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2 years ago

Tree in Oakland has a mass of approximately 3 x 10 kilograms. A tree in Mapleville has a m

Mathematics

1 answer:

soldier1979 [14.2K]2 years ago

6 0

Honestly don’t know but when in doubt pick c

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2. A highway makes an angle of 6 with the horizontal. This angle is maintained for a horizontal distance of 5 miles. To the near

hoa [83]

Given:
angle of 6 : i'm assuming this is 6 degrees.
distance of 5 miles.

tan(6) = rise/5 mi
rise = 5 mi * tan(6)
rise = 5 mi * 0.10510423526
rise = 0.52552117632

rise = 0.53 miles.

7 0

3 years ago

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

7 0

2 years ago

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$

Volume of the box at x= 1.59:

$V = \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\$

Volume of the box at x= 0.53:

$V = \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3$

The volume of the box is maximized when x = 0.53 cm

Therefore,

x = 0.53 cm

Maximum volume = 1.75 cm³

7 0

2 years ago

Draw a line through each point using the given slope. What do you notice about the two lines?

morpeh [17]

I think the answer is that they are parallel

5 0

3 years ago

three tanks are capable of holding 108 litres, 168 litres and 180 litres of milk. determine the capacity of the greatest vessel

SpyIntel [72]

A vessel of 7,560 liters

Step-by-step explanation:

This means finding the Least Common Multiple for;

108 168 180

We begin by finding the multiples of each number;

Multiples of 108:

108, 216, 324, 432, 540, 648, 756, 864, 972, 1080, 1188, 1296, 1404, 1512, 1620, 1728, 1836, 1944, 2052, 2160, 2268, 2376, 2484, 2592, 2700, 2808, 2916, 3024, 3132, 3240, 3348, 3456, 3564, 3672, 3780, 3888, 3996, 4104, 4212, 4320, 4428, 4536, 4644, 4752, 4860, 4968, 5076, 5184, 5292, 5400, 5508, 5616, 5724, 5832, 5940, 6048, 6156, 6264, 6372, 6480, 6588, 6696, 6804, 6912, 7020, 7128, 7236, 7344, 7452, 7560, 7668, 7776

Multiples of 168:

168, 336, 504, 672, 840, 1008, 1176, 1344, 1512, 1680, 1848, 2016, 2184, 2352, 2520, 2688, 2856, 3024, 3192, 3360, 3528, 3696, 3864, 4032, 4200, 4368, 4536, 4704, 4872, 5040, 5208, 5376, 5544, 5712, 5880, 6048, 6216, 6384, 6552, 6720, 6888, 7056, 7224, 7392, 7560, 7728, 7896

Multiples of 180:

180, 360, 540, 720, 900, 1080, 1260, 1440, 1620, 1800, 1980, 2160, 2340, 2520, 2700, 2880, 3060, 3240, 3420, 3600, 3780, 3960, 4140, 4320, 4500, 4680, 4860, 5040, 5220, 5400, 5580, 5760, 5940, 6120, 6300, 6480, 6660, 6840, 7020, 7200, 7380, 7560, 7740, 7920

The Least Common Multiple for the 3 number is;

= 7560

Therefore a vessel of 7,560 liters fills each one of them an exact number of times with no remainder.

Learn More:

For more on Least Common Multiple check out;

brainly.com/question/11978021

#LearnWithBrainly

5 0

3 years ago