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irina1246 [14]
2 years ago
6

In a large​ city, 6060​% of people pass the​ drivers' road test. suppose that every​ day, 300300 people independently take the t

est. complete parts​ (a) through​ (d) below.
a. what is the number of people who are expected to​ pass? the expected number is 180180. ​(round to the nearest whole number as​ needed.)
b. what is the standard deviation for the number expected to​ pass? the standard deviation is nothing. ​(round to the nearest whole number as​ needed.)
c. after a great many​ days, according to the empirical​ rule, on about​ 95% of these​ days, the number of people passing will be as low as​ _____ and as high as​ _____. (hint: find two standard deviations below and two standard deviations above the​ mean.) after a great many​ days, according to the empirical​ rule, on about​ 95% of these​ days, the number of people passing will be as low as nothing and as high as nothing. ​(round to the nearest whole number as​ needed.)
d. if you found that one​ day, 134134 out of 300300 passed the​ test, would you consider this to be a very lowlow ​number? ▼ yes, no, because 134134 is ▼ less than 1 standard deviation between 1 and 2 standard deviations between 2 and 3 standard deviations more than 3 standard deviations belowbelow the mean.
Mathematics
1 answer:
erma4kov [3.2K]2 years ago
8 0
A) 180 would be expected to pass.
B) The standard deviation is 4.
C) 95% of people would fall between 172 and 188.
D) Yes, this is more than 3 standard deviations below the mean.

Explanation
A) Multiply the probability by the sample size:
0.6(300) = 180

B) Standard deviation is found by:
√n(p)(1-p)
For our data, we have:
√300(0.6)(0.4) = 4

C) Two standard deviations below the mean is 180-2(4) = 172; two standard deviations above the mean is 180+2(4)= 188.

D) Three standard deviations below the mean is 180-3(4) = 168; 134 is more than this below the mean.  99.7% of data fall within 3 standard deviations of the mean; 0.15% fall below this point, so yes, this is unusually low.
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The average sea level in 1900 at London bridge was 33 feet. In 1990 it was 33.08 feet. Use linear interpolation or extrapolation
eduard

Answer:

In 1956, the height will be 33.05 ft

Step-by-step explanation:

Let +x+ = number of years after 1900

Calll 1900 +x+=+0+

Then 1990 is +x+=+90+

Let +y+ = height in feet

--------------------------

You are gven the points

( 0, 33 )

( 90, 33.08 )

Use general point-slope formula

+%28+y+-+33+%29+%2F+%28+x+-+0+%29+=+%28+33.08+-+33+%29+%2F+%28+90+-+0+%29+

Multiply both sides by +90x+

+90%2A%28+y+-+33+%29+=+.08x+

+90y+-+2970+=+.08x+

+90y+=+.08x+%2B+2970+

+y+=+.000889x+%2B+33+ ( equation to use )

--------------------------

Check:

does it go through ( 90, 33.08 ) ?

+33.08+=+.000889%2A90+%2B+33+

+33.08+=+.08001+%2B+33+

+33.08+=+33.08001+

close enough

-------------------------------

(a)

+x+=+61%0D%0A%7B%7B%7B+y+=+.000889x+%2B+33+

+y+=+.000889%2A61+%2B+33+

+y+=+.0542+%2B+33+

+y+=+33.0542+

In 1961, the height will be 33.05 ft

8 0
2 years ago
For the month of June, Mae Green budgeted the following amounts: $180 for food, $475 for rent, $15 for transportation, $50 for i
ycow [4]
Question 1

Total budget = 180 + 475 + 15 + 50 + 65 + 25 + 150 + 30 = $990
Real spending = 182 + 475 + 12 + 65 + 68 + 12.50 + 150 + 36  = $1000.5

Mae Green didn't stay within the allocated budget

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Question 2

Eleanor:
Income = 380.48
Spending = 16.50

Peter:
Income = 120 + 13.65 + 100 = 233.65

Total income = 233.65 + 380.48 - 16.50 = 597.63

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Question 3

Total spent = 540 + 48.55 + 34.15 + 12.80 + 18.95 + 38.60 + 2 + 6.50 = 701.55

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Question 4 

Marie's new balance = 250.65 - [21.95+48.50+75.60] + 55 = $159.50

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Question 5

[235/825] × 100 = 28.48%
4 1
2 years ago
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what are the coordinates of the point (2, -3) after a counterclockwise rotation of 90 about the origin
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It should be. (-2,-3)
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2 years ago
Colin drove 45 minutes to the airport. He arrived 90 minutes before his flight departed, and then he spent 70 minutes in the air
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Answer: The numbers can be added in any order.

Step-by-step explanation:

To calculate the total time that Colin spent traveling from his house to the hotel the numbers can be added in any order because time is a scalar quantity. It cannot be negative.

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A child who is trick-or-treating chooses a lollipop at random from a bowl of lollipops with mean weight 1.5 oz and standard devi
son4ous [18]

Answer:

The expected combined weight of the child's lollipop and two chocolate bars is 5.5 oz.

Step-by-step explanation:

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.

Addition of normal variables:

The mean is the sum of the means.

Find the expected combined weight of the child's lollipop and two chocolate bars.

Child lollipop: Mean 1.5 oz

Chocolate bars: Mean 2 oz

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The mean will be 1.5 + 2*2 = 5.5 oz

The expected combined weight of the child's lollipop and two chocolate bars is 5.5 oz.

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2 years ago
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