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White raven [17]
3 years ago
6

Oscar has $1.00 in quarters and dimes. He has 7 coins altogether. How many coins of each kind does he have?

Mathematics
1 answer:
tigry1 [53]3 years ago
6 0

Answer:

2 quarters and 5 dimes

Step-by-step explanation:

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Which make a triangle 3 4 5. 3 5 7. 1 3 4. 2 8 9.
adelina 88 [10]

i don't understand this question but I will say 1/3/4

8 0
3 years ago
Represent Real-world Problems If one store is selling of a bushel of apples for $9, and another store is selling of a bushel of
Firlakuza [10]
Better deal normlly means the rate that cost less
rate=cost/amount

9=cost and 3/4 bushel = amount for one store
3/4=0.75
9/0.75=12/1 aka $12/1bushel

9 =cost and 2/3 bushel =amount for another
2/3=about 0.66666
9/0.66666=13.5/1 aka $13.5/1 bushell

compare
12 dollars per bushe vs 13.5 dollar per bushel
12 is cheaper

therefor $9 For 3/4 bushel is better deal
5 0
3 years ago
What is 45/90= to <br> i need help
Sati [7]

Answer:

1/2

Step-by-step explanation:

45/90 ÷ 45/45 = 1/2

5 0
3 years ago
Read 2 more answers
Two trains on opposite tracks leave the same station at the same
nikdorinn [45]

Answer:

20 minutes( 0.3333 hours) after they leave the station will they be 50 km apart, if the Two trains on opposite tracks leave the same station at the same time.

Step-by-step explanation:

One train travels at an average speed of 80 km/hr and the other travels at an average speed of 70 km/hr.

4 0
2 years ago
Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f
castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean (\mu) = 115 \times 20

= 2300

Standard deviation = 20 \times \sqrt{115}

Standard deviation (SD) = 214.4761

TO find:

a) P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})

P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})

P(x > 3500)= P(Z > \dfrac{1200}{214.4761})

P(x > 3500)= P(Z >5.595)

From the Z-table, since 5.595 is > 3.999

P(x > 3500)=1-0.9999

P(x > 3500) = 0.0001

b)

Here, the replacement time for the mean (\mu) = \dfrac{0+0.5}{2}

= 0.25

Replacement time for the Standard deviation \sigma = \dfrac{0.5-0}{\sqrt{12}}

\sigma = 0.1443

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = \sqrt{(115\times 20^2) +(114\times (0.1443)^2)}

= \sqrt{(115\times 400) +(114\times 0.02082249}

= \sqrt{(46000) +2.37376386}

= \sqrt{(46000) +(2.37376386)}

= \sqrt{46002.374}

= 214.482

Now; the required probability:

P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})

P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})

P(x > 4125) = P(Z >8.376)

P(x > 4125) =1-  P(Z

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001

7 0
2 years ago
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