Help i need to ppooooppp

Help me with this problem show your work please

Akimi4 [234]

It cannot be simplified...

5 0

3 years ago

What is the sum of an infinite geometric series if the first term is 156 and the common ratio is 2⁄3?

Zanzabum

Answer:

Step-by-step explanation:

We have

$a_1=156,\ r=\dfrac{2}{3}$

If |r| < 1, then the formula of a sum of an infinite geometric sequence is:

$S=\dfrac{a_1}{1-r}$

Substitute:

$S=\dfrac{156}{1-\frac{2}{3}}=\dfrac{156}{\frac{1}{3}}=156\cdot\dfrac{3}{1}=468$

4 0

3 years ago

A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life

Tasya [4]

Answer:

(a) H₀: μ ≤ 10. vs. Hₐ: μ > 10.

(b) The test statistic value is 1.86.

(c) The critical value to test the phone manufacturer's claim is 1.301.

(d) The battery life of the new touch screen phone is not more than 10 hours.

Step-by-step explanation:

In this case we need to determine the significance of the claim made by the phone manufacturer, that the battery life of the new touch screen phone is more than twice as long as that of the leading product.

The information provided is:

$\bar x=10.5\\s=1.8\\n=45$

(a)

The hypothesis can be defined as follows:

H₀: The battery life of the new touch screen phone is not more than 10 hours, i.e. μ ≤ 10.

Hₐ: The battery life of the new touch screen phone is more than 10 hours, i.e. μ > 10.

(b)

As the population standard deviation is not provided, we will use a t-test for single mean.

Compute the test statistic value as follows:

$t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{10.5-10}{1.8/\sqrt{45}}=1.86$

The test statistic value is 1.86.

(c)

The significance level of the test is, α = 0.10.

The degrees of freedom will be:

$df=n-1=45-1=44$

Compute the critical value as follows:

$t_{0.10, 44}=1.301$

*Use a t-table for the value.

Thus, the critical value to test the phone manufacturer's claim is 1.301.

(d)

Decision rule:

If the test statistic value is less than the critical value then the null hypothesis will be rejected and vice-versa.

t = 1.86 > t₀.₁₀, ₄₄ = 1.30

The calculated t-value of the test is more than the critical value.

The null hypothesis will not be rejected.

Thus, it can be concluded that the battery life of the new touch screen phone is not more than 10 hours.

6 0

3 years ago

Tript Corporation has a process costing system and uses the weighted-average method. The company had 3,000 units in work in proc

ycow [4]

Answer:

11,000 units were started into production which were 100 % complete for materials and 89.1 % complete for Conversion Costs.

Step-by-step explanation:

With the help of Equivalent Units of Production we find out how many units were started.

11,000 units were started into production.

Tript Corporation

Physical Units % of Completion Equivalent Units

w.r.t Conversion Costs Mat Conversion C

Units Started 11,000 89.1% 11,000 9,800

Beginning WIP 3000 60% 3000 1800

Total units accounted For 14,000 11600

We compare the Equivalent units and then find the required balance amounts

Physical Units % of Completion Equivalent Units

w.r.t Conversion Costs Mat Conversion C

Units Completed 10,000 100 10,000 10,000

WIP Ending 4000 40% 4000 1600

Total units accounted For 14,000 11,600

4 0

3 years ago

10. Determine the open interval(s) on which the function f(x) = 12x - x^3 is increasing.

Gre4nikov [31]

Answer:

$\large\boxed{x\in(-2,\ 2)}$

Step-by-step explanation:

$f(x)=12x-x^3$

Calculate the derivative of the function:

$f'(x)=\left(12x-x^3\right)'=(12x)'-\left(x^3\right)'=12-3x^2$

Calculate the zeros of the derivative:

$f'(x)=0\Rightarrow12-3x^2=0\qquad\text{add}\ 3x^2\ \text{to both sides}\\\\3x^2=12\qquad\text{divide both sides by 3}\\\\x^2=4\to x=\pm\sqrt4\\\\x=\pm2$

Let's check the derivative sign:

$f'(x)>0\iff12-3x^2>0\qquad\text{use the distributive property}\\\\-3(x^2-4)>0\\\\-3(x^2-2^2)>0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\-3(x-2)(x+2)>0\\^{\qquad x=2\qquad x=-2$

Sketch a parabola open down

(before the brackets there is the number -3 < 0)

Look at the picture.

Where the derivative is negative, the function decreases.

Where the derivative is positive, the function increases.

Therefore your answer is:

$x\in(-2,\ 2)$

3 0

3 years ago