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3 years ago

Help solving this math

Mathematics

1 answer:

9966 [12]3 years ago

6 0

Answer:

24 cm²
6400 -1600π ≈ 1373.45 square units

Step-by-step explanation:

1. You can look at this several ways.

It can be considered to be a stack of 1+2+3 = 6 squares that are 2 cm on a side, so have a total area of 6 × (2 cm)² = 24 cm².

You can lop off one of the squares sticking out and put it into the empty corner to form a 6 cm by 4 cm rectangle with an area of (6 cm)(4 cm) = 24 cm².

You can consider it to be a 6 cm square with 3 squares cut out of it that are 2 cm on a side: (6 cm)² -3(2 cm)² = (36 -12) cm² = 24 cm².

2. The shaded area is the difference in areas of a square 80 units on a side and a circle of 40 units radius. If we let d represent the side dimension of the square, then the area formulas are ...

Square area = d²

Circle area = π(d/2)²

So the difference area is ...

shaded area = d² -(π/4)d² = d²(1 -π/4)

For d=80, the shaded area is ...

shaded area = 80²(1 -π/4) = 6400 -1600π ≈ 1373.45 square units

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

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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

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

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

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

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

= [3.4125 , 3.4155]

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

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Answer:

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Step-by-step explanation:

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