1 2.5n + 3.5p = 825.5

SOMEONE HELPPPPPPPPP PLEEEASSEEEEEE IMMMM STUCCCCKKKKK PLLEASSE ILL GIVE YOU 24 POINTs: On Monday, Jan made withdrawals of $25,

kolezko [41]

Just write down the negative values of each withdrawal and add up those values for the first two.
1. -25 + -45 + -75 = -145
2.-35 + -55 + -65 = -155
As for the third question, the brother took out $10 more than Julie, so he withdrew $165. A possible equation with possible values could be:
3. -15 + -65 + -85 = -165
Keep in mind that there are multiple ways to answer the third question and that's only one of them. Hope this helped!

8 0

2 years ago

Read 2 more answers

How did business leaders in the late nineteenth century utilize vertical integration? a. They maintained control of the raw mate

bulgar [2K]

Business leaders in the late nineteenth century utilized vertical integration by maintaining control of production and distribution of their products.

Answer: Option C

<u>Explanation: </u>

Vertical integration is a competitive strategy that gives the company full control over one or more stages of product production or distribution. Rockefeller tirelessly tried to take full control of business 'oil refinery'. While other business people were flooding the area in search of quick fortune, Rockefeller was thinking of destroying his rivals and creating a real monopoly in the refining industry.

Looking for even more control, Rockefeller saw the benefits of organizing the transportation to his products. Then, he began to develop his business through vertical integration, in which the company analyses all aspects of the product life cycle, from raw material extraction, through the production process, to the final delivery of the product.

Other industrialists quickly followed, including Gustavus Swift, who at the end of the 19th century used vertical integration to dominate the American meat packaging industry.

3 0

2 years ago

an open rectangular cardboard box with a square base is to have a volume of 256cm^3. Find the dimensions of the box if the area

Anit [1.1K]

Let
x------> the length side of the square base of the box
y-------> the height of the box

we know that
volume of the box=b²*h
b=x
h=y
volume=256 cm³
so
256=x²*y------>y=256/x²--------> equation 1

<span>The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.</span>

surface area of the cardboard=area of the base+perimeter of base*height
area of the base=x²
perimeter of the base=4*x
height=y
surface area=x²+4x*y-----> equation 2
substitute equation 1 in equation 2
SA=x²+4x*[256/x²]-----> SA=x²+1024/x

step 1
find the first derivative of SA and equate to zero
2x+1024*(-1)/x²=0------> 2x=1024/x²----> x³=512--------> x=8 cm
y=256/x²------> y=256/8²-----> y=4 cm

the answer is
the length side of the square base of the box is 8 cm
the height of the box is 4 cm

5 0

2 years ago

A company hauls its 1 cubic foot boxes in the truck as shown.

Alenkinab [10]

Answer:

96

Step-by-step explanation:

3*4*8=96

4 0

2 years ago

Read 2 more answers

A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa

Norma-Jean [14]

Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.

c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation

$\bar X=18.3$ represent the sample mean

$s=3.72$ represent the sample standard deviation

$n=25$ sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b: P-value and conclusion

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.

Part c

Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0

2 years ago