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3 years ago

Find the median, mode, and range for

1 answer:

7 0

To find the median we put the numbers from least to greatest, and cross out the smallest number with the biggest number, until we can't any longer.
54/96, 55/95, 60/89, 66/88, 69/83, 72/82, 73/75, 74/75.
Since we ended with 74 and 75, pick the decimal number that's in between those two numbers, 74.5.
74.5 is your median.
The mode is the number that's most frequent.
There are two 75's, so 75 is your mode.
Range is the difference of the highest number and the lowest number.
96-54=42.
The range is 42.

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Solve 10/12 - 3/8 show your solution as an equation

elena55 [62]

Answer:

11/24

Step-by-step explanation:

The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator. Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1. The two fractions now have like denominators so you can subtract the numerators.

This fraction cannot be reduced.

Therefore the answer is 11/24

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3 years ago

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Anne has three more dollars than Freddy. Together they have $99. How many dollars does each have?

Tamiku [17]

Anne has 51. Freddy has 48.

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3 years ago

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Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

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3 years ago

Is 5 1/2 yds greater than 192 ins?

iren2701 [21]

Answer:

Step-by-step explanation:

there is 12 inches in a foot

there are 3 feet in a yard

3 x 12 = 36 in in a yard

36 x 5 1/2 is 198

198 is greater than 192

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3 years ago

A researcher interested in Springfield citizens' shopping habits surveys a randomly selected group of 200 Walmart shoppers. 76%

luda_lava [24]

Answer and Explanation: The results of the survey would not be accurate to the whole of springfield, as the sample was not a random sample. A random sample is where each person in a given area (Springfield in this case) has an equal chance of being surveyed. In this case, only walmart shoppers were surveyed. Therefor, the results can only apply to shoppers at walmart, not the whole of Springfield.

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3 years ago

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