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gtnhenbr [62]
4 years ago
10

bob eats 6 apples in 2 days. how many days will it takes to eat a basket of apples containing 51 apples?

Mathematics
2 answers:
Strike441 [17]4 years ago
5 0
51/6=8.5 8.5*2=17 It would take 17 days for Bob to eat a basket of apples containing 51 apples.
Ilia_Sergeevich [38]4 years ago
3 0
If he eat 6 apples in 2 days so we can say he eat 3 apple in 1 day so if we want to know in how many days he can eat 51 apple you can say 51 ÷ 3= 17 also we can say 1day / 3 apples = ? days / 51 apples ⇒ 51×1 / 3 = 17 so he can eat 51 apples in 17 days :))))
i hope this be helpful 
* happy new year * 
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Answer:

99

Step-by-step explanation:

1.  rectangular pyramid formula:

2.  V=lwh/3

3.  99 = (3)(9)(11)/3

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The figure shows a construction completed by hand.
Tresset [83]

Based on the construction, we can logically deduce that: A. yes; the compass was kept at the same width to create the arcs for points C and D.

<h3>What is a line segment?</h3>

A line segment can be defined as the part of a line in a geometric figure such as a triangle, circle, quadrilateral, etc., that is bounded by two (2) distinct points and it typically has a fixed length.

In Geometry, a line segment can be measured by using the following measuring instruments:

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  • A compass

<h3>What is an arc?</h3>

In Geometry, an arc can be defined as a trajectory that is generally formed when the distance from a given point has a fixed numerical value.

Based on the construction with arcs created above and below the line segment from points A, we can infer and logically deduce that it is true that the compass was kept at the same width to create the arcs for points C and D.

In conclusion, yes, the construction demonstrated how to bisect a line segment correctly by hand.

Read more on arcs here: brainly.com/question/11126174

#SPJ1

Complete Question:

The construction has a given segment AB. Arcs have been created above and below the segment from points A that are equidistant from point A. The compass was kept at the same distance, placed on point B, and two additional arcs were created above and below the segment that intersect with the first arcs created. The intersection of the arcs above the segment created point C. The intersection of the arcs below the segment created point D. A line was drawn from point C to D through the segment.

Does the construction demonstrate how to bisect a segment correctly by hand?

A. Yes; the compass was kept at the same width to create the arcs for points C and D.

B. Yes; a straightedge was used to create segment CD.

C. No; the compass was not kept at the same width to create the arcs for points C and D.

D. No; a straightedge was used to create segment CD.

7 0
2 years ago
The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that m = 100, \mu = \frac{1}{100} = 0.01

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

P(X > x) = e^{-\mu x}

This is P(X > 190). So

P(X > 190) = e^{-0.01*190} = 0.1496

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

P(X > x) = 0.08

So

e^{-0.01x} = 0.08

\ln{e^{-0.01x}} = \ln{0.08}

-0.01x = \ln{0.08}

x = -\frac{\ln{0.08}}{0.01}

x = 252.6

Capacity of 252.6 cubic feet per second

5 0
3 years ago
What is the sum of the first five terms of a geometric series with a1=15 and r=1/3
Solnce55 [7]

Answer:

15, 5, 5/3, 5/9, 5/27

Step-by-step explanation:

Recall, that the general form of a geometric series is

a, ar, ar², ar³, .........

where a is the first term and r is the ratio

It is given that the first time is 15, hence a = 15

also given that ratio is 1/3

hence the first 5 terms are:

15, (15)(1/3), (15)(1/3)^2, (15)(1/3)^3, (15)(1/3)^4

simplifying each term gives

15, 5, 5/3, 5/9, 5/27

6 0
3 years ago
Read 2 more answers
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