2 years ago

Evaluate 10b + 3c, if b=3 and c = 11.

Mathematics

1 answer:

kozerog [31]2 years ago

4 0

Answer:

Step-by-step explanation:

10 × 3 = 30

3× 11 = 33

30 + 33 = 63

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3 years ago

David & Brad are competing in a 20-mile bicycle race. David rides 6.2 miles in 1/4 hour. Brad rides 11.9 miles in 1/2 hour.

Elodia [21]

David will finish first, because if you divide Brad's miles by 2 to his miles in a 1/4 of an hour are 5.95 miles. This means David went 6.2 miles in the same time that Brad went 5.95. Therefore, David will finish first.

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3 years ago

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,

REY [17]

Answer:

a) 217

b) 1351

c) 5403

Step-by-step explanation:

Given that:

confidence interval (c) = 0.95

$\alpha =1-0.95=0.05\\\frac{\alpha }{2} =\frac{0.05}{2}=0.025$

The Z score of $\frac{\alpha }{2}$ is from the z table is given as:

$Z_{\frac{\alpha }{2} }=Z_{0.025}=1.96$

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The standard deviation (σ) is given as:

$\sigma=\frac{Range}{4} =\frac{15000}{4}=3750$

Sample size (n) is given as:

$n=(\frac{Z_{\frac{\alpha}{2} }\sigma}{E} )^2$

a) E = $500

$n=(\frac{Z_{\frac{\alpha}{2} }\sigma}{E} )^2= (\frac{1.96*3750}{500} )^2$ ≈ 217

b) $n=(\frac{Z_{\frac{\alpha}{2} }\sigma}{E} )^2= (\frac{1.96*3750}{200} )^2$ ≈ 1351

c) $n=(\frac{Z_{\frac{\alpha}{2} }\sigma}{E} )^2= (\frac{1.96*3750}{100} )^2$ ≈ 5403

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3 years ago

Read 2 more answers

At a used bookstore, Cindy bought 2 paperback books for $1.75 and some children's books for $0.75 each,

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C. 5 bc u jus gotta add it on the calculator

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3 years ago

What is the value of r of the geometric series n=11.3(0.8)n-1

jeyben [28]

the value of r of the geometric series n=11.3(0.8)n-1

$an=11.3(0.8)^{n-1}$

General formula for nth term of any geometric series is $a_n=a_1(r)^{n-1}$

Here 'r' is the common ratio

a_1 is the first term of the series

Now we compare the given formula with general formula

Compare $an=11.3(0.8)^{n-1}$ with $a_n=a_1(r)^{n-1}$

The value of r= 0.8

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3 years ago