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4 years ago

1.

Mathematics

1 answer:

Debora [2.8K]4 years ago

8 0

LIFO method

Units Available for sale = 25 + 20 + 30 + 15 = 90 units

Ending Inventory = 24 units

Units Sold = (90 – 24) = 66 units

Cost of Goods Sold

Sales from September = 15 units x $45 = $675

Sales from June = 30 units x $38 = $1140

Sales from March = 20 units x $42 = $900

Sales from Beginning Inventory = 1 unit x $25 = $25

Total: 66 units = $2740

Ending Inventory from beginning inventory = 24 units x 35 = $840

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Find f(4) when f(x) = 5x^2 - 5x + 7. A) 107 B) 53 C) 67 D) 3

fgiga [73]

I think the answer is 67

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4 years ago

Which reason best describes why you can divide any number by 1,000

dezoksy [38]

Answer:

1 one

Step-by-step explanation:

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3 years ago

A number is chosen at random from 1 to 10. Find the probability of not selecting

PilotLPTM [1.2K]

Answer:

0.3

Step-by-step explanation:

Number of ways of selecting r elements from a set of n different elements is given by

C(n,r)=(n r)=n!/(r!(n−r)!)

No of ways of selecting one number out of ten numbers from one to 10 is 10

It can also be calculated using

C(n,r)=(n r)=n!/(r!(n−r)!)

where n = 10 and r = 1

C(10,1)=(10 1)=10!(1!(10−1)!) = 10*9!/ 1!*9! = 10

multiples of 2 in range 1 to 10 are 2, 4, 6, 8 , 10

multiples of 3 in range 1 to 10 are 3, 6, 9

Therefore number which are multiple of 2 and 3 are

2, 3,4,6,8,9,10 ( 7 numbers)

therefore no of ways of selecting multiple of 2 and 3 is 7 ways

number which are not multiple of 2 and 3 in range 1 to 10

1,5,7 (3 numbers)

no of ways of not selecting multiple of 2 and 3 is 3 ways

It can also be calculated using

C(3,1)=(3 1)=3!/(1!(3−1)!)

where n = 3 and r = 1

C(3,1)=(3 1)=3!/(1!(3−1)!) = 3!/(1!(2)!)= 3*2!/ 1!*2! = 2

Therefore no of ways of not selecting multiple of 2 and 3 is 3 ways

probability of not selecting a multiple of 2 or a multiple of 3 = no of ways of not selecting multiple of 2 and 3 /No of ways of selecting one number out of one to 10 = 3/10 = 0.3

4 0

3 years ago

We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.

Juli2301 [7.4K]

Answer:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: $p < 0.8$

Since is a left tailed test the p value would be:

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

$\hat p=\frac{97}{110}=0.882$ estimated proportion of students who completed the modules before class

$p_o=0.8$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%: Null hypothesis:[tex]p\geq 0.8$

Alternative hypothesis: $p < 0.8$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z$

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

5 0

4 years ago

2+4447=8ii what is ii

Anvisha [2.4K]

Answer:

$ii = 556.125$

Step-by-step explanation:

$4449 = 8ii$

divide both side by 8

$4449 \div 8 = 8ii \div 8$

$ii = 556.125$

8 0

3 years ago