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3 years ago

Guys help me what is the answer?

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

7 0

Step-by-step explanation:

C. represent the number line for x

where x ix

3x<u>></u>6
x<u>></u><u> </u>6/3
x<u>></u>2

<h2>hope it helps.</h2><h2>stay safe healthy and happy.</h2>

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Lisa Company uses the periodic inventory system to account for inventories. Information related to Lisa Company's inventory at O

scZoUnD [109]

Answer:

The value of ending inventory will be items of latest purchase.

Step-by-step explanation:

Given that,

Lisa Company uses the periodic inventory system to account for inventories.

Information related to Lisa Company's inventory at October 31 is given,

Suppose, find the value of ending inventory using the FIFO cost assumption if 500 units remains on hand at october 31

We need to calculate the value of ending inventory

Using FIFO method

$value\ of\ ending\ inventory =305\ units\times\$10.6+195\ units\times\$11.60$

$value\ of\ ending\ inventory = 3233+2262$

$value\ of\ ending\ inventory =\$5495$

Hence, The value of ending inventory will be items of latest purchase.

6 0

3 years ago

What is the common difference of the arithmetic sequence 1/5, 6/5 , 11/5 ,16/5 , ...?

stepladder [879]

+ 5/5 each time

hope this helps

8 0

3 years ago

Read 2 more answers

Molly and Ryan both need to answer this question on their test in math class. 34×45=? They both got 12/20 as their answer but do

Neporo4naja [7]

It would be 3/5 because if you simplify it you would bet 3/5

4 0

3 years ago

Describe the transformations of f(x) = x3 represented by g(x) = (x - 6) - 2.

devlian [24]

Answer:

Step-by-step explanation:

4 0

3 years ago

A fair coin is flipped twelve times. What is the probability of the coin landing tails up exactly nine times?

seraphim [82]

Answer:

$P\left(E\right)=\frac{55}{1024}$

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

2¹² = 4096

We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

$nCr=\frac{n!}{r!\left(n-r\right)!}$

Plug in n = 12 and r = 9

$=\frac{12!}{9!\left(12-9\right)!}$

$=\frac{12!}{9!\cdot \:3!}$

$=\frac{12\cdot \:11\cdot \:10}{3!}$ ∵ $\frac{12!}{9!}=12\cdot \:11\cdot \:10$

$=\frac{1320}{6}$ ∵ $3!\:=\:3\times 2\times 1=6$

$=220$

Thus, the probability will be:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$=\frac{220}{4096}$

$=\frac{55}{1024}$

Thus, the probability of the coin landing tails up exactly nine times will be:

$P\left(E\right)=\frac{55}{1024}$

4 0

3 years ago