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3 years ago

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helPpppPPppPPPppppPppppPPppppppPPPPPPPPpppppppPPPPpppppppPPPPppppppPPPPppppppPPPPPpppppPPPPPPPPPPPPPPPPPPppppPPPPPPPppppPPPppppp

PPpp THANK YOU

1 answer:

Snezhnost [94]3 years ago

7 0

Answer:

1) rectangle

2) rectangle

Step-by-step explanation:

Any cross section of a rectangular prism is a rectangle.

<em>Hope it helps <3</em>

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If a dart board was thrown randomly at the dartboard shown below, what is the probability that it will land in the bull’s-eye? T

vampirchik [111]

Answer:

Correct option: C -> 2%

Step-by-step explanation:

To find this probability we just need to divide the area of the bull's-eye circle by the total area of the outer circle.

The area of a circle is given by the equation:

$area = \pi * radius^2$

The outer radius is 14 cm, so:

$outer\ area = \pi * 14^2 = 196\pi\ cm^2$

The inner radius is 2 cm, so:

$inner\ area = \pi * 2^2 = 4\pi\ cm^2$

So the probability is:

$inner\ area / outer\ area = 4\pi / 196\pi = 0.02 = 2\%$

Correct option: C

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3 years ago

Simplify in Exponential form <br> First correct answer gets BRAINLIEST

svlad2 [7]

Answer:

Steps to simplifying fractions

Therefore, 9/9 simplified to lowest terms is 1/1.

Reduce 9/8 to lowest terms

9/8 is already in the simplest form. It can be written as 1.125 in decimal form (rounded to 6 decimal places).Step-by-step explanation:

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3 years ago

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3 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

3 years ago

How to solve y for this problem:

True [87]

$16x+9=9y-2x\ \ |Add\ 2x\ to\ both\ sides\\\\
16x+9+2x=9y-2x+2x\\\\
18x+9=9y\ \ \ |Divide\ by\ 9\\\\
\frac{18x}{9}+\frac{9}{9}=\frac{9y}{9}\\\\
2x+1=y\\\\
Solution\ is\ y=2x+1.$

6 0

3 years ago