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Lera25 [3.4K]
2 years ago
9

How much change will MRS Sloan receive....

Mathematics
1 answer:
NISA [10]2 years ago
6 0
15. 75 s̶i̶n̶c̶e̶ 20.00 p̶o̶u̶n̶d̶s̶ - 4a̶n̶d̶ 25 p̶o̶u̶n̶d̶s̶ e̶q̶u̶a̶l̶ 15.75 p̶o̶u̶n̶d̶s̶
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What is the least common multiple of 8 and 11
tino4ka555 [31]

Answer:

88

Step-by-step explanation:

Find the prime factorization of both 8 and 11.

8 → 4 × 2 → 2 × 2 × 2

11 → 11

Then, multiply each factor the greater number of times it occurs. Since no factors are duplicated between 8 and 11 (as in, 2 is not a factor of 11, nor is 11 a factor of 8), LCM will equal 2 × 2 × 2 × 11.

2 × 2 × 2 × 11 = 88, so the LCM is 88.

It's a little hard to explain without visuals. A video could be useful, though I can't put links here.

7 0
3 years ago
Suppose a tank contains 400 gallons of salt water. If pure water flows into the tank at the rate of 7 gallons per minute and the
Strike441 [17]

Answer:

Step-by-step explanation:

This is a differential equation problem most easily solved with an exponential decay equation of the form

y=Ce^{kt}. We know that the initial amount of salt in the tank is 28 pounds, so

C = 28. Now we just need to find k.

The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is \frac{dy}{dt}. Thus, the change in the concentration of salt is found in

\frac{dy}{dt}= inflow of salt - outflow of salt

Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

3(\frac{y}{400})

Therefore,

\frac{dy}{dt}=0-3(\frac{y}{400}) or just

\frac{dy}{dt}=-\frac{3y}{400} and in terms of time,

-\frac{3t}{400}

Thus, our equation is

y=28e^{-\frac{3t}{400} and filling in 16 for the number of minutes in t:

y = 24.834 pounds of salt

6 0
2 years ago
What divides into 78?? Thanks for your help if your trying to solve it <3
Mademuasel [1]
6 divides into 78 which gives a quotient of 13.
Hope i'm the brainliest!
5 0
2 years ago
A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat
zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

Z = \frac{X - \mu}{s}

X = 205

Z = \frac{X - \mu}{s}

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

Z = \frac{X - \mu}{s}

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

Z = \frac{X - \mu}{s}

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

Z = \frac{X - \mu}{s}

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0
3 years ago
Lamar is considering two loans.
Varvara68 [4.7K]

Answer:

a.

Loan I's increase was 0.03 percentage points greater than Loan H's

8 0
3 years ago
Read 2 more answers
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