Ask question

Ask question

All categories

3 years ago

15

The Stock Market Alexander is a stockbroker. He earns 11% commission each week. Last week, he sold $6,400 worth of stocks. Ho

w much did he make last week in commission? If he averages that same amount each week, how much did he make in commission in 2011?
Last week, he made ____ in commission.

Giving brainliest

2 answers:

kaheart [24]3 years ago

6 0

Answer:

Last week, he made $704 in commission. In a year he makes $36,608.

Step-by-step explanation:

You simply take 11% of $6,400 and for the year, multiply $704 by 52 (there are 52 weeks in a year).

Anna007 [38]3 years ago

5 0

Answer:I believe 704 is the answer

Step-by-step explanation:

Hope this helped

You might be interested in

Whats the volume of a pyramid

Fantom [35]

Answer:

The volume of a pyramid is 1/3 x Base Area x Height.

Step-by-step explanation:

3 0

3 years ago

Help pleaseeeeeeeeee

Alex73 [517]

Reflecting across the y axis followed by rhe x atation 180 degrees clockwise

Rotation 180 counterclowise

4 0

3 years ago

Complete the equation describing how x<br> and y are related.

Natali5045456 [20]

Answer: -5

Step-by-step explanation:

4 0

3 years ago

A random sample of 77 fields of corn has a mean yield of 26.226.2 bushels per acre and standard deviation of 2.322.32 bushels pe

PSYCHO15rus [73]

Answer:

Therefore the 95% confidence interval is (25,707.480 < E < 26,744.920)

Step-by-step explanation:

n = 77

mean u = 26,226.2 bushels per acre

standard deviation s = 2,322.32

let E = true mean

let A = test statistic

Find 95% Confidence Interval

so

let A = (u - E) * ( $\sqrt{n}$ / s) be the test statistic

we want P( average_l < A < average_u ) = 95%

look for lower 2.5% and the upper 97.5% Because I think this is a 2-tail test

average_l = -1.96 which corresponds to the 2.5%

average_u = 1.96

P( -1.96 < A < 1.96) = 95%

P( -1.96 < (u - E) * ( $\sqrt{n}$ / s) < 1.96) = 95%

Solve for the true mean E ok

E < u + 1.96* (s / $\sqrt{n}$ )

from -1.96 < (u - E) * ( $\sqrt{n}$ / s)

E < 26,226.2 + 1.96*( 2,322.32 / $\sqrt{77}$ )

E < 26,226.2 + 1.96*( 2,322.32 / $\sqrt{77}$ )

E < 26,226.2 + 518.7197348105429466

upper bound is 26,744.9197

or

u - 1.96* (s / $\sqrt{n}$ ) < E

26,226.2 - 518.7197348105429466 < E

25,707.48026519 < E

lower bound is 25,707.48026519

Therefore the 95% confidence interval is (25,707.480 < E < 26,744.920)

7 0

3 years ago

13 POINTS- please help me

allsm [11]

Answer:

See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures $(6x+20)^{\circ}$ and $(x+100)^{\circ}$ are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

$6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16$

Then

$(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}$

17. Two parallel lines are cut by transversal. Angles with measures $(2x+12)^{\circ}$ and $(3x-22)^{\circ}$ are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

$2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34$

Then

$(2x+12)^{\circ}=(2\cdot 34+12)^{\circ}=80^{\circ}\\ \\(3x-22)^{\circ}=(3\cdot 34-22)^{\circ}=80^{\circ}$

18. Two parallel lines are cut by transversal. Angles with measures $(6x-7)^{\circ}$ and $(5x+10)^{\circ}$ are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

$6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17$

Then

$(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}$

19. The diagram shows two complementary angles with measures $2x^{\circ}$ and $56^{\circ}$ . The measures of complementary angles add up to $90^{\circ},$ then

$2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17$

Hence,

$2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}$

Check:

$34^{\circ}+56^{\circ}=90^{\circ}$

20. Angles $\angle 1$ and $\angle 2$ are vertical angles. By vertical angles theorem, vertical angles are congruent, so

$m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4$

Hence,

$m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}$

21. $\angle 5$ and $\angle 8$ are supplementary. The measures of supplementary angles add up to $180^{\circ},$ so

$m\angle 5+m\angle 8=180^{\circ}\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34$

Therefore,

$m\angle 5=(3x-40)^{\circ}=(3\cdot 34-40)^{\circ}=62^{\circ}\\ \\m\angle 8=(7x-120)^{\circ}=(7\cdot 34-120)^{\circ}=118^{\circ}\\ \\62^{\circ}+118^{\circ}=180^{\circ}$

7 0

3 years ago