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aleksklad [387]
3 years ago
6

Nadia is a stockbroker. She earns 14​% commission each week. Last​ week, she sold ​$7 comma 500 worth of stocks. How much did sh

e make last week in​ commission? If she averages that same amount each​ week, how much did she make in commission in​ 2011?
Mathematics
1 answer:
Artyom0805 [142]3 years ago
5 0

Answer:

$1050

$54,600

Step-by-step explanation:

In this question, we are looking at calculating two things. We want to calculate the amount of commission made in a week and the amount made in a year.

Firstly, to calculate the amount of commission she made last week, we simply find the percentage.

Mathematically, this is equal simply to:

14% of $7,500 = 14/100 * 7500 = $1050

She made a commission of $1050 last week

Now, we are told this is the average she makes per week. We now want to calculate how much she makes per year. To do this, what we do is to multiply what she had made in a week by the number of weeks in a year.

The number of weeks in a year is 52. The amount she made is just 52 * $1050 = $54,600

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Ad libitum [116K]

Answer:

(i)unique solution

Explanation:

We solve for x1 thus in the first equation:

x1-3x3=-3

x1-9=-3

x1=-3+9

x1= 6

We solve for a thus in the second equation:

2x1+ax2-x3=-2

2+a2-x3=-2

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3 years ago
The average amount of money spent for lunch per person in the college cafeteria is $6.35 and the standard deviation is $2.31. Su
alexdok [17]

Answer:

(6.35, 5.34)

(6.35, 0.99)

0.10253

0.3984

Step-by-step explanation:

Given that :

μ = 6.35

σ = 2.31

n = 41

What is the distribution of X?X ~ N(,)

X :

Mean of distribution = μ= 6.35

The variance = σ^2 = 2.31^2 = 5.3361 = 5.34

X ~ N = (6.35, 5.34)

What is the distribution of x¯? x¯ ~ N(,)

The mean = μ = 6.35

The standard deviation of the mean = σ/sqrt(n) = 6.35/sqrt(41) = 0.9917 = 0.99

X ~N = (6.35, 0.99)

find the probability that this patron's lunch cost is between $6.1605 and $6.757.

P(6.1605 < x < 6.757)

Obtain the standardized scores:

Z = (x - μ) / σ ; (6.1605 - 6.35) / 2.31 = - 0.082

P(Z < - 0.082) = 0.46732 (Z probability calculator)

(6.757 - 6.35) / 2.31 = 0.176

P(Z < 0.176) = 0.56985 (Z probability calculator)

P(Z < 0.176) -P(Z < - 0.082)

0.56985 - 0.46732 = 0.10253

For the group of 17 patrons, find the probability that the average lunch cost is between $6.1605 and $6.757.

P(6.1605 < x < 6.757) ; n = 17

Obtain the standardized scores:

Z = (x - μ) / σ/sqrt(n) ; (6.1605 - 6.35) / 2.31/sqrt(17) = - 0.338

P(Z < - 0.338) = 0.36768 (Z probability calculator)

(6.757 - 6.35) / 2.31/sqrt(17) = 0.726

P(Z < 0.726) = 0.76608 (Z probability calculator)

P(Z < 0.726) -P(Z < - 0.338)

0.76608 - 0.36768 = 0.3984

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