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mote1985 [20]
3 years ago
14

Please solve with explanation I’ve been asking all day (this is not a multiple choice question)

Mathematics
1 answer:
Anastaziya [24]3 years ago
3 0

Answer:

a) SA = 522.9~cm^2

b) V_{cone} = 670.2~cm^3

c) V_{empty} = 1340.4~cm^3

Step-by-step explanation:

a)

For a cone,

SA = \pi r (L + r)

where L = slant height

L = \sqrt{r^2 + h^2}

We have r = 8 cm; h = 10 cm

L = \sqrt{(8~cm)^2 + (10~cm)^2}

L = \sqrt{164~cm^2}

SA = (\pi)(8~cm)(\sqrt{164~cm^2} + 8~cm)

SA = 522.9~cm^2

b)

V_{cone} = \dfrac{1}{3}\pi r^2 h

V_{cone} = \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)

V_{cone} = 670.2~cm^3

c)

V_{cylinder} = \pi r^2 h

empty space = volume of cylinder - volume of cone

V_{empty} = V_{cylinder} - V_{cone}

V_{empty} = \pi r^2 h - \dfrac{1}{3}\pi r^2 h

V_{empty} = (\pi)(8~cm)^2(10~cm) - \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)

V_{empty} = 1340.4~cm^3

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Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = <u><em>thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.</em></u>

So, X ~ Normal(\mu=5.0,\sigma^{2} =1.7^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean thickness = 5.0 mm

           \sigma = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

    P(X < 3.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{3.0-5.0}{1.7} ) = P(Z < -1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

    P(X > 7.0 mm) = P( \frac{X-\mu}{\sigma} > \frac{7.0-5.0}{1.7} ) = P(Z > 1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X \leq 3.0 mm)

    P(X < 7.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{7.0-5.0}{1.7} ) = P(Z < 1.18) = 0.881

    P(X \leq 3.0 mm) = P( \frac{X-\mu}{\sigma} \leq \frac{3.0-5.0}{1.7} ) = P(Z \leq -1.18) = 1 - P(Z < 1.18)

                                                           = 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

4 0
4 years ago
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