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VladimirAG [237]
3 years ago
15

Suppose that 16 inches of wire costs 96 cents at the same rate how much in cents will 13 inches of wire cost

Mathematics
1 answer:
Rom4ik [11]3 years ago
8 0

Answer:

78 cents

Step-by-step explanation:

96/16 = 6/1

6/1 is the unit rate so you input 13 into the bottom and multiply

13×6=78

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A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the card drawn is neither a red card
Reil [10]
Number of red cards including 2 red queens = 26
Number of black queens = 2
Therefore, number of red cards including 2 red queens and 2 black queens = 26 + 2 = 28
Number of cards neither a red card nor a queen = 52 - 28 = 24

P  =  \frac{ Number \: of \: favourable \: outcomes}{ Total \: numer \: of \: possible \: outcomes}

P (neither \: a \: red \: nor \: a \: queen \: card \: ) =  \frac{24}{52}  =  \frac{6}{13}
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George desposited 2000 in his bank, which offers 5% compoud interest annually. What would be the principla amount available in g
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George would have 2200
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3 years ago
Write the following percents as both fractions and decimals.
wolverine [178]

Answer:

1/4, 0.25

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1/25, 0.04

3/4, 0.75

13/200, 0.065

1 1/4, 1.25

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107/100, 1.07

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4 0
3 years ago
The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs a minimum number o
SCORPION-xisa [38]

Answer:

The significance level is \alpha=0.01 and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

z_{\alpha/2}= 2.326

So we reject the null hypothesis is z>2.326

Step-by-step explanation:

For this case we define the random variable X as the number of entry-level swimmers and we are interested about the true population mean for this variable . On specific we want to test this:

Null hypothesis: \mu \leq 15

Alternative hypothesis: \mu > 15

And the statistic is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

The significance level is \alpha=0.01 and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

z_{\alpha/2}= 2.326

So we reject the null hypothesis is z>2.326

7 0
3 years ago
Maths<br> 1. 24÷ (-6)<br><br> 2. -15 ÷ 0.3<br><br> 3. -4 ÷ (-20)
Lesechka [4]

Answer:

hope it helps

Step-by-step explanation:

1) -72

2) -50

3) 0.2

8 0
3 years ago
Read 2 more answers
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