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muminat
3 years ago
15

1.8.2

Mathematics
1 answer:
VLD [36.1K]3 years ago
3 0

Answer:

5,

4.66

4

-0.97

-3/2

-20

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Which value is equivalent to 7 multiplied by 5 multiplied by 2 whole over 7 multiplied by 3, the whole raised to the power of 2
marissa [1.9K]
This is a little difficult but it might be 14,000,000
7 0
3 years ago
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A two-factor study with two levels of factor A and three levels of factor B uses a separate group of n = 5 participants in each
Mashcka [7]

Answer: There are 30 participants are needed for the entire study.

Step-by-step explanation:

Since we have given that

Number of levels of factor A = 2

Number of levels of factor B = 3

Number of participants in each treatment condition = 5

So, the number of participants are needed for the entire study is given by

2\times 3\times 5\\\\=6\times 5\\\\=30

Hence, there are 30 participants are needed for the entire study.

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3 years ago
6x+5y=-15 into intercept form
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General\ equation\ for\ line\ in\ slope\ intercept\ form:\\\\y=ax+b\\\\6x+5y=-15\\
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y=-3-\frac{6}{5}x\\\\
In\ slope\ intercept\ form\ y=-3-\frac{6}{5}x.
5 0
3 years ago
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If the volume of the pyramid shown is 108 inches cubed, what is the area of its base? A pyramid with a height of 8 inches. 4 inc
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Answer:

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Step-by-step explanation:

7 0
3 years ago
A lifeguard is in an observation chair and spots a person who needs help. The angle of depression to the person is 22°. The eye
balu736 [363]

Answer:

The horizontal distance between the lifeguard and the person is approximately 24.751 feet.

Step-by-step explanation:

At first we include a geometrical representation of the statement, which is shown in the image attached below. We can determine the horizontal distance between the lifeguard and the person by means of the following trigonometrical relationship:

\tan \theta = \frac{OL}{OP} (1)

Where:

\theta - Angle of depression to the person, measured in sexagesimal degrees.

OL - Height of the lifeguard above the ground, measured in feet.

OP - Horizontal distance between the lifeguard and the person, measured in feet.

If we know that OL = 10\,ft and \theta = 22^{\circ}, then the horizontal distance between the lifeguard and the person is:

OP = \frac{OL}{\tan \theta}

OP = \frac{10\,ft}{\tan 22^{\circ}}

OP \approx 24.751\,ft

The horizontal distance between the lifeguard and the person is approximately 24.751 feet.

4 0
3 years ago
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