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2 years ago

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A small business owner is determining her profit for one month. Her expenses were $230.21 for utilities, $2,679.82 for rent, and

$3,975.00 for employee salaries. She made $11,449.27 in sales for the month. What is her profit?

1 answer:

ss7ja [257]2 years ago

6 0

She made 9,604.21 sorry if I’m wrong I really tried

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Margo bought 7 pens from a bookstore. Some of the pens cost $3 each, and the rest cost $4 each. If she paid a total of $23 for t

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Answer: 5 Pens!

Step-by-step explanation: 5 Pens Would Be The Answer.

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2 years ago

on a piece of paper, graph this system of inequalities. Then determine which region contains the solution to the system. a pictu

slava [35]

C. Region D

Step-by-step explanation:

Just plug the equations into desmos the calculator will graph it and show you with colors where the solution is.

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2 years ago

Statistics can help decide the authorship of literary works. Sonnets by a certain Elizabethan poet are known to contain an avera

tiny-mole [99]

Answer:

We conclude that the sonnets were written by by a certain Elizabethan poet.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 8.9

Sample mean, $\bar{x}$ =10.2

Sample size, n = 6

Alpha, α = 0.05

Population standard deviation, σ = 2.5

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 8.88\\H_A: \mu > 8.88$

We use One-tailed z test to perform this hypothesis.

a) Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{10.2 - 8.9}{\frac{2.5}{\sqrt{6}} } = 1.28$

Now, $z_{critical} \text{ at 0.05 level of significance } = 1.64$

b) We calculate the p value with the help of z-table.

P-value = 0.1003

The p-value is greater than the significance level which is 0.05

c) Since the p-value is greater than the significance level, there is not enough evidence to reject the null hypothesis and accept the null hypothesis.

Thus, we conclude that the sonnets were written by by a certain Elizabethan poet.

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2 years ago

se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y

alexandr1967 [171]

Answer:

It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$ exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$ exists.

To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$ where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0

2 years ago

In your sock drawer you have 5 blue, 7 gray, and 2 black socks. Half asleep one morning you grab 2 socks at random and put them

pentagon [3]

The question is incomplete! The complete question along with answers and explanation is provided below!

In your sock drawer you have 5 blue, 7 gray, and 2 black socks. Half asleep one morning you grab 2 socks at random and put them on. Find the probability you end up wearing the following socks. (Round your answers to four decimal places.)

a) 2 blue socks

b) no gray socks

c) at least 1 black sock

d) a green sock

e) matching socks

Answer:

a) P(2 blue socks) = 0.1099 = 10.99%

b) P(no gray socks) = 0.2307 = 23.07%

c) P(at least 1 Black sock) = 0.2748 = 27.48%

d) P(green sock) = 0%

e) P(Matching socks) = 0.3516 = 35.16%

Step-by-step explanation:

Given Information:

5 Blue socks

7 Gray socks

2 black socks

Total socks = 5 + 7 + 2 = 14

a) The probability of wearing 2 blue socks

P(2 blue socks) = P(B1 and B2)

P(B1) = no. of blue socks/total no. of socks

P(B1) = 5/14 = 0.3571

Now there are 4 blue socks remaining and total 13 socks remaining

P(B2|B1) = 4/13 = 0.3077

P(B1 and B2) = 0.3571*0.3077 = 0.1099 = 10.99%

b) The probability of wearing no gray socks

5 Blue socks + 2 black socks = 7 socks are not gray

P(no gray socks) = P(Not G1 and Not G2)

P(Not G1) = no. socks that are not grey/ total no. of socks

P(Not G1) = 7/14 = 0.5

Now there are 6 socks remaining that are not gray and total 13 socks remaining

P(Not G2 | Not G1) = 6/13 = 0.4615

P(Not G1 and Not G2) = 0.5*0.4615 = 0.2307 = 23.07%

c) The probability of wearing at least 1 black sock

5 Blue socks + 7 Gray socks = 12 socks are not black

P(at least 1 Black) = 1 - P(Not B1 and Not B2)

P(Not B1) = no. socks that are not black/ total no. of socks

P(Not B1) = 12/14 = 0.8571

Now there are 11 socks remaining that are not black and total 13 socks remaining

P(Not B2 | Not B1) = 11/13 = 0.8461

P( Not B1 and Not B2) = 0.8571*0.8461 = 0.7252

P(at least 1 Black) = 1 - P( Not B1 and Not B2)

P(at least 1 Black) = 1 - 0.7252 = 0.2748 = 27.48%

d) The probability of wearing a green sock

There are 0 green socks, therefore

P(Green) = 0/14 = 0%

e) The probability of wearing matching socks

P(Matching socks) = P(2 Blue socks) + P(2 Gray socks) + P(2 Black socks)

P(2 Blue socks) already calculated in part a

P(2 Blue socks) = P(B1 and B2) = 0.1099

For Gray socks

P(G1) = no. of gray socks/ total no. of socks

P(G1) = 7/14 = 0.5

Now there are 6 gray socks remaining and total 13 socks remaining

P(G2 | G1) = 6/13 = 0.4615

P(2 Gray socks) = P(G1 and G2) = 0.5*0.4615 = 0.2307

For Black socks

P(B1) = no. of black socks/ total no. of socks

P(B1) = 2/14 = 0.1428

Now there is 1 black sock remaining and total 13 socks remaining

P(B2 | B1) = 1/13 = 0.0769

P(2 Black socks) = P(B1 and B2) = 0.1428*0.0769 = 0.0110

P(Matching socks) = P(2 Blue socks) + P(2 Gray socks) + P(2 Black socks)

P(Matching socks) = 0.1099 + 0.2307 + 0.0110 = 0.3516 = 35.16%

7 0

2 years ago