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3 years ago

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Determine if the sequence is geometric. If it is, find the common ratio, the 8th term, and the explicit formula. -2, 6, -18, 54,

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1 answer:

velikii [3]3 years ago

7 0

Answer:4374

Step-by-step explanation: multipy each number by -3

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What expression is equivalent to [x1/4*y16]^1/2

tigry1 [53]

Answer:

The given expression is ${{{\bf x}^{\frac{\bf 1}{\bf 4}}\times y^{\bf 16}}} ^ {\frac{1}{2}} =x^{\frac {\bf 1}{\bf 8}}\times y^{\bf 4}$ .

Step-by-step explanation:

The given expression is ${{x^{\frac{1}{4}}\times y^{16}}}^\frac{1}{2}$

${{x^{\frac{1}{4}}\times y^{16}}}^{\frac{1}{2}}=\sqrt{x^{\frac{1}{4}}\times y^{16}}$

${{x^{\frac{1}{4}}\times y^{16}}}^{\frac{1}{2}}= \sqrt{x^{\frac{1}{4}}}\times \sqrt{y^{16}}$

${{x^{\frac{1}{4}}\times y^{16}}}^{\frac{1}{2}}=x^{\frac{1}{4}}^\frac{1}{2}} \times y^{4}$

${{x^{\frac{1}{4}}\times y^{16}}}^{\frac{1}{2}}= x^{\frac{1}{8}}\times y^{4}$ by using ${a^{m}} ^{n}=a^ {mn}$ property.

5 0

4 years ago

Help me please and thank you!

lorasvet [3.4K]

Answer:

-4/3

Step-by-step explanation:

-2 - 1 is -3

6-2 is 4

6 0

3 years ago

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each nig

brilliants [131]

Answer:

Confidence interval

$8.98-2.2\frac{1.29}{\sqrt{12}}=8.16$

$8.98+2.2\frac{1.29}{\sqrt{12}}=9.80$

And the margin of error would be:

$ME=2.2\frac{1.29}{\sqrt{12}}=0.819$

Step-by-step explanation:

For this case we have the followig dataset:

DATA: 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5

We can calculate the mean and the deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$\bar X= 8.98[/tex[tex]s = 1.29$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are given by:

$df=n-1=12-1=11$

The Confidence level is 0.95 or 95%, the value of significance is $\alpha=0.05$ and $\alpha/2 =0.025$ , and the critical value would be $t_{\alpha/2}=2.20$

Repplacing the info we got:

$8.98-2.2\frac{1.29}{\sqrt{12}}=8.16$

$8.98+2.2\frac{1.29}{\sqrt{12}}=9.80$ And the margin of error would be:

$ME=2.2\frac{1.29}{\sqrt{12}}=0.819$

7 0

3 years ago

A youth group is setting up camp. Rain is predicted, so the campers decide to build a fly, or rain cover, over their tent. The f

inessss [21]

Most of the information's required for finding the answer is already given in the question.
Height of the fly = 12 feet
Width of the fly = 16 feet
Now we can find the length of the poles using Pythagoras theorem.

Length of the poles = square root [(12)^2 + (8)^2]
= square root (144 + 64)
= square root (208)
= 14.4 feet
From the above deduction, we can conclude that the height of the poles needed to be 14.4 feet. I hope the answer has helped you.

4 0

3 years ago

Imagine you have some workers and some handheld computers that you can use to take inventory at a warehouse. There are diminishi

Svet_ta [14]

Answer:

a. $1.03

b. $0.93

c. 0.98

d. 2 workers

Step-by-step explanation:

a. Given that:

1 computer : 1 worker : inventory 150 items per hour
1 computer : 2 workers : inventory 200 items per hour
1 computer : 3 workers : inventory 220 items per hour
1 computer : 4+ workers : fewer than 235 items per hour
Cost: $100 per computer ; $25 per worker

The fixed production factor in the warehouse is the computer used:

-One computer used, but the number of users is varied to inventory a specified number of items.

-The variable production factor is the number of workers assigned per one computer.

#The cost of inventorying a single item by one worker is:

$Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_1=\frac{125+30}{150}\\\\\\=1.03$

Hence, the cost of inventorying a single item is $1.03

b. Using the information provided above, the cost of inventorying a single item when two workers are assigned is :

$Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_2=\frac{125+2\times30}{200}\\\\\\=0.925$

Hence, the cost of inventorying a single item is $0.93

c.Using the information provided above, the cost of inventorying a single item when three workers are assigned is :

$Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_3=\frac{125+3\times30}{220}\\\\\\=0.98$

Hence, the cost of inventorying a single item is $0.98

d. To determine the most cost-effective job assignment, we calculate the cost of 4+ workers.

Take any number less than 235(say 234) as the inventory units:

$Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_4=\frac{125+4\times30}{234}\\\\\\=1.05$

From our calculations, it's clear that two workers per computer costs the least amount($0.93) per unit item. Hence, it is best to assign two workers per computer.

4 0

4 years ago