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Aliun [14]
3 years ago
12

Mike has a rowboat for crossing the river. The rowboat can hold only 200 pounds without sinking. If mike has his two little brot

hers with him, Jerry and David, who weigh 100 and 80 pounds respectively, and Mike weighs 160 pounds, how can all three boys get across the river?
This is not a Joke btw.
Mathematics
1 answer:
Elina [12.6K]3 years ago
4 0

Something that could be done is to tie a rope long enough to cross the river and send the two younger brothers across first, then have Mike pull the boat back across the river with the rope and him get in the boat and row it across the river with just himself and the gear.

Another option is you can unload some gear, put the boys in the boat, row them across the river, unload the two younger boys, go back for the rest of gear and load it into the boat then come back.


Hope this helps!

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What is the perimeter of the parallelogram?
Leni [432]

Answer:

P= 2 (a+b)

Step-by-step explanation:

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3 years ago
Use the triangles Name one pair of congruent angles.
Likurg_2 [28]
Your correct answer is option B
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3 years ago
Why is the number of operators per shift multiplied by approximately 4.5 to obtain the total number of operators required to run
4vir4ik [10]

I believe you meant "why is the number of shifts multiplied by approximately 4.5 to obtain the total number of operators required to run the plant"

Answer and Explanation:

There are 3 shifts per day, 49 weeks per year and 5 shifts per operator per week

To get total number of operators required to run the plant, we multiply number of shifts in a year by number if operators per shift.

49 weeks×5 shifts= 245 shifts per operator per year

365×3 shifts= 1095 shifts per year

1095/245=4.5 operators per shift

total number of operators required to run the plant(per day) = 4.5×3= 13.5 approximately 14

total number of operators required to run the plant(per year) =4.5×1095=4927.5 approximately 4928

3 0
3 years ago
A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq
Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

s=16.6 represent the sample deviation

\bar X=12.8 represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

\alpha=1-0.99=0.01 represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448

12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

3 0
3 years ago
Select all the conditions for which it is possible to construct a triangle. (7.G.1.2) Group of answer choices a. A triangle with
saw5 [17]

Answer:

  b, d, e, f

Step-by-step explanation:

Here are the applicable restrictions:

  • The sum of angles in a triangle is 180°, no more, no less.
  • The sum of the lengths of the two shortest sides exceeds the longest side.
  • When two sides and the angle opposite the shortest is given, the sine of the given angle must be at most the ratio of the shortest to longest sides.

a. A triangle with angle measures 60°, 80°, and 80° (angle sum ≠ 180°, not OK)

b. A triangle with side lengths 4 cm, 5 cm, and 6 cm (4+5 > 6, OK)

c. A triangle with side lengths 4 cm, 5 cm, and 15 cm (4+5 < 15, not OK)

d. A triangle with side lengths 4 cm, 5 cm, and a 50° angle across from the 4 cm side (sin(50°) ≈ 0.766 < 4/5, OK)

e. A triangle with angle measures 30° and 60°, and an included 3 cm side length (OK)

f. A triangle with angle measures 60°, 20°, and 100° (angle sum = 180°, OK)

_____

<em>Additional comment</em>

In choice "e", two angles and the side between them are specified. As long as the sum of the two angles is less than 180°, a triangle can be formed. The length of the side is immaterial with respect to whether a triangle can be made.

__

The congruence postulates for triangles are ...

  SSS, SAS, ASA, AAS, and HL

These essentially tell you the side and angle specifications necessary to define <em>a singular triangle</em>. As we discussed above, the triangle inequality puts limits on the side lengths specified in SSS. The angle sum theorem puts limits on the angles when only two are specified (ASA, AAS).

In terms of sides and angles, the HL postulate is equivalent to an SSA theorem, where the angle is 90°. In that case, the angle is opposite the longest side (H). In general, SSA will specify a singular triangle when the angle is opposite the <em>longest</em> specified side, regardless of that angle's measure. However, when the angle is opposite the <em>shortest</em> specified side, the above-described ratio restriction holds. If the sine of the angle is <em>less than</em> the ratio of sides, then <em>two possible triangles are specified</em>.

4 0
2 years ago
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