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nignag [31]
3 years ago
5

7 kids on the basketball team return sick from a weekend playing games at Lakenheath High school with a virus. The aggressive vi

rus spreads at 130% each day. How long many days till 500 students are sick at school?
Mathematics
1 answer:
Lilit [14]3 years ago
8 0

Answer:

It would take 6 days for 500 students to be affected.

Step-by-step explanation:

Given that:

1. Seven kids returned with the virus.

2. The aggressive virus spreads at 130% each day.

We want to find how many days until 500 students are infected by the virus.

First day:

7 students have the virus, 130% of 7 student will contract the virus again.

130% of 7 = (130/100) × 7

= 1.3 × 7 = 9.1 ≈ 9

9 new students are affected

New total affected = 7 + 9 = 16 students.

Day 2:

Affected = 1.3 × 16 = 20.8 ≈ 20

New total affected = 20 + 16 = 36

Day3

Affected = 1.3 × 36 = 46.8 ≈ 46

New total affected = 36 + 46 = 82

Day4

Affected = 1.3 × 82 = 106.6 ≈ 106

New total affected = 106 + 82 = 188

Day5

Affected = 1.3 × 188 = 244.4 ≈ 244

New total affected = 244 + 188 = 432

Day6

Affected = 1.3 × 432 = 561.6 ≈ 561

New total affected = 561 + 432 = 993

Therefore, it would take 6 days for 500 students to be affected.

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Step-by-step explanation:

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A sample of 49 observations is taken from a normal population with a standard deviation of 10. the sample mean is 55. determine
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10a. The y-intercept of the estimated line of best fit is at (0,b). Enter the approximate value of b. Round your estimate to the
vredina [299]

Answer:

The answer is below

Step-by-step explanation:

The equation of a straight line is given as:

y = mx + b; where m is the slope of the line and b is the y intercept (that is value of y at x = 0).

a) From the graph, we can see that the straight line touches the y axis at $100 in which the temperature is 0 degrees. Therefore the y intercept is:

(0, 100)

b) The slope of the line can be gotten by taking two points that the line passes through. Using the points (0, 100) and (12, 500):

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4 0
3 years ago
A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using 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.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

s = \frac{\sigma}{\sqrt{n}}

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean

3 0
3 years ago
I have a ribbon 5/7 of a foot. I want to cut it in 5/12 pieces. How many pieces can I cut 5/7 divided by 5/12= 5/7 x5/12= 15/7 p
Otrada [13]
100 cm = 1 meter 
<span>4.5 meter x 100 cm = 450 cm </span>

<span>450 cm / 4.5 cm = number of pieces of ribbon </span>

<span>450/4.5 = 100</span>
5 0
3 years ago
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