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Kay [80]
3 years ago
15

In the number 15.2201 how does the value of 2 in the tenths place compare to the value of the 2 in the hundredths place

Mathematics
1 answer:
lakkis [162]3 years ago
6 0

Answer:

The 2 in the hundreths is 10% of the 2 in the tenths

Step-by-step explanation:

We are given the number 15.2201

#The value of 2 in the tenths:

0.2

#The value of 2 in the hundredths:

0.02

The difference of the 2 in the tenths and in the hundredths is:

=2_t-2_h\\\\=0.2-0.02\\\\=0.18\\\\\#2_h \ as \ a \ \% \ of \ 2_t\\=\frac{2_h}{2_t}\times 100\%\\\\=\frac{0.02}{0.2}\times100\%\\\\2_h=10\% \ of\  2_t

Hence, the difference between the 2 in the tenths  minus 2 in the hundreths is 0.18, 2 in the hundreths is 10% the 2 in the tenths.

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SAT verbal scores are normally distributed with a mean of 433 and a standard deviation of 90. Use the Empirical Rule to determin
laila [671]

34% of the scores lie between 433 and 523.

Solution:

Given data:

Mean (μ) = 433

Standard deviation (σ) = 90

<u>Empirical rule to determine the percent:</u>

(1) About 68% of all the values lie within 1 standard deviation of the mean.

(2) About 95% of all the values lie within 2 standard deviations of the mean.

(3) About 99.7% of all the values lie within 3 standard deviations of the mean.

$Z(X)=\frac{x-\mu}{\sigma}

$Z(433)=\frac{433-\ 433}{90}=0

$Z(523)=\frac{523-\ 433}{90}=1

Z lies between o and 1.

P(433 < x < 523) = P(0 < Z < 1)

μ = 433 and μ + σ = 433 + 90 = 523

Using empirical rule, about 68% of all the values lie within 1 standard deviation of the mean.

i. e. ((\mu-\sigma) \ \text{to} \ (\mu+\sigma))=68\%

Here μ to μ + σ = \frac{68\%}{2} =34\%

Hence 34% of the scores lie between 433 and 523.

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A jar contains 8 large red marbles, 5 small red marbles, 9 large blue marbles, and 7 small blue marbles. If a marble is chosen a
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A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses
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Answer:

The original duration of the tour = 20 days

Step-by-step explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for x days.

So, for x days the total expense = $360

<em>Thus the daily expense in dollars can be given by</em> = \frac{360}{x}

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

<em>Tour duration now</em> = (x+4) days

On extension, his daily expense is cut by $3

<em>New daily expense in dollars </em>= (\frac{360}{x}-3)

Total expense in dollars can now be given as:  (x+4)(\frac{360}{x}-3)

Simplifying by using distribution (FOIL).

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So, we have the equation as:

348-3x+\frac{1440}{x}=360

Multiplying each term with x to remove fractions.

348x-3x^2+1440=360x

Subtracting 348x both sides

348x-348x-3x^2+1440=360x-348x

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Dividing each term with -3.

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Adding 4x both sides.

x^2+4x-480=-4x+4x

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x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Plugging in values from the equation we got.

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