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3 years ago

15

Randy elks purchased 3000 shares of a juice company stock at $50.10 per share. His broker charges $0.03 per share for comission.

What is the total cost of the stock?

2 answers:

Talja [164]3 years ago

8 0

$150,390 because you would need to add $0.03 to $50.10 and multiply it by 3000

Fed [463]3 years ago

6 0

59.850239520958083832335329341317
is what I got

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F(x)=3x means the same thing as y=3x but is written in function notation.<br> A. true <br> B. false

Darya [45]

False I think it is if I’m right

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3 years ago

Tell whether the given number is a solution of each question. ( Must show work for credit)

guajiro [1.7K]

Answer/Step-by-step explanation:

1. $9 = 2a + 3$ ; 3

Let's solve for a.

$9 - 3 = 2a$ (subtraction property of equality)

$6 = 2a$

$\frac{6}{2} = a$ (division property of equality)

$3 = a$

$a = 3$

Therefore, 3 is a solution to 9 = 2a + 3

2. $5n - (-30) = 5$ ; 7

$5n + 30 = 5$ (- × - = +)

$5n = 5 - 30$ (Subtraction property of equality)

$5n = -25$

$\frac{5n}{5} = \frac{-25}{5}$ (division property of equality)

$n = -5$

Therefore, 7 is not a solution of $5n - (-30) = 5$

3. $4.4r - 2.8 = 1.6$ ; 1

$4.4r = 1.6 + 2.8$

$4.4r = 4.4$

$\frac{4.4r}{4.4} = \frac{4.4}{4.4}$

$r = 1$

Therefore, 1 is a solution of $4.4r - 2.8 = 1.6$ .

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3 years ago

Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif

dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

<u>Step:-1</u>

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals $t_{0.01} =2.82$ (from tabulated value).

$The mean value = 0.0365$

$x±t_{0.01} \frac{S}{\sqrt{n-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{22} }$

$0.0365±2.82 \frac{0.0046}{4.690 }$

using calculator

$0.0365±0.00276$

Confidence interval is

$(0.0365-0.00276,0.0365+0.00276)$

$(0.0338,0.0392)$

the mean value is lies between in this confidence interval

(0.0338,0.0392).

<u>Answer:-</u>

<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>

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3 years ago

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Contact [7]

You worded this weird, but each bag of sugar is 8 pounds...

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3 years ago

100 POINTS!! A company has 400 employees. There are 300 female employees. There are 80 employees who are under the age of 25. Th

Fittoniya [83]

Answer:

probability of randomly selecting an employee who is female and under the age of 25= 0.15

In percentage= 15%

Step-by-step explanation:

There are 300 female employees. There are 80 employees who are under the age of 25

Probabilty of choosing a female employees= 300/400

Probabilty of choosing a female employees= 0.75

Probabilty of choosing an employees under the age of 25

= 80/400

Probabilty of choosing an employees under the age of 25

= 0.2

probability of randomly selecting an employee who is female and under the age of 25= 0.2*0.75

probability of randomly selecting an employee who is female and under the age of 25= 0.15

In percentage= 0.15*100

In percentage= 15%

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3 years ago