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2 years ago

Perform the indicated operation. (y4 - 1) ÷ (y + 1)

2 answers:

7 0

$The\ domain:\\D:y\neq-1\\\\(y^4-1)\div(y+1)=\dfrac{(y^2)^2-1^2}{y+1}=\dfrac{(y^2-1)(y^2+1)}{y+1}\\\\=\dfrac{(y-1)(y+1)(y^2+1)}{y+1}=(y+1)(y^2+1)=y^3+y^2+y+1\\\\Used:\\(a^n)^m=a^{nm}\\\\a^2-b^2=(a-b)(a+b)$

Dmitry_Shevchenko [17]2 years ago

3 0

Answer: y^3 - y^2 + y - 1

Step-by-step explanation: Factor the numerator and denominator and cancel the common factors. This should leave you with y^3-y^2+y-1.

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A company currently has 200 units of a product on hand that it orders every 2 weeks when the salesperson visits the premises. De

disa [49]

Answer: 289 units

Step-by-step explanation:

Given the following :

Inventory (I) = 180

Lead time (L) = 7 days

Review time (T) = 2 weeks = 14 days

Demand (D) = 20

Standard deviation (σ) = 5

Zscore for 95% probability = 1.645

Units to be ordered :

D(T + L) + z(σT+L)

(σT+L) = √(T + L)σ²

= √(14 + 7)5²

= √(21)25

= 22.9

D(T + L) + z(σT+L) - I

20(14 + 7) + 1.645(22.9 + 7) - I

= 420 + 49.1855 - 180

= 289.1855

= 289 quantities

5 0

2 years ago

A web-based company has a goal of processing 90 percent of its orders on the same day they are received. If 434 out of the next

Kamila [148]

Answer:

We conclude that a web-based company are not exceeding their goal of 90%.

Step-by-step explanation:

We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.

434 out of the next 471 orders are processed on the same day.

Let p = proportion of orders processing on the same day they are received.

SO, Null Hypothesis, $H_0$ : p $\leq$ 0.90 {means that they are not exceeding their goal of 90%}

Alternate Hypothesis, $H_0$ : p > 0.90 {means that they are exceeding their goal of 90%}

The test statistics that would be used here One-sample z test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of orders that are processed on the same day = $\frac{434}{471}$ = 0.92

n = sample of orders = 471

So, the test statistics = $\frac{0.92-0.90}{\sqrt{\frac{0.92(1-0.92)}{471} } }$

= 1.599

The value of z test statistics is 1.599.

Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.

Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that a web-based company are not exceeding their goal of 90%.

8 0

1 year ago

Four people—Rob, Sonja, Jack, and Ang—enter their names into a drawing. The winner receives either a t-shirt or a mug, and which

stira [4]

Answer:

this is the answer 25

Step-by-step explanation:

i guessed

4 0

1 year ago

The rate of transmission in a telegraph cable is observed to be proportional to x2ln(1/x) where x is the ratio of the radius of

sergij07 [2.7K]

Answer:

The value of x that gives the maximum transmission is 1/√e ≅0.607

Step-by-step explanation:

Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.

$f'(x) = k*((x^2)'*ln(1/x) + x^2*(ln(1/x)')) = k*(2x\,ln(1/x)+x^2*(\frac{1}{1/x}*(-\frac{1}{x^2})))\\= k * (2x \, ln(1/x)-x)$

We need to equalize f' to 0

k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
ln(1/x) = x/2x = 1/2 ------- we send the natural logarithm as exp
1/x = e^(1/2)
x = 1/e^(1/2) = 1/√e ≅ 0.607

Thus, the value of x that gives the maximum transmission is 1/√e.

7 0

2 years ago

Explain how you used the bar model in exercise 2 to solve the problem

lutik1710 [3]

We are given that Grandma has 14 red roses and 7 pink roses.

In order to represent them on bar modal, we need to make a bar with 14 identical sections for 14 red roses.

And for pink roses, we need to make a bar with 7 identical sections for 7 pink roses.

From the bar graph, we can see that bar for red roses have 7 more boxes than bar of pink roses.

<h3>Therefore, she have 7 more red roses than pink roses.</h3>

3 0

1 year ago