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3 years ago

Put the real numbers in order from greatest to least

Mathematics

1 answer:

kozerog [31]3 years ago

8 0

Reverse the order they are in now, and you have your answer

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On a field trip, there are 8 adults and 24 students. What is the ratio of students to the total number of people on the field tr

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24:32 is the ratio. Just add up he total number and then put the total number of student over the total number of people

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3 years ago

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Calculate the next three terms of the following sequence.

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A is the correct answer

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Use the "rule of 72" to estimate the doubling time (in years) for the interest rate, and then calculate it exactly. (Round your

Law Incorporation [45]

Answer:

According to the rule of 72, the doubling time for this interest rate is 8 years.

The exact doubling time of this amount is 8.04 years.

Step-by-step explanation:

Sometimes, the compound interest formula is quite complex to be solved, so the result can be estimated by the rule of 72.

By the rule of 72, we have that the doubling time D is given by:

$D = \frac{72}{Interest Rate}$

The interest rate is in %.

In our exercise, the interest rate is 9%. So, by the rule of 72:

$D = \frac{72}{9} = 8$ .

According to the rule of 72, the doubling time for this interest rate is 8 years.

Exact answer:

The exact answer is going to be found using the compound interest formula.

$A = P(1 + \frac{r}{n})^{nt}$

In which A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

So, for this exercise, we have:

We want to find the doubling time, that is, the time in which the amount is double the initial amount, double the principal.

is double the initial amount, double the principal.

$A = 2P$

$r = 0.09$

The interest is compounded anually, so $n = 1$

$A = P(1 + \frac{r}{n})^{nt}$

$2P = P(1 + \frac{0.09}{1})^{t}$

$2 = (1.09)^{t}$

Now, we apply the following log propriety:

$\log_{a} a^{n} = n$

So:

$\log_{1.09}(1.09)^{t} = \log_{1.09} 2$

$t = 8.04$

The exact doubling time of this amount is 8.04 years.

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3 years ago

The bottom of the ladder must be 3 feet from a wall the latter is 12feet long how far above the ground does the Ladder touch the

Oksanka [162]

Question :
The bottom of the ladder must be 3 feet from a wall the latter is 12feet long how far above the ground does the Ladder touch the wall

Answer :
11.6ft

I hope it helps and please press brain
(^_-) Rie

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3 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$