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Effectus [21]
3 years ago
12

Stephanie has a total of 27 nickels and dimes. The total value of the coins is $1.95. Which system of equations can be used to f

ind the number of nickels(n), and the number of dimes(d), that Stephanie has?
A. n+d=27
0.05n+0.10d=1.95
B. n+d=27
5n+10d=1.95
C. n+d=27
0.5n+0.10d=1.95
D. n+d=27
0.15(n+d)=1.95
Mathematics
1 answer:
Genrish500 [490]3 years ago
5 0
The correct equation to use is A. n+d=27 0.05n+0.10d=1.95
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Answer:

11/45

Step-by-step explanation:

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Number of customers that go 1 or 2 days:

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Determine which postulate can be used to prove that the triangles are congruent. If it is not possible to prove congruence, writ
irakobra [83]

Answer:

The triangles are congruent by

SSS

Step-by-step explanation:

Given:

First Label the Diagram:

AB ≅ CD

AD ≅ BC

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Δ ABC ≅ Δ CDA

Proof:

In  Δ ABC and Δ CDA

AB ≅ CD     ....……….{Given}

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8 0
3 years ago
According to an article in Newsweek, the rate of water pollution in China is more than twice that measured in the US and more th
jeyben [28]

Answer:

(a) 0.119

(b) 0.1699

Solution:

As per the question:

Mean of the emission, \mu = 11.7 million ponds/day

Standard deviation, \sigma = 2.8 million ponds/day

Now,

(a) The probability for the water pollution to be at least 15 million pounds/day:

P(X\geq 15) = P(\frac{X - /mu}{\sigma} \geq \frac{15 - 11.7}{2.8})

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P(X\geq 15) = 1 - P(Z < 1.178)

Using the Z score table:

P(X\geq 15) = 1 - 0.881 = 0.119

The required probability is 0.119

(b) The probability when the water pollution is in between 6.2 and 9.3 million pounds/day:

P(6.2 < X < 9.3) = P(\frac{6.2 - 11.7}{2.8} < \frac{X - \mu}{\sigma} < \frac{9.3 - 11.7}{2.8})

P(6.2 < X < 9.3) = P(- 1.96 < Z < - 0.86)

P(6.2 < X < 9.3) = P(- 1.96 < Z < - 0.86)

P(Z < - 0.86) - P(Z < - 1.96)

Now, using teh Z score table:

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4 0
3 years ago
The following table shows the length and width of two rectangles:
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Rectangle B: (Still with all the variables already doubled)
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So it would be C.
5 0
3 years ago
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