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OLEGan [10]
3 years ago
13

Tamika makes a 4% commission selling electronics. How much commission does she make if she sells a flat-screen TV for $8,000?

Mathematics
1 answer:
Vesna [10]3 years ago
8 0

Answer:

D.) $320

Step-by-step explanation:

4% of 8,000

so

0.04 x 8,000

= 320

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The mean student loan debt for college graduates in Illinois is $30000 with a standard deviation of $9000. Suppose a random samp
Nataly [62]

Answer:

the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

Step-by-step explanation:

Given that:

Mean = 30000

Standard deviation = 9000

sample size = 100

The probability that the mean student loan debt for these people is between $31000 and $33000 can be computed as:

P(31000 < X < 33000) = P( X \leq 33000) - P (X \leq 31000)

P(31000 < X < 33000) = P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )- P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )

P(31000 < X < 33000) = P( Z \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )- P(Z \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}}    )

P(31000 < X < 33000) = P( Z \leq \dfrac{3000}{\dfrac{9000}{10}}}) -P(Z \leq \dfrac{1000}{\dfrac{9000}{10}}})

P(31000 < X < 33000) = P( Z \leq 3.33)-P(Z \leq 1.11})

From Z tables:

P(31000 < X

P(31000 < X

Therefore; the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

8 0
3 years ago
If a+b+c =0 show that a³+b³+c³= 3abc
slega [8]

Answer:

Step-by-step explanation:

a+b+c=0, a+b=-c,a+c=-b, b+c=-a

(a+b+c)^3=(a+b+c)^2*(a+b+c)=(a^2+b^2+c^2+2ab+2ac+2bc)*(a+b+c)=

a^3+ab^2+ac^2+2a^2b+2a^2c+2abc+a^2b+b^3+bc^2+2ab^2+2abc+2b^2c+a^2c+b^2c+c^3+2abc+2ac^2+2bc^2=a^3+b^3+c^3+3a^2b+3a^2c+3ac^2+3ab^2+3bc^2+3b^2c+6abc=

a^3+b^3+c^3+3a^2*(b+c)+3c^2(a+b)+3b^2(a+c)+6abc=

a^3+b^3+c^3+3a^2*(-a)+3c^2*(-c)+3b^2*(-b)+6abc=

a^3+b^3+c^3-3a^3-3c^3-3b^3+6abc=

6abc-2a^3-2b^3-2c^3=2(3abc-a^3-b^3-c^3)=

2*[3abc-(a^3+b^3+c^3)]=0

so 3abc-(a^3+b^3+c^3)=0

so a^3+b^3+c^3=3abc

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