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2 years ago

A box containing 90 tiles weighs 23.4 lbs. How much would 15 tiles weigh?

Mathematics

1 answer:

andrew-mc [135]2 years ago

7 0

Answer:

3.9

x/15 = 23.4/90

15 X 23.4= 351

351/90=3.9

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Rewrite the following without an exponent:<br><br> Two to the negative fourth power.

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Answer:

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Victor Malala has a net income of $1240.00 per month. If he spends $150.00 on food, $244.00 on a car payment, $300.00 on rent, a

Nady [450]

Net Income: I=$1,240.00
Food: F=$150.00
Car payment: C=$244.00
Rent: R=$300.00
Saving: S=$50.00
Other things: O=?

F+C+R+S+O=I
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$744.00+O=$1240.00
$744.00+O-$744.00=$1240.00-$744.00
O=$496.00

Percent of his net income Victor can spend on other things: P=?
P=(O/I)*100%
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2 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

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Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

2 years ago