 d = 7

q + d = 22, so if d = 7, then q = 22-7, or q=15.

There are 15 quarters and 7 dimes.
Ad libitum [116K]2 years ago
4 0

Let

q------> the number of quarters coins

d------> the number of dollars coins

we know that

1\ quarter=\$0.25

q+d=22

q=22-d ------> equation A

0.25q+d=10.75 -----> equation B

substitute equation A in equation B

0.25[22-d]+d=10.75

therefore

<u>the answer is the option</u>

0.25[22-d]+d=10.75

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Answer:

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(b) E (Z) = 0 and SD (Z) = 1

Step-by-step explanation:

The time of the finishers in the New York City 10 km run are normally distributed with a mean,<em>μ</em> = 61 minutes and a standard deviation, <em>σ</em> = 9 minutes.

(a)

The random variable <em>X</em> is defined as the finishing time for the finishers.

Then the expected value of <em>X</em> is:

<em>E </em>(<em>X</em>) = 61 minutes

The variance of the random variable <em>X</em> is:

<em>V</em> (<em>X</em>) = (9 minutes)²

Then the standard deviation of the random variable <em>X</em> is:

<em>SD</em> (<em>X</em>) = 9 minutes

(b)

The random variable <em>Z</em> is the standardized form of the random variable <em>X</em>.

It is defined as:Z=\frac{X-\mu}{\sigma}

Compute the expected value of <em>Z</em> as follows:

E(Z)=E[\frac{X-\mu}{\sigma}]\\=\frac{E(X)-\mu}{\sigma}\\=\frac{61-61}{9}\\=0

The mean of <em>Z</em> is 0.

Compute the variance of <em>Z</em> as follows:

V(Z)=V[\frac{X-\mu}{\sigma}]\\=\frac{V(X)+V(\mu)}{\sigma^{2}}\\=\frac{V(X)}{\sigma^{2}}\\=\frac{9^{2}}{9^{2}}\\=1

The variance of <em>Z</em> is 1.

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Step-by-step explanation:

We have a mean \mu = 105 and a standard deviation \sigma = 7. For a value x we compute the z-score as (x-\mu)/\sigma, so, for x = 94.5 the z-score is (94.5-105)/7 = -1.5, and for x = 115.5 the z-score is (115.5-105)/7 = 1.5. We are looking for P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668 = 0.8664. Therefore, the proportion of student heights that are between 94.5 and 115.5 is 86.64%

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<h2>Answer:</h2>

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<h2>Step-by-step explanation:</h2>

Range of a function--

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