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2 years ago

14

The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives 22.4 years; the standard deviation

is 2.7 years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than 14.31 years.

2 answers:

oksano4ka [1.4K]2 years ago

7 0

P(27.8 < C < 30.5) = P(X>27.8) - P(X> 30.5) =0.025 - 0.0015 = 0.0235 hope this helps:)

liq [111]2 years ago

5 0

Answer:

ok

Step-by-step explanation:

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A community swimming pool is a rectangular prism that is 30 feet long, 12 feet wide, and 5 feet deep. The wading pool is half as

topjm [15]

Answer:

The volume of the community swimming pool is 4 times greaters than the volume of the wading pool.

Step-by-step explanation:

By definition of rectangular prism, we get the respective formulas for the volumes of the community swimming pool and the wadling pool, respectively:

Community swimming pool

$V_{c} = l\cdot w\cdot h (1)$

Wading pool

$V_{w} = \left(\frac{1}{2}\cdot l \right)\cdot \left(\frac{1}{2}\cdot h\right)\cdot w (2)$

Where:

l – Length of the swimming pool, measured in feet.

h – Depth of the swimming pool, measured in feet.

w – Width of the swimming pool, measured in feet.

$V_{c}$ – Volume of the community swimming pool, measured in cubic feet.

$V_{w}$ – Volume of the wading swimming pool, measured in cubic feet.

The ratio of the volume of the community swimming pool to the volume of the wadling pool is:

$\frac{V_{c}}{V_{w}} = \frac{l\cdot w \cdot h}{\left(\frac{1}{2}\cdot l \right)\cdot \left(\frac{1}{2}\cdot h\right)\cdot w} (3)$

$\frac{V_{c}}{V_{w}} = \frac{1}{\frac{1}{4} }$

$\frac{V_{c}}{V_{w}} = 4$

The volume of the community swimming pool is 4 times greaters than the volume of the wading pool.

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1 year ago

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Points D, E, and F are on circle C and EF ≅ DF. What is the measure of ∠CDF?

antiseptic1488 [7]

The answer is

26 degrees

I just did the test hope this helps :)

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2 years ago

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How to Graph of y=1/2x-4

Advocard [28]

Answer and explanation:

We know that this is in slope-intercept form. ( y = mx + b )

1/2 would be the slope (you would use this to graph the next point)

-4 would by your y-intercept (where the line crosses the x-axis)

With this information you could graph any point using the slope given, starting at -4 (x-axis) going down or up.

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2 years ago

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For what values of k the relation R:{(k2 , 5), (3k, 6)} is not a function?

Nastasia [14]

Answer:

$k=0$ and $k=3$ .

Step-by-step explanation:

To this relation be a function, the horizontal coordinates can't be equal. So, let's find the value of $k$ that makes those coordinates equal.

$k^{2}=3k\\ k^{2}-3k=0\\k(k-3)=0\\$

Using the null factor property, we have

$k=0\\k-3=0 \implies k=3$

Therefore, the given relation is not a function when $k=0$ and $k=3$ .

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2 years ago

A bucket that weighs 4 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f

jonny [76]

Answer:

2580 ft-lb

Step-by-step explanation:

Water leaks out of the bucket at a rate of $\frac{0.15 \mathrm{lb} / \mathrm{s}}{1.5 \mathrm{ft} / \mathrm{s}}=0.1 \mathrm{lb} / \mathrm{ft}$

Work done required to pull the bucket to the top of the well is given by integral

$W=\int_{a}^{b} F(x) dx$

Here, function $F(x)$ is the total weight of the bucket and water $x$ feet above the bottom of the well. That is,

$F(x)=4+(42-0.1 x)$

$=46-0.1x$

$a$ is the initial height and $b$ is the maximum height of well. That is,

$a=0 \text { and } b=60$

Find the work done as,

$W=\int_{a}^{b} F(x) d x$

$=\int_{0}^{60}(46-0.1 x) dx$

$&\left.=46x-0.05 x^{2}\right]_{0}^{60}$

$=(2760-180)-0[$

$=2580 \mathrm{ft}-\mathrm{lb}
$

Hence, the work done required to pull the bucket to the top of the well is $2580 \mathrm{ft}- \mathrm{lb}$

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2 years ago