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lozanna [386]
2 years ago
15

A physical fitness association is including the mile run in its secondary school fitness test. The time for this event for boys

in secondary school is known to possess a normal distribution with a mean of 440 seconds and a standard deviation of 40 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 348 seconds.A) 0.0107 B) 0.9893 C) 0.5107 D) 0.4893
Mathematics
1 answer:
Ierofanga [76]2 years ago
6 0

Answer:

A) 0.0107

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 440 seconds and a standard deviation of 40 seconds.

This means that \mu = 440, \sigma = 40

Find the probability that a randomly selected boy in secondary school can run the mile in less than 348 seconds.

This is the p-value of Z when X = 348. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{348 - 440}{40}

Z = -2.3

Z = -2.3 has a p-value of 0.0107, and thus, the correct answer is given by option A.

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The surface (call it S) is a triangle with vertices at the points

x=0,y=0\implies z=2\implies(0,0,2)

x=0,y=2\implies z=-2\implies(0,2,-2)

x=2,y=0\implies z=-8\implies(2,0,-8)

Parameterize S by

\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)

with 0\le u\le1 and 0\le v\le1. Take the normal vector to S to be

\vec s_v\times\vec s_u=(20v,8v,4v)

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv

=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}

8 0
3 years ago
2 - 2n = 3n + 17 solve n
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Answer:

n = -3

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-3n    -3n

<u>-5n = 15</u>

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Hope this helps you!!! :)

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3 years ago
Read 2 more answers
!!Plz urgent help!!
Firdavs [7]

Answer:

890 beads can be fitted in the triangular prism.

Step-by-step explanation:

If we can fill the spherical beads completely in the triangular prism,

Volume of the triangular prism = Volume of the spherical beads

Volume of triangular prism = Area of the triangular base × Height

From the picture attached,

Area of the triangular base = \frac{1}{2}(\text{Base})(\text{Height})

                                             = \frac{1}{2}(AB)(BC)

By applying Pythagoras theorem in the given triangle,

AC² = AB² + BC²

(13)² = 5² + BC²

169 = 25 + BC²

BC² = 144

BC = 12

Area of the triangular base = \frac{1}{2}(5)(12)

                                             = 30 cm²

Height of the triangular prism = 18 cm

Volume of the triangular prism = 30 × 18

                                                   = 540 cm³

Volume of one spherical bead = \frac{4}{3}\pi r^{3}

                                                   = \frac{4}{3}\pi (0.525)^{3}

                                                   = 0.606 cm³

Let there are 'n' beads in the triangular prism,

Volume of 'n' beads = Volume of the prism

540 = 0.606n

n = 890.90

n ≈ 890

Therefore, 890 beads can be fitted in the triangular prism.

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2 years ago
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