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3 years ago

N his history class, Han's homework scores are: 100% 100% 100% 100% 95% 100% 90% 100% 0% What is the mode of Han's scores?

Mathematics

1 answer:

Basile [38]3 years ago

3 0

Answer:

There is no mode

Step-by-step explanation:

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The solution to the equation x3 = 45

Darina [25.2K]

Answer:
x=15

Because 3x15=45

Hope that helped!

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2 years ago

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Solve the system of linear equations. <br>- 3x + 5y = 13 <br> x + 4y = -10

Molodets [167]

Answer:

(-6,-1)

Step-by-step explanation:

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3 years ago

An animal shelter provides a bowl with 1.45 liters of water for 3 cats.

Kay [80]

Answer:

n animal shelter provides a bowl with 1.45 liters of water for 3 cats.

About how much water will be left after the cats drink their average daily amount of water?

Canada Goose 0.24

Cat 0.15

Mink 0.10

Opossum 0.30

Bald Eagle 0.16

Step-by-step explanation:

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3 years ago

A homeowner puts a patio in the northwest corner of a square backyard. The area of the patios is 750

miskamm [114]

Answer:

17.6621ft²

Step-by-step explanation:

First we need to find the side length of the patio x

SInce the area of the patio = 750

Area of patio = x * x

Area of patio = x²

750 = x²

Swap

x² = 750

x = √750

x = 27.39ft

Length of the backyard = 27.39ft - 25ft = 2.39ft

Width of the backyard = 27.39ft - 20ft = 7.39ft

Area of the backyard = 2.39 * 7.39

Area of the backyard = 17.6621ft²

7 0

3 years ago

The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas

Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is $= \frac{(n+1)}{2}$ .

$\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5$ position orders

$10^{th} \ and \ 11^{th}$ place an average of observations so, the

$Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430$

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

$Q_1$ often falls throughout the ordered BELOW average is $= \frac{(n+1)}{2}$ place.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered position

Average $5^{th} \ and \ 6^{th}$ location findings

$Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5$

In $= \frac{(n+1)}{2}$ the ordered place, Q3 always falls ABOVE the median.

$\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}$ ordered At median ABOVE

Consequently $Q_3$ falls between $5^{th} \ and \ 6^{th}$ ABOVE the median position

$15^{th}\ and \ 16^{th}$ place average of observations

$\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2} = 502$

$\to IQR = Q_3 - Q_1= 502-358.5= 143.5$

$\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25$

4 0

2 years ago