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katovenus [111]

3 years ago

6

A child is sitting in a tree 8 ft off the ground. A raccoon is sighted 6.5 ft away from the base of the tree. What is the angle

of depression from where the child is sitting to the raccoon? Round to the nearest hundredth.

2 answers:

Katen [24]3 years ago

7 0

)

<span>tanx = 3/10 </span>
<span>x=16.7 B </span>

<span>2) </span>

<span>tan 64 = 173/x </span>

<span>x = 84.38 D </span>

<span>3) </span>

<span>tan21 = x/442 </span>
<span>x= 169.67 B </span>

<span>4) </span>

<span>tanx = 16.2/48.3 </span>
<span>x = 18.54 A </span>

<span>5) </span>

<span>tanx = 8/6.5 </span>
<span>x = 50.91 C </span>

<span>6) </span>
<span>tan89 = 1149/x </span>
<span>x=20.06 A </span>

<span>7) </span>
<span>tanx = 21/12 </span>
<span>x= 60.26 B </span>

<span>8) B </span>

<span>9) </span>
<span>tanx = 25/63 </span>
<span>x = 21.64 C </span>

<span>10) </span>
<span>tanx = 60/15 </span>
<span>x= 75.96 D</span>

Vitek1552 [10]3 years ago

6 0

Tanx = 8/6.5
<span>x = 50.91 </span>

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If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal

lorasvet [3.4K]

Answer:

a

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

b

The probability is $P(Z > 2.5 ) = 0.0062097$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 800$

The variance is $var(x) = 1600 \ kg$

The range consider is $x_1 = 778 \ kg \ x_2 = 834 \ kg$

The value consider in second question is $x = 900 \ kg$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{var (x)}$

substituting value

$\sigma = \sqrt{1600}$

$\sigma = 40$

The percentage of a cucumber give the crop amount between 778 and 834 kg is mathematically represented as

$P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )$

Generally $\frac{X - \mu }{ \sigma } = Z (standardized \ value \ of \ X)$

So

$P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )$

$P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)$

From the z-table the value for $P(z_1 < 0.85) = 0.80234$

and $P(z_1 < -0.55) = 0.29116$

So

$P(x_1 < X < x_2 ) = 0.80234 - 0.29116$

$P(x_1 < X < x_2 ) = 0.51118$

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

$P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )$

substituting values

$P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )$

$P(X > x ) = P(Z >2.5 )$

From the z-table the value for $P(Z > 2.5 ) = 0.0062097$

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