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2 years ago

NEED HELP ASAP

1 answer:

8 0

Starts out as 27,000 cm^3. then after one hour shrinks to 21,952 cm^3 and then 17,576cm^3 and so on. The ice is losing volume exponentially, there is no set rate. However it seems that you have said yourself it is losing volume at 2cm^3 per hour.

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Help me plzzzzzzzzzzzzzzzz

irina [24]

Answer:

Step-by-step explanation:

4 0

2 years ago

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Solve for b A= \frac{a+b}{2}

lina2011 [118]

A = (a+b)/2
2*A = 2*(a+b)/2 ... multiply both sides by 2
2A = a+b
2A-a = a+b-a ... subtract 'a' from both sides to isolate b
2A-a = b
b = 2A-a

Answer: b = 2A-a

6 0

3 years ago

Three students earned $48.76 at the bake sell. The students split the earnings evenly, how much did each student recieve?

Sveta_85 [38]

Answer

Find out the how much did each student recieve .

To prove

Let us assume that each student earned in the bake sell be x .

As given

Three students earned $48.76 at the bake sell.

The students split the earnings evenly i.e earning is divided into three equal parts .

Than the equation becomes

$x = \frac{48.76}{3}$

Solving the above

x = $16.25(approx)

Therefore the each student recieve $16.25(approx) .

Hence proved

4 0

3 years ago

Read 2 more answers

What is the sum of the first 50 terms of the series 2 + 17 + 32 + 47+ ...?

Alex

Answer:

$\huge\boxed{18,475}$

Step-by-step explanation:

The terms of this sum make the arithmetic sequence.

The fomula of a sum of <em>n</em> terms of an arithmetic sequence:

$S_n=\dfrac{[2a_1+(n-1)d]\cdot n}{2}$

We have

$a_1=2\\d=17-2=32-12=47-32=15\\n=50$

Substitute:

$S_{50}=\dfrac{[2\cdot2+(50-1)\cdot15]\cdot50}{2}=(4+49\cdot15)\cdot25=(4+735)\cdot25\\\\=739\cdot25=18,475$

3 0

2 years ago

The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o

frosja888 [35]

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 36, \sigma = 5$

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40 - 36}{5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

8 0

2 years ago