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Setler79 [48]
2 years ago
10

Sam ran 3 1/2 miles in 25 minutes. If he continues running at this pace how far will he run in 1 hour

Mathematics
2 answers:
zheka24 [161]2 years ago
5 0

Answer: 8.4 or 8 2/5

Step-by-step explanation:

3 1/2 for 25 minutes and 0.14 miles per minute so 0.14 * 60 since there are 60 minutews in a hour and you get ur answer.

11111nata11111 [884]2 years ago
3 0
He will run 8.4 miles in an hour. 3.5/25=x/60
3.5*60=210
210/25=8.4
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An investor invested a total of $2500 in two mutual funds. One fund earned a 5% profit while the other earned a 3% profit. If th
7nadin3 [17]

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$500 and $2000

Step-by-step explanation:

Let x represent the total investment = $2500

also, this total is split into two different funds

Lets represent these funds as a and b, such that fund a yields a profit of 3% and fund b yield a profit of 5%

So,

a + b = x

a + b =  2500   ......eq 1

Profit from each fund gives;

0.03 a + 0.05b = 115     ....eq 2

Solve simultaneously using substitution method

From eq 1;

b = 2500 - a

Slot in this value in eq 2

0.03a + 0.05 (2500 - a) = 115

expand

0.03a + 125 - 0.05a = 115

collect like terms

0.03a - 0.05a = 115 - 125

-0.02a = -10

Divide both sides by -0.02

a =  $500

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500 + b =  2500

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Answer:

Step-by-step explanation:

Let C be the number of correct answers

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You have been doing research for your statistics class on the prevalence of severe binge drinking among teens. You have decided
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Answer:

(1.218 ; 1.322)

the confidence interval is appropriate

Step-by-step explanation:

The confidence interval :

Mean ± margin of error

Sample mean = 1.27

Sample standard deviation, s = 0.80

Sample size, n = 914

Since we are using tbe sample standard deviation, we use the T table ;

Margin of Error = Tcritical * s/√n

Tcritical at 95% ; df = 914 - 1 = 913

Tcritical(0.05, 913) = 1.96

Margin of Error = 1.96 * 0.80/√914 = 0.05186

Mean ± margin of error

1.27 ± 0.05186

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Upper boundary = 1.27 + 0.05186 = 1.322

(1.218 ; 1.322)

According to the central limit theorem, sample means will approach a normal distribution as the sample size increases. Hence, the confidence interval is valid, the sample size of 914 gave a critical value at 0.05 which is only marginally different from that will obtained using a normal distribution table. Hence, the confidence interval is appropriate

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